Homework 1

1. For A := ( \begin{matrix}−8 & −10 & −1 \\ 7 & 9 & 1 \\ 3 & 2 & 0 \end{matrix} ), we have that c_{A}(X) = (X − 1)^{2}(X + 1). Since u_{1} :=

( \begin{matrix}2 \\ −1 \\ −4 \end{matrix} ) is an eigenvector corresponding with -1, we have that U_{1} := ⟨u_{1}⟩ is an A-

submodule as

(∑_{j=0}^{r} α_{j}T^{j}) (βu_{1}) = (β ∑_{j=0}^{r} α_{j}(−1)^{j}) u_{1} ∈ U_{1}.

Linear algebra gives that

N (A + 1)^{2} = ⟨( \begin{matrix}−1 \\ 1 \\ 1 \end{matrix} ) , ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} )⟩ .

is the generalized eigenspace for the eigenvector 1, with the first vector listed being an eigenvector for 1. Now taking

u_{3} := ( \begin{matrix}0 \\ 0 \\ 1 \end{matrix} ) , and u_{2} := (A + 1)u_{3} = ( \begin{matrix}−1 \\ 1 \\ 1 \end{matrix} ) ,

we have that U_{2} := ⟨u_{2}, u_{3}⟩ = ⟨u_{3}⟩ is a A-submodule as Au_{2} = u_{2} ∈ U_{2} and Au_{3} = u_{2} + u_{3} ∈ U_{2}.

Also, {v_{1}, v_{2}, v_{3}} is a basis for V_{1} so that V_{1} = ⟨u_{1}⟩ ⊕ ⟨u_{3}⟩ = U_{1} ⊕ U_{2}.

However, U_{2} is not irreducible as ⟨u_{2}⟩ is a proper A-submodule of U_{2}. Also U_{2} is not completely reducible as the only possible compliment of ⟨u_{2}⟩ would be ⟨u_{3}⟩ , which is not a submodule as Au_{3} = u_{2} ∉ ⟨u_{3}⟩. This also means that V_{1} is not completely reducible.

We compute the Jordan form for A for our amusment: define

S_{A} = (u_{1}, u_{2}, u_{3}) = ( \begin{matrix}2 & −1 & 0 \\ −1 & 1 & 0 \\ −4 & −1 & 1 \end{matrix} )

and

J_{A} = (Au_{1}, Au_{2}, Au_{3}) = ( \begin{matrix}−1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix} ) ,

so that A = SJS^{−1}.

Continuing, let B := ( \begin{matrix}−3 & 2 & −4 \\ 4 & −1 & 4 \\ 4 & −2 & 5 \end{matrix} ). Then c_{B}(X) = c_{A}(X) = (X − 1)^{2}(X + 1)

and the Jordan form for B is J_{B} = ( \begin{matrix}−1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) with conjugating matrix S_{B} =

( \begin{matrix}−1 & 1 & 0 \\ 1 & 2 & 2 \\ 1 & 0 & 1 \end{matrix} ). In particular, this means that B has 3 independent eigenvectors

v_{1} := ( \begin{matrix}−1 \\ 1 \\ 1 \end{matrix} ) , v_{2} := ( \begin{matrix}1 \\ 2 \\ 0 \end{matrix} ) , and v_{3} := ( \begin{matrix}0 \\ 2 \\ 1 \end{matrix} )

so that ⟨v_{1}⟩,⟨v_{2}⟩, and ⟨v_{3}⟩ are all submodules of V_{2} and V_{2} = ⟨v_{1}⟩ ⊕ ⟨v_{2}⟩ ⊕ ⟨v_{3}⟩ so that V_{2} is completely reducible. We have, however, that V_{1} is not completely reducible so that V_{2} is not isomorphic with V_{1}.

2. Let V = ⟨e_{1}, e_{2} ⋯ e_{m}⟩. We have that

e_{j_{1}} ⊗ e_{j_{2}} ⊗ ⋯ ⊗ e_{j_{n}} = e_{σj_{1}} ⊗ e_{σj_{2}} ⊗ ⋯ ⊗ e_{σj_{n}}

in S^{n}V for any σ ∈ S_{n} since if σ = τ_{1} ∘ τ_{2} ∘ ⋯ ∘ τ_{r} for transposition s τ_{j}, then

e_{j_{1}} ⊗e_{j_{2}} ⊗⋯⊗e_{j_{n}} = e_{τ_{r}j_{1}} ⊗e_{τ_{r}j_{2}} ⊗⋯⊗e_{τ_{r}j_{n}} = e_{τ_{r−1}τ_{r}j_{1}} ⊗e_{τ_{r−1}τ_{r}j_{2}} ⊗⋯⊗e_{τ_{r−1}τ_{r}j_{n}} = ⋯ =

e_{τ_{1}···τ_{r−1}τ_{r}j_{1}} ⊗ e_{τ_{1}···τ_{r−1}τ_{r}j_{2}} ⊗ ⋯ ⊗ e_{τ_{1}···τ_{r−1}τ_{r}j_{n}} = e_{σj_{1}} ⊗ e_{σj_{2}} ⊗ ⋯ ⊗ e_{σj_{n}}.

This shows that two tensors are equal in the symmetric product if their component vectors are permuted. We can then assume that any tensor can be expressed with its components in non-decreasing order as follows:

\underbrace{\overbrace{e_{1} ⊗ e_{1} ⊗ ⋯ ⊗ e_{1}}^{r_{1}} ⊗ \overbrace{e_{2} ⊗ e_{2} ⊗ ⋯ ⊗ e_{2}}^{r_{2}} ⊗ ⋯ \overbrace{e_{m} ⊗ e_{m} ⊗ ⋯ ⊗ e_{m}}^{r_{m}}}_{n}

with r_{j} ≥ 0. Then two tensors will be equal if they have the same number of e_{1}’s, the same number of e_{2}’s, etc., so that the problem is reduced to selecting r_{1}, r_{2} … r_{m} with r_{j} ≥ 0 and r_{1} + r_{2} + ⋯ r_{m} = n.

Now we all know that there are \binom{n + m − 1}{r} ways to put n balls into m baskets, for,

you see, the m baskets, when placed side by side, have m − 1 interior walls, so that we only have to count the number of ways to select n of n + m − 1 objects to be balls (and not walls), giving the formula above. Now, returning to the symmetric product, we see that selecting r_{j} ≥ 0 with ∑_{j} r_{j} = n

is equivalent to putting n balls into m baskets so that there will be \binom{n + m − 1}{r} ways

to do so. Hence, dim S^{n}V = \binom{n + m − 1}{r}.

(⋀^{n} ρ)(e_{1} ∧ e_{2} ∧ ⋯ ∧ e_{n}) = ρe_{1} ∧ ρe_{2} ∧ ⋯ ∧ ρe_{n} =

∑_{i} a_{i,1}e_{i} ∧ ∑_{i} a_{i,2}e_{i} ∧ ⋯ ∧ ∑_{i} a_{i,n}e_{i} =

∑_{σ∈S_{n}} sgn(σ)(a_{σ(1),1}a_{σ(2),2} ⋯ a_{σ(n),n})(e_{1} ∧ e_{2} ∧ ⋯ ∧ e_{n}) = det A(e_{1} ∧ e_{2} ∧ ⋯ ∧ e_{n}).

4. If gh = k, then h = g^{−1}k so that

g (∑_{h} α_{h}e_{h}) = ∑_{h} α_{h}e_{gh} = ∑_{k} α_{g^{−1}k}e_{k}.

We identify the vector ∑_{h} α_{h}e_{h} with the C-valued function ∑_{h} α_{h}I_{h} and make the collection R_{1} of C-valued function into a G-module by defining

gφ(k) := φ(g^{−1}k)

for φ ∈ R_{1} and g, k ∈ G. Then

g (∑_{h} α_{h}I_{h}) (k) = (∑_{h} α_{h}I_{h}) (g^{−1}k) = α_{g^{−1}k}

so that

g (∑_{h} α_{h}I_{h}) = ∑_{k} α_{g^{−1}k}I_{k}.

Then under the identification above, g (∑_{h} α_{h}e_{h}) corresponds with g (∑_{h} α_{h}I_{h}) as required.

Let R_{2} be the space of functions on G made into a module by the action

gφ(k) := φ(kg)

for φ ∈ R_{2} and k, g ∈ G.

Then

g (∑_{h} α_{h}I_{h}) (k) = (∑_{h} α_{h}I_{h}) (kg) = α_{kg}

so that

g (∑_{h} α_{h}I_{h}) = ∑_{k} α_{kg}I_{k}.

Define Φ : R_{1} ⟶ R_{2} by

Φ (∑_{h} α_{h}I_{h}) := ∑_{h} α_{h^{−1}}I_{h}.

Then Φ is clearly a C-vector space isomorphism. To see that Φ is a G-module homomorphism,

Φ (g ∑_{h} α_{h}I_{h}) = Φ (∑_{k} α_{g^{−1}k}I_{k}) = ∑_{k} α_{k^{−1}g}I_{k} = g ∑_{h} α_{h^{−1}}I_{h} = gΦ (∑_{h} α_{h}I_{h}) .

Cheers!

Homework 10

Marcus

1. Let V be a vector space of dimension k over ℂ. Show that V^{⊗3} is the direct sum of 4 irreducible GL(V) = GL(k, ℂ)-modules of dimensions

\frac{k(k + 1)(k + 2)}{6}, \frac{k(k − 1)(k − 2)}{6}, \frac{k(k^{2} − 1)}{3}, \frac{k(k^{2} − 1)}{3}.

𝕊_{λ}V is irreducible for all λ ⊢ 3 with fewer than 4 parts. Assuming that k ≥ 3, all the partitions of 3 have fewer than 4 parts. For λ = 1^{3}, we compute

dim 𝕊_{(1^{3})}V = ∏_{1≤i<j≤k} \frac{λ_{i} − λ_{j} + j − i}{j − i}

= (\frac{1 − 1 + 2 − 1}{2 − 1}) (\frac{1 − 1 + 3 − 1}{3 − 1}) (\frac{1 − 0 + 4 − 1}{4 − 1}) ⋯ (\frac{1 − 0 + k − 1}{k − 1}) ·

(\frac{1 − 1 + 3 − 2}{3 − 2}) (\frac{1 − 0 + 4 − 2}{4 − 2}) (\frac{1 − 0 + 5 − 2}{5 − 2}) ⋯ (\frac{1 − 0 + k − 2}{k − 2}) ·

(\frac{1 − 0 + 4 − 3}{4 − 3}) (\frac{1 − 0 + 5 − 3}{5 − 3}) (\frac{1 − 0 + 6 − 3}{6 − 3}) ⋯ (\frac{1 − 0 + k − 3}{k − 3}) ·

(\frac{0 − 0 + 5 − 4}{5 − 4}) (\frac{0 − 0 + 6 − 4}{6 − 4}) (\frac{0 − 0 + 7 − 4}{7 − 4}) ⋯ (\frac{0 − 0 + k − 4}{k − 4}) ·

⋯ (\frac{0 − 0 + k − (k − 1)}{k − (k − 1)})
= 1 · 1 · \frac{4}{3} · \frac{5}{4} ⋯ \frac{k}{k − 1} ·

1 · \frac{3}{2} · \frac{4}{3} ⋯ \frac{k − 1}{k − 2} ·

\frac{2}{1} · \frac{3}{2} ⋯ \frac{k − 2}{k − 3} ·

1 · 1 ⋯ 1 ·

1

= \frac{k}{3} · \frac{k − 1}{2} · \frac{k − 2}{1}
= \frac{k(k − 1)(k − 2)}{6}.

We similarly compute that

dim 𝕊_{(2,1)}V = \frac{k(k^{2} − 1)}{3} and dim 𝕊_{(3)}V = \frac{k(k − 1)(k − 2)}{6}

and that

V^{⊗3} = 𝕊_{(1^{3})}V ⊕ (𝕊_{(2,1)}V )^{⊗2} ⊕ 𝕊_{(3)}V

2. Let k = 3 and d = 3 as above. Let {e_{1}, e_{2}, e_{3}} be a basis of V. Let λ = 2, 1.

(a) Verify that 𝕊_{λ}V has dimension 8 and that is has a basis as given in the Semistandard Basis Theorem. We compute that the semi-standard tableux are

1 1 2 , 1 1 3 , 1 2 2 , 1 2 3 , 1 3 2 , 1 3 3 , 2 2 3 , and 2 3 3 .

Now for 1 1 2 , we have

(e_{1} ⊗ e_{2} ⊗ e_{1})c_{λ} = (e_{1} ⊗ e_{2} ⊗ e_{1})(1 + (12) − (13) − (132))
= (e_{1} ⊗ e_{2} ⊗ e_{1}) + (e_{2} ⊗ e_{1} ⊗ e_{1}) − (e_{1} ⊗ e_{2} ⊗ e_{1}) − (e_{1} ⊗ e_{1} ⊗ e_{2})
= (e_{2} ⊗ e_{1} ⊗ e_{1}) − (e_{1} ⊗ e_{1} ⊗ e_{2}).

Similarly, we have

(e_{1} ⊗ e_{3} ⊗ e_{1})c_{λ} = (e_{3} ⊗ e_{1} ⊗ e_{1}) − (e_{1} ⊗ e_{1} ⊗ e_{3}),
(e_{1} ⊗ e_{2} ⊗ e_{2})c_{λ} = (e_{1} ⊗ e_{2} ⊗ e_{2}) − (e_{2} ⊗ e_{2} ⊗ e_{1}),
(e_{1} ⊗ e_{3} ⊗ e_{2})c_{λ} = (e_{1} ⊗ e_{3} ⊗ e_{2}) + (e_{3} ⊗ e_{1} ⊗ e_{2}) − (e_{2} ⊗ e_{3} ⊗ e_{1}) − (e_{2} ⊗ e_{1} ⊗ e_{3}),
(e_{1} ⊗ e_{2} ⊗ e_{3})c_{λ} = (e_{1} ⊗ e_{2} ⊗ e_{3}) + (e_{2} ⊗ e_{1} ⊗ e_{3}) − (e_{3} ⊗ e_{2} ⊗ e_{1}) − (e_{3} ⊗ e_{1} ⊗ e_{2}),
(e_{1} ⊗ e_{3} ⊗ e_{3})c_{λ} = (e_{1} ⊗ e_{3} ⊗ e_{3}) − (e_{3} ⊗ e_{3} ⊗ e_{1}),
(e_{2} ⊗ e_{3} ⊗ e_{2})c_{λ} = (e_{3} ⊗ e_{2} ⊗ e_{2}) − (e_{2} ⊗ e_{2} ⊗ e_{3}),

and

(e_{2} ⊗ e_{3} ⊗ e_{3})c_{λ} = (e_{2} ⊗ e_{3} ⊗ e_{3}) − (e_{3} ⊗ e_{3} ⊗ e_{2})

We notice that no simple tensor e_{i}⊗e_{j} ⊗e_{k} appears more than once as a summand of any of the 8 vectors above so that if

0 = α_{1}(e_{2} ⊗ e_{1} ⊗ e_{1} − e_{1} ⊗ e_{1} ⊗ e_{2}) + α_{2}(e_{3} ⊗ e_{1} ⊗ e_{1} − e_{1} ⊗ e_{1} ⊗ e_{3}) + ⋯
= 0 · (e_{1} ⊗ e_{1} ⊗ e_{1}) − α_{1}(e_{1} ⊗ e_{1} ⊗ e_{2}) + ⋯

then by the independence of {e_{i} ⊗ e_{j} ⊗ e_{k}|1 ≤ i, j, k ≤ 3}, we have that α_{j} = 0

for all 1 ≤ j ≤ 8 so that the above vectors are independent. Thus these vectors form a basis for 𝕊_{λ}V.

(b) Let g = (\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}). Verify the character formula given in Fulton and Harris

in Theorem 6.3

We compute the character of 𝕊_{V} at g = (\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) directly by determining

the images of the basis vectors computed in the previous problem under the action of g. We have that

(\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) · (e_{2} ⊗ e_{1} ⊗ e_{1} − e_{1} ⊗ e_{1} ⊗ e_{2}) = x_{2}e_{2} ⊗ x_{1}e_{1} ⊗ x_{1}e_{1} − x_{1}e_{1} ⊗ x_{1}e_{1} ⊗ x_{2}e_{2}
= x_{1}^{2}x_{2}(e_{2} ⊗ e_{1} ⊗ e_{1} − e_{1} ⊗ e_{1} ⊗ e_{2})

and similarly for the other basis elements so that the matrix corresponding with this action is

(\begin{matrix}x_{1}^{2}x_{2} & 0 & ⋯ \\ 0 & x_{1}^{2}x_{3} \\ ⋮ & & ... \end{matrix})

and the trace of this matrix is

x_{2}^{2}x_{3} + x_{1}x_{2}^{2} + x_{3}^{2}x_{2} + 2x_{1}x_{2}x_{3} + x_{1}^{2}x_{2} + x_{1}x_{3}^{2} + x_{1}^{2}x_{3}

which corresponds with the Schur polynomial I computed with Maple.

We further corroborate the claim that the Schur polynomials S_{λ} are in fact the characters of 𝕊_{λ} by computing the trace of (g, (12)) on V^{⊗3} and verifying that this is ∑_{λ∈Λ(3,3)} S_{λ}(x_{1}, x_{2}, x_{3})χ_{λ}(12), the trace of (g, (12) on ⨁_{λ∈Λ(3,3)} 𝕊_{λ}V ⊗ V_{λ}. To compute the trace on V^{⊗3}, we note that only the basis vectors e_{j} ⊗ e_{j} ⊗ e_{k} will be fixed under the right action of (12) and that in this situation,

((\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) , (12)) · e_{j} ⊗ e_{j} ⊗ e_{k} = ((\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) · e_{j} ⊗ e_{j} ⊗ e_{k}) · (12)
= (x_{j}e_{j} ⊗ x_{j}e_{j} ⊗ x_{k}e_{k}) · (12)
= x_{j}e_{j} ⊗ x_{j}e_{j} ⊗ x_{k}e_{k}
= x_{j}^{2}x_{k}(e_{j} ⊗ e_{j} ⊗ e_{k}).

where · generically represents all actions involved. Thus,

Trace of ((\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) , (12)) on V^{⊗3} = Tr (x_{1}^{3} 0  0  0 ⋯; 0 x_{1}^{2}x_{2} 0  0 ⋯; 0 0 x_{1}^{2}x_{3} 0 ⋯; 0 0  0  0 ⋯; ⋮ ; 0 0  0 x_{1}^{2}x_{2} )

= ∑_{1≤j,k≤3} x_{j}^{2}x_{k}
= (x_{1} + x_{2} + x_{3})(x_{1}^{2} + x_{2}^{2} + x_{3}^{2})

Maple gives that

S_{1^{3}}(x_{1}, x_{2}, x_{3}) = x_{1}x_{2}x_{3}
S_{2,1}(x_{1}, x_{2}, x_{e}) = x_{2}^{2}x_{3} + x_{1}x_{2}^{2} + x_{3}^{2}x_{2} + 2x_{1}x_{2}x_{3} + x_{1}^{2}x_{2} + x_{1}x_{3}^{2} + x_{1}^{2}x_{3}
S_{3}(x_{1}, x_{2}, x_{3}) = x_{2}^{3} + x_{2}^{2}x_{3} + x_{1}x_{2}^{2} + x_{3}^{2}x_{2} + x_{1}x_{2}x_{3} + x_{1}^{2}x_{2} + x_{3}^{3} + x_{1}x_{3}^{2} + x_{1}^{2}x_{3} + x_{1}^{3}.

Substituting these polynomials for χ_{𝕊_{1^{3}}V}(x_{1}, x_{2}, x_{3}), χ_{𝕊_{2,1}V}(x_{1}, x_{2}, x_{3}), and χ_{𝕊_{3}V}(x_{1}, x_{2}, x_{3}), we have

Trace of ((\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) , (12))
on ⨁_{λ∈Λ(3,3)} 𝕊_{λ}V ⊗ V_{λ} = ∑_{λ∈Λ(3,3)} χ_{𝕊_{λ}V}(x_{1}, x_{2}, x_{3})χ_{λ}(12)
= χ_{𝕊_{1^{3}}V}(x_{1}, x_{2}, x_{3})χ_{1^{3}}(12) + χ_{𝕊_{2,1}V}(x_{1}, x_{2}, x_{3})χ_{2,1}(12)
+χ_{𝕊_{3}V}(x_{1}, x_{2}, x_{3})χ_{3}(12)
= χ_{𝕊_{1^{3}}V}(x_{1}, x_{2}, x_{3}) · (−1) + χ_{𝕊_{2,1}V}(x_{1}, x_{2}, x_{3})χ_{2,1}(12) · 0
+χ_{𝕊_{3}V}(x_{1}, x_{2}, x_{3})χ_{3}(12) · 1
= x_{2}^{3} + x_{2}^{2}x_{3} + x_{1}x_{2}^{2} + x_{3}^{2}x_{2}
+x_{1}^{2}x_{2} + x_{3}^{3} + x_{1}x_{3}^{2} + x_{1}^{2}x_{3} + x_{1}^{3}
= (x_{1} + x_{2} + x_{3})(x_{1}^{2} + x_{2}^{2} + x_{3}^{2})
= Trace of ((\begin{matrix}x_{1} & 0 & 0 \\ 0 & x_{2} & 0 \\ 0 & 0 & x_{3} \end{matrix}) , (12)) on V^{⊗3}

Homework 2

2. Let W be some representation of D_{4} = ⟨r, s|r^{4} = s^{2} = 1, srs = r^{3}⟩. When we allow the abelian subgroup ⟨r⟩ ≤ D_{4} to act on W, we have by Maschke that W = ⨁_{j=1}^{m} V_{j} for irreducible ⟨r⟩-submodules V_{j}, but by Schur, V_{j} = ⟨v_{j}⟩ for eigenvectors v_{j} of r. Moreover, the eigenvalue for v_{j} is i^{k_{j}}, k_{j} = 0, 1, 2, 3.

1. Now suppose v_{j} is an eigenvector for r with eigenvalue i. Then

rsv_{j} = sr^{3}v_{j} = si^{3}v_{j} = −isv_{j}

so that sv_{j} is also an eigenvector for r with eigenvalue −i. Then v_{j} and sv_{j} are independent, having different eigenvalues, so that W_{1} = ⟨v_{j}, sv_{j}⟩ is a D_{4}-submodule of W.

Now W_{1} is irreducible since any proper subspace ⟨αv_{j} + βsv_{j}⟩ ≤ W_{1}, is not D_{4}-stable as s(αv_{j} +βsv_{j}) = βv_{j} +αsv_{j} ∈ ⟨αv_{j} +βsv_{j}⟩ so that α = ±β and we can assume α = 1 and β = ±1, but r(v_{j} ± sv_{j}) = iv_{j} ∓ isv_{j} ∉ ⟨v_{j} ± sv_{j}⟩.

Thus W_{1} is a irreducible two-dimensional D_{4}-module.

2. If v_{j} is an eigenvector for r with eigenvalue −1, then by a similar argument, we have that sv_{j} is also an eigenvector for r with eigenvalue −1.

(a) If v_{j} and sv_{j} are not independent, then sv_{j} = cv_{j} for some constant c ∈ C. Then

applying s to both sides,

v_{j} = csv_{j} = c^{2}v_{j}

so that c = ±1.

i. If sv_{j} = v_{j}, then let W_{2} = ⟨v_{j}⟩. W_{2} is an irreducible D_{4}-module.

ii. If sv_{j} = −v_{j}, then let W_{3} = ⟨v_{j}⟩. W_{3} is also an irreducible D_{4}-module.

(b) If v_{j} and sv_{j} are independent, then ⟨v_{j}, sv_{j}⟩ is a D_{4}-submodule, but is not irreducible as ⟨v_{j} + sv_{j}⟩ and ⟨v_{j} − sv_{j}⟩ are proper D_{4}-submodules of ⟨v, sv_{j}⟩ with ⟨v_{j} + sv_{j}⟩ ⊕ ⟨v_{j} − sv_{j}⟩ = ⟨v_{j}, sv_{j}⟩. These modules are isomorphic with W_{2} and W_{3} above.

3. If v_{j} is an eigenvector for r with eigenvalue 1, then as above, sv_{j} is also an eigenvector for r with eigenvalue 1.

(a) If v_{j} and sv_{j} are not independent, then as above, sv_{j} = ±v_{j}

i. If sv_{j} = v_{j}, then W_{4} = ⟨v_{j}⟩ is an irreducible submodule and is isomorphic

with the trivial module. ii. If sv_{j} = −v_{j}, then W_{5} = ⟨v_{j}⟩ is a irreducible submodule.

(b) If v_{j} and sv_{j} are independent, then ⟨v_{j}, sv_{j}⟩ is a D_{4}-submodule, but is not irreducible as ⟨v_{j} + sv_{j}⟩ and ⟨v_{j} − sv_{j}⟩ are proper D_{4}-submodules of ⟨v, sv_{j}⟩ with ⟨v_{j} + sv_{j}⟩ ⊕ ⟨v_{j} − sv_{j}⟩ = ⟨v_{j}, sv_{j}⟩. These modules are isomorphic with W_{4} and W_{5} above.

We have thus computed that the only irreducible D_{4}-modules are W_{1}, which is two-dimensional, and W_{2}, W_{3}, W_{4}, and W_{5}, which are one-dimensional.

1.11 S^{2}V has basis {α^{2}, αβ, β^{2}} with

τα^{2} = τα · τα = ωα · ωα = ω^{2}α^{2}.

We similarly compute

\begin{matrix}τα^{2} = ω^{2}α^{2} & σα^{2} = β^{2} \\ ταβ = αβ & σαβ = αβ \\ τβ^{2} = ωβ^{2} & σβ^{2} = α^{2}. \end{matrix}

We have, by arguments similar to those given in problem 2 above, that ⟨αβ⟩ and ⟨α^{2}, β^{2}⟩ are irreducible submodules and

S^{2}V = ⟨αβ⟩ ⊕ ⟨α^{2}, β^{2}⟩ ≅ U ⊕ V.

Similarly, {α^{3}, α^{2}β, αβ^{2}, β^{3}} is a basis for S^{3}V. We compute

\begin{matrix}τα^{3} = α^{3} & σα^{3} = β^{3} \\ τα^{2}β = ωα^{2}β & σα^{2}β = αβ^{2} \\ ταβ^{2} = ω^{2}αβ^{2} & σαβ^{2} = α^{2}β \\ τβ^{3} = β^{3} & σβ^{3} = α^{3}. \end{matrix}

Then

S^{3}V = ⟨α^{2}β, αβ^{2}⟩ ⊕ ⟨α^{3} + β^{3}⟩ ⊕ ⟨α^{3} − β^{3}⟩.

1.12 This approach is extremely direct. You will hate every word of it. We use matrices to compute the eigenvectors for τ.

Eigenvector for τ Eigenvalue
v_{1} 1 + τ + τ^{2}  1
v_{2} σ + στ + στ^{2}  1
v_{3} 1 + ω^{2}τ + ωτ^{2}  ω
v_{4} σ + ωστ + ω^{2}στ^{2} ω
v_{5} 1 + ωτ + ω^{2}τ^{2}  ω^{2}
v_{6} σ + ω^{2}στ + ωστ^{2} ω^{2}

and note that v_{3} and σv_{3} = v_{6} are independent, being eigenvectors for τ which correspond with different eigenvalues. Similarly, v_{4} and σv_{4} = v_{5} are independent. Moreover, ⟨v_{3}, v_{6}⟩ and ⟨v_{4}, v_{5}⟩ are irreducible as demonstrated in problem 2 above. Moreover,

⟨v_{3}, v_{6}⟩ ≅ ⟨v_{4}, v_{5}⟩ ≅ V

where V is the standard module.

Finally, we can all see that v_{1} and σv_{1} = v_{2} are independent. However, ⟨v_{1}, v_{2}⟩ is not irreducible as ⟨v_{1} + v_{2}⟩ ⊕ ⟨v_{1} − v_{2}⟩ = ⟨v_{1}, v_{2}⟩. Moreover,

⟨v_{1} + v_{2}⟩ ≅ U and ⟨v_{1} − v_{2}⟩ ≅ U′

where U is the trivial module and U′ is the alternating module.

Hence, we have that R ≅ U ⊕ U′ ⊕ V^{2}.

Homework 3

1. (a) Let V_{j} := ⟨e_{1}, e_{2} … e_{j}⟩ where e_{j} = (0, 0 … 1 … 0) with the 1 in the jth position. Then V_{j} is a B-submodule. Now as V_{j}/V_{j−1} = ⟨e_{j}⟩ + V_{j−1} ≅ ⟨e_{j}⟩, we have that V_{j}/V_{j−1} is irreducible and that

V ≥ V_{n−1} ≥ V_{n−1} ≥ ⋯ ≥ V_{2} ≥ V_{1} ≥ 0

Write ρ_{j} for the representation of B induced by V_{j}/V_{j−1}. Then for M ∈ B we have

M ( v_{1} ; v_{2} ; ⋮ ; v_{j−1} ; v_{j} ; 0 ; ⋮ ; 0 ) = ( ♠ ; ♠ ; ⋮ ; ♠ ; m_{j,j}v_{j} ; 0 ; ⋮ ; 0 )

so that ρ_{j}(M) : v_{j}e_{j} + V_{j−1} ↦ m_{j,j}v_{j}e_{j} + V_{j−1}. Thus, M acts as multiplication by m_{j,j}.

(b) Write E_{i,j} for the matrix with 1 in the (i, j) position and 0’s elsewhere. Define B_{j} = ⟨E_{1,j}, E_{2,j} … E_{j,j}⟩. Then B_{j} is a B-submodule with B = ⨁_{j=1}^{n} B_{j}. Write ρ_{k} for the representation of B induced by B_{k}. To describe ρ_{k}, it suffices to describe the transformations ρ_{k}(E_{i,j}) for E_{i,j}, i ≤ j, the generators of B. Now in general, left multiplication by E_{i,j} replaces the the jth row with the ith row and clobbers everything else. Then for M ∈ B_{k}, where only the kth column is non-zero, we have E_{i,j}M = m_{j,k}E_{i,k}, that is, ρ_{k}(E_{i,j}) is the transformation which extracts the (j, k) entry and moves it to position (i, k), and clobbers everything else.

2. (a) Let ρ_{g} : h ↦ gh for all h ∈ G. Then ρ_{a} : a^{j} ↦ a^{j+1} and this transformation has matrix

( \begin{matrix}0 & 0 & ⋯ & 0 & 1 \\ 1 & 0 & & 0 & 0 \\ 0 & 1 & & 0 & 0 \\ ⋮ & & ... \\ 0 & & & 1 & 0 \end{matrix} )

with respect to the basis G of KG.

(b) Computation gives that the characteristic polynomial for A is c_{A}(X) = (X − 1)^{p}. To find the Jordan form of A, we locate a basis for ℂ^{n} consisting of (A−1)-strings. There are at least two good ways to do this. First, we know by the theorem in Professor Radford’s book about the number o (A − 1)-strings that there are

Rank(A − 1)^{p+1} + Rank(A − 1)^{p−1} − 2Rank(A − 1)^{p} = Rank(A − 1)^{p−1}

(A − 1)-strings of length p in any basis of (A − 1)-strings. But the range of (A − 1)^{p−1} is the nullspace of A − 1 as (A − 1) [(A − 1)^{p−1}v] = (A − 1)^{p}v = 0 by Caley-Hamilton. Thus, the nullity of A − 1 is 1 as

N (A − 1) = N ( \begin{matrix}−1 & 0 & ⋯ & 0 & 1 \\ 1 & −1 & & 0 & 0 \\ 0 & 1 & & 0 & 0 \\ ⋮ & & ... \\ 0 & & & 1 & −1 \end{matrix} ) = ⟨( 1 1 1 ⋮ 1 )⟩

so that there exits an (A − 1)-string of length p. Alternately, if we know that (A − 1)^{p−1} ≠ 0 for some other reason, then we can produce an (A − 1)-string of length p by defining v_{n} := e_{j}, where e_{j} is a standard basis element of ℂ^{p} and j is the column of (A−1)^{p−1} which has a non-zero element. Define also v_{k} := (A − 1)v_{k−1}. Then {v_{k}} forms an (A − 1)-string of length p. Now, to show that (A − 1)^{p−1} is not 0, we note that if

0 = (A − 1)^{p−1} = ∑_{j=0}^{p−1} \binom{p − 1}{j}A^{j}I^{p−1−j} = ∑_{j=0}^{p−1} \binom{p − 1}{j}A^{j},

then applying this transformation to the group-algebra element 1, we have

0 = ∑_{j=0}^{p−1} \binom{p − 1}{j}a^{j},

but since \{a_{j}\} forms a basis for ℂG, we have that \binom{p−1}{j} = 0 for all j, a contradiction. Incidently, taking v_{n} := e_{1} and v_{k} = (A − 1)^{k−1}e_{1}, we can easily compute v_{k} for all 1 ≤ k ≤ p, since v_{k} is the first column of

(A − 1)^{k−1} = ∑_{j=0}^{k−1} \binom{k − 1}{j}A^{j}

and we can easily compute powers of A. For example, the first column of (A − 1)^{2} = A^{2} − 2A + I is

( 0 0 1 ⋮ 0 ) + ( 0 ; −2 ; 0 ; ⋮ ; 0 ) + ( 1 0 0 ⋮ 0 ) = ( 1 ; −2 ; 1 ; ⋮ ; 0 ) .

Note also that v_{p} = (A − 1)^{p−1} = ∑_{j=0}^{p−1} \binom{p−1}{j}A^{j} ≠ 0 as the first columns of I, A, A^{2}, … A^{p−1} have entries in different positions and the \binom{p−1}{j} are not all zero in 𝔽_{p}.

This basis of ℂ^{p} yields the basis {1, 1 − 2a + a^{2}, −1 + 3a − 3a^{2} + a^{4} …} of ℂG. Now taking one of the (A − 1)-strings generated above as a basis, we have that

A ∼ ( \begin{matrix}1 & 1 & 0 & ⋯ & 0 \\ 0 & 1 & 1 & & 0 \\ 0 & 0 & 1 & & 0 \\ ⋮ & & & ... \\ 0 & & & & 1 \\ 0 & & & & 1 \end{matrix} )

so that defining V_{j} = ⟨e_{1}, e_{2} ⋯ e_{j}⟩ as in the previous problem, we have that V_{j} are submodules and that

V ≥ V_{p−1} ≥ V_{p−1} ≥ ⋯ ≥ V_{2} ≥ V_{1} ≥ 0

with V_{j}/V_{j−1} ≅ ⟨e_{j}⟩ irreducible.

3. (a) Let V = ⟨v_{1}, v_{2}⟩ be the two-dimensional irreducible ℂG-module with sv_{1} = v_{2}, sv_{2} = v_{1}, rv_{1} = iv_{1}, and rv_{2} = −iv_{2}. Then V induces a representation ρ of ℂG on V defined by

ρ (∑_{j} α_{j}g_{j}) (v) = (∑_{j} α_{j}g_{j}) · v

where · represents the action of group ring on the module described above. In particular, this means that

ρ : r ↦ ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) and ρ : s ↦ ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} )

with respect to the basis {v_{1}, v_{2}} of V. Using this information, we compute

ρ (−\frac{1}{2}ir − \frac{1}{2}r^{2}) = ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) ,

ρ (\frac{1}{2}isr + \frac{1}{2}s) = ( \begin{matrix}0 & 1 \\ 0 & 0 \end{matrix} ) ,

and similarly for the other two generators of M_{2}(ℂ) , so that ρ is surjective.

(b) The procedure employed in this computation is justified by the following: Lemma. Let N ◁ G. If ∑_{gN∈G/N} α_{gN}gN is idempotent in ℂ (G/N), then \frac{1}{|N|} ∑_{g∈G} α_{π(g)}g is idempotent in ℂG where π(g) = gN is the cannonical projection of G onto G/N. Proof. Since

∑_{gN∈G/N} α_{gN}gN = ( ∑_{gN∈G/N} α_{gN}gN)^{2} = ∑_{gN∈G/N} ( ∑_{h_{1}Nh_{2}N=gN} α_{h_{1}N}α_{h_{2}N}) gN,

we have that for each gN,

α_{gN} = ∑_{h_{1}Nh_{2}N=gN} α_{h_{1}N}α_{h_{2}N}.

But for each fixed pair h_{1}N, h_{2}N with h_{1}h_{2}N = gN, we have that

h_{1,j}Nh_{2,k}N = gN

for all |N| of the elements h_{1,j} ∈ h_{1}N and for all |N| of the elements h_{2,j} ∈ h_{2}N. But of the |N|^{2} pairs h_{1,j}, h_{2,k}, exactly |N| of them satisfy h_{1,j}h_{2,k} = g and the product α_{π(h_{1,j})}α_{π(h_{2,k})} = α_{h_{1}N}α_{h_{2}N} remains fixed for each such pair. Thus

∑_{h_{1}h_{2}=g} α_{π(h_{1})}α_{π(h_{h})} = |N| ∑_{h_{1}Nh_{2}N=gN} α_{h_{1}N}α_{h_{2}N} = |N|α_{gN}

so that

(\frac{1}{|N|} ∑_{g∈G} α_{πg}g)^{2} = \frac{1}{|N|^{2}} ∑_{g∈G} ( ∑_{h_{1}h_{2}=g} α_{π(h_{1})}α_{π(h_{h})}) g = \frac{1}{|N|} ∑_{g∈G} α_{πg}g.

Now getting down to buisness, we take the normal subgroup ⟨r⟩ of D_{8}. Then D_{8}/⟨r⟩ ≅ ℤ_{2} and the idempotents for the quotient are

ẽ_{1} = \frac{1}{2}(⟨r⟩ + s⟨r⟩) and ẽ_{2} = \frac{1}{2}(⟨r⟩ − s⟨r⟩)

Lifting as in the lemma above, we have

e_{1} := \frac{1}{8} (1 + r + r^{2} + r^{3} + s + sr + sr^{2} + sr^{3})

and

e_{2} := \frac{1}{8} (1 + r + r^{2} + r^{3} − s − sr − sr^{2} − sr^{3}) .

Repeating the above procedure for the subgroups ⟨s, r^{2}⟩ and ⟨sr, r^{2}⟩, we generate

e_{3} := \frac{1}{8} (1 − r + r^{2} − r^{3} + s − sr + sr^{2} − sr^{3})

and

e_{4} := \frac{1}{8} (1 − r + r^{2} − r^{3} − s + sr − sr^{2} + sr^{3}) .

Finally, to construct the fifth idempotent f, note that since ∑_{j=1}^{5} e_{j} = 1 by the definition of the e_{j}, we have

1 − ∑_{j=1}^{4} e_{j} = \frac{1}{2} (1 − r^{2}) .

(a) Subject to further consideration.

(b) i. Take v to be an eigenvector for a with eigenvalue i. Then as in the proceeding homework, the computation

a(bv) = ba^{3}v = −i(bv)

shows both that bv is an eigenvector for a with eigenvalue −i and that v and bv are independent. This together with the observation that

b(bv) = a^{2}v = −v

proves that W := ⟨v, bv⟩ is a ℂG submodule. W is irreducible since if ⟨αv + βbv⟩ is any proper subspace, then

a(αv + βbv) = iαv − iβbv ∉ ⟨αv + βbv⟩

so that ⟨αv + βbv⟩ is not G-stable. Now if ρ is the representation induced by this module, then by the above computations,

ρ(a) = ( \begin{matrix}i & 0 \\ 0 & −i \end{matrix} ) and ρ(b) = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) .

ii. If ρ is a one-dimensional representation, then ρ(a^{4}) = ρ(1) = (1) which we will write as 1, so that ρ(a) = i^{j}, j = 0, 1, 2, or 3. If ρ(a) = i, then since

ρ(b)^{−1}iρ(b) = ρ(b^{−1})ρ(a)ρ(b) = ρ(b^{−1}ab) = ρ(a^{−1}) = ρ(a)^{−1} = −i,

we have that

iρ(b) = ρ(b)(−i) = −iρ(b)

which is impossible, so we conclude that ρ(a) ≠ i. Similarly, ρ(a) ≠ −i. This means that ρ(a) = ±1 and in both cases, ρ(b) is easily seen to be ±1.

iii. Knowing that

\frac{1}{2}(⟨a⟩ + b⟨a⟩) and \frac{1}{2}(⟨a⟩ − b⟨a⟩),

are idempotent in ℂ(Q_{8}/⟨a⟩), we have that

e_{1} := \frac{1}{8}(1 + a + a^{2} + a^{3} + b + ba + ba^{2} + ba^{3})

and

e_{2} := \frac{1}{8}(1 + a + a^{2} + a^{3} − b − ba − ba^{2} − ba^{3})

are idempotent in ℂQ_{8}. By replacing ⟨a⟩ in the above argument with ⟨b⟩ and ⟨ba⟩, we similarly generate

e_{3} := \frac{1}{8}(1 − a + a^{2} − a^{3} + b − ba + ba^{2} − ba^{3})

and

e_{4} := \frac{1}{8}(1 − a + a^{2} − a^{3} − b + ba − ba^{2} + ba^{3}).

Finally, as ∑_{j=1}^{5} e_{j} = 1, we have that

e_{5} = \frac{1}{2}(1 − a^{2}).

Multiplication (not shown here for sanity) confirms that e_{j}^{2} = e_{j} for all j and that e_{j}e_{k} = 0 for e_{j} ≠ e_{k} so that the e_{j} are mutually orthogonal.

iv. Assume that ρ is a representation of G over ℝ and assume also (why?) that

ρ(a) = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) .

Now if

ρ(b) = ( \begin{matrix}x & y \\ z & w \end{matrix} )

for x, y, z, w ∈ ℝ, we compute

ρ(ab) = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ( \begin{matrix}x & y \\ z & w \end{matrix} ) = ( \begin{matrix}−z & −w \\ x & y \end{matrix} )

and

ρ(ba^{−1}) = ( \begin{matrix}x & y \\ z & w \end{matrix} ) ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ) = ( \begin{matrix}−y & x \\ −w & z \end{matrix} ) .

Then, as ρ(ab) = ρ(ba^{−1}), we must have x = −w and y = z. Then write

ρ(b) = ( \begin{matrix}x & y \\ y & x \end{matrix} )

and note that since

( \begin{matrix}x^{2} + y^{2} & 2xy \\ 2xy & x^{2} + y^{2} \end{matrix} ) = ρ(b^{2}) = ρ(a^{2}) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) ,

we have that x^{2} + y^{2} = −1 which contradicts x, y ∈ ℝ.

(c) i. Since

(x − x_{1})(x − x_{2}) ⋯ (x − x_{n}) = x^{n} − E_{1}x^{n−1} + E_{2}x^{n−2} + ⋯ + (−1)^{n}E_{n}, we have all of the following:

0 = (x_{1} − x_{1})(x_{1} − x_{2}) ⋯ (x_{1} − x_{n}) = x_{1}^{n} − E_{1}x_{1}^{n−1} + E_{2}x_{1}^{n−2} + ⋯ + (−1)^{n}E_{n}
0 = (x_{2} − x_{1})(x_{2} − x_{2}) ⋯ (x_{2} − x_{n}) = x_{2}^{n} − E_{1}x_{2}^{n−1} + E_{2}x_{2}^{n−2} + ⋯ + (−1)^{n}E_{n}

⋮

0 = (x_{n} − x_{1})(x_{n} − x_{2}) ⋯ (x_{n} − x_{n}) = x_{n}^{n} − E_{1}x_{n}^{n−1} + E_{2}x_{n}^{n−2} + ⋯ + (−1)^{n}E_{n}

so that adding we have

0 = p_{n} − p_{n−1}E_{1} + p_{n−2}E_{2} + ⋯ + (−1)^{n}nE_{n}

which is the nth equation. Now for the (n − 1)st equation, factor out the x_{j}:

0 = x_{1}(x_{1}^{n−1} − E_{1}x_{1}^{n−2} + E_{2}x_{1}^{n−3} + ⋯ + (−1)^{n−1}E_{n−1} + \frac{1}{x_{1}} (−1)^{n}E_{n})
0 = x_{2}(x_{2}^{n−1} − E_{1}x_{2}^{n−2} + E_{2}x_{2}^{n−3} + ⋯ + (−1)^{n−1}E_{n−1} + \frac{1}{x_{2}} (−1)^{n}E_{n})
⋮
0 = x_{n}(x_{n}^{n−1} − E_{1}x_{n}^{n−2} + E_{2}x_{n}^{n−3} + ⋯ + (−1)^{n−1}E_{n−1} + \frac{1}{x_{n}} (−1)^{n}E_{n})

but as x_{j} ≠ 0, begin indeterminants, we have

0 = x_{1}^{n−1} − E_{1}x_{1}^{n−2} + E_{2}x_{1}^{n−3} + ⋯ + (−1)^{n−1}E_{n−1} + \frac{1}{x_{1}} (−1)^{n}E_{n}
0 = x_{2}^{n−1} − E_{1}x_{2}^{n−2} + E_{2}x_{2}^{n−3} + ⋯ + (−1)^{n−1}E_{n−1} + \frac{1}{x_{2}} (−1)^{n}E_{n}
⋮
0 = x_{n}^{n−1} − E_{1}x_{n}^{n−2} + E_{2}x_{n}^{n−3} + ⋯ + (−1)^{n−1}E_{n−1} + \frac{1}{x_{n}} (−1)^{n}E_{n}.

Note that the sum of the last terms of each equation is

(−1)^{n}E_{n}(\frac{1}{x_{1}} +\frac{1}{x_{2}} +⋯+\frac{1}{x_{n}} ) = (−1)^{n}E_{n} \frac{x_{2}x_{3} ⋯ x_{n} + x_{1}x_{3} ⋯ x_{n} + ⋯ x_{1}x_{2} ⋯ x_{n−1}}{x_{1}x_{2} ⋯ x_{n}}

= (−1)^{n}E_{n} \frac{E_{n−1}}{E_{n}} = (−1)^{n}E_{n−1}.

Now, adding the equations,

0 = p_{n−1} − p_{n−2}E_{1} + p_{n−2}E_{2} + ⋯ + (−1)^{n−1}nE_{n−1} + (−1)^{n}E_{n−1}
= p_{n−1} − p_{n−2}E_{1} + p_{n−2}E_{2} + ⋯ + (−1)^{n−1}E_{n−1}(n − 1),

which is the (n − 1)st equation. Similarly for the others. ii. Suppose that we are given some character χ of G of dimension n and let g

be some fixed element of G. Write x_{1}, x_{2} … x_{n} for the eigenvalues of g. Then p_{1} = χ(g), p_{2} = χ(g^{2}), etc, since, for example

χ(g) = tr(ρ(g)) = tr(Sρ(g)S^{−1}) = tr(J) = ∑_{j=1}^{n} x_{j} = p_{1}

where J is the Jordan form for ρ(g). Thus, the p_{j} are known for j = 1, 2, … n. We can then inductively determine the E_{j}. For example, E_{1} = p_{1}, E_{2} = \frac{1}{2} (E_{1}p_{1} − p_{2}), etcetera. Then, the characteristic polynomial for g is given by

c_{g}(X) = (X−x_{1})(X−x_{2}) ⋯ (X−x_{n}) = X^{n}−E_{1}X^{n−1}+E_{2}X^{n−2}+⋯+(−1)^{n}E_{n}

so that the eigenvalues x_{j} are determined provided we can find the roots of this polynomial, and that’s not my problem.

Homework 5

0. We begin with the trivial and alternating characters

1 (12) (12)(34) (123) (123)(45) (1234) (12345)
χ_{U} 1 1  1  1  1  1  1
χ_{U′} 1 −1  1  1  −1  −1  1

We directly compute the character for

V = {∑_{j} α_{j}v_{j}| ∑_{j} α_{j} = 0} = ⟨v_{1} − v_{2}, v_{2} − v_{3}, v_{3} − v_{4}, v_{4} − v_{5}⟩

where v_{j} are a basis for the permutation module. For example, to compute χ_{V} (12), we compute that

(12)(v_{1} − v_{2}) = −(v_{1} − v_{2})
(12)(v_{2} − v_{3}) = v_{1} − v_{3} = (v_{1} − v_{2}) + (v_{2} − v_{3})
(12)(v_{3} − v_{4}) = v_{3} − v_{4}
(12)(v_{4} − v_{5}) = v_{4} − v_{5}

so that

ρ_{V} (12) = ( \begin{matrix}−1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{matrix} )

and so χ_{V} (12) = 2. Similarly for the other elements. The character for V′ := U′ ⊗ V is given by multiplication.

Next, we compute the character for ⋀^{2} V directly. A basis for ⋀^{2} V is

v_{1} − v_{2} ∧ v_{2} − v_{3} = v_{1} ∧ v_{2} − v_{1} ∧ v_{3} + v_{2} ∧ v_{3}
v_{1} − v_{2} ∧ v_{3} − v_{4} = v_{1} ∧ v_{3} − v_{2} ∧ v_{3} − v_{1} ∧ v_{4} + v_{2} ∧ v_{4}
v_{1} − v_{2} ∧ v_{4} − v_{5} = v_{1} ∧ v_{4} − v_{2} ∧ v_{4} − v_{1} ∧ v_{5} + v_{2} ∧ v_{5}
v_{2} − v_{3} ∧ v_{3} − v_{4} = v_{2} ∧ v_{3} − v_{2} ∧ v_{4} + v_{3} ∧ v_{4}
v_{2} − v_{3} ∧ v_{4} − v_{5} = v_{2} ∧ v_{4} − v_{3} ∧ v_{4} − v_{2} ∧ v_{5} + v_{3} ∧ v_{5}
v_{3} − v_{4} ∧ v_{4} − v_{5} = v_{3} ∧ v_{4} − v_{3} ∧ v_{5} + v_{4} ∧ v_{5}

Now to compute the value of the character at (12), we compute

(12)(v_{1} ∧ v_{2} − v_{1} ∧ v_{3} + v_{2} ∧ v_{3}) = −(v_{1} ∧ v_{2} − v_{1} ∧ v_{3} + v_{2} ∧ v_{3})
(12)(v_{1} ∧ v_{3} − v_{2} ∧ v_{3} − v_{1} ∧ v_{4} + v_{2} ∧ v_{4}) = −(v_{1} ∧ v_{3} − v_{2} ∧ v_{3} − v_{1} ∧ v_{4} + v_{2} ∧ v_{4})
(12)(v_{1} ∧ v_{4} − v_{2} ∧ v_{4} − v_{1} ∧ v_{5} + v_{2} ∧ v_{5}) = −(v_{1} ∧ v_{4} − v_{2} ∧ v_{4} − v_{1} ∧ v_{5} + v_{2} ∧ v_{5})
(12)(v_{2} ∧ v_{3} − v_{2} ∧ v_{4} + v_{3} ∧ v_{4}) = v_{2} ∧ v_{3} − v_{2} ∧ v_{4} + v_{3} ∧ v_{4} +
v_{1} ∧ v_{3} − v_{2} ∧ v_{3} − v_{1} ∧ v_{4} + v_{2} ∧ v_{4}
(12)(v_{2} ∧ v_{4} − v_{3} ∧ v_{4} − v_{2} ∧ v_{5} + v_{3} ∧ v_{5}) = v_{2} ∧ v_{4} − v_{3} ∧ v_{4} − v_{2} ∧ v_{5} + v_{3} ∧ v_{5} +
v_{1} ∧ v_{4} − v_{2} ∧ v_{4} − v_{1} ∧ v_{5} + v_{2} ∧ v_{5}
(12)(v_{3} ∧ v_{4} − v_{3} ∧ v_{5} + v_{4} ∧ v_{5}) = v_{3} ∧ v_{4} − v_{3} ∧ v_{5} + v_{4} ∧ v_{5}

so that

ρ_{⋀^{2} V} (12) = ( \begin{matrix}−1 & 0 & 0 & 0 & 0 & 0 \\ 0 & −1 & 0 & 1 & 0 & 0 \\ 0 & 0 & −1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} )

and so χ_{⋀^{2} V} (12) = 0. Similarly for the other elements. Compiling this information, we have

\begin{matrix}S_{5} & 1 & (12) & (12)(34) & (123) & (123)(45) & (1234) & (12345) \\ χ_{U} & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{U′} & 1 & −1 & 1 & 1 & −1 & −1 & 1 \\ χ_{V} & 4 & 2 & 0 & 1 & −1 & 0 & −1 \\ χ_{V′} & 4 & −2 & 0 & 1 & 1 & 0 & −1 \\ χ_{⋀^{2} V} & 6 & 0 & −2 & 0 & 0 & 0 & 1 \end{matrix}

We also verify that χ_{⋀^{2} V} is irreducible as

⟨χ_{⋀^{2} V} , χ_{⋀^{2} V} ⟩ = 1.

We recall a portion of the character table for S_{4}:

\begin{matrix}S_{4} & 1 & (12) & (12)(34) & (123) & (1234) \\ χ_{U} & 1 & 1 & 1 & 1 & 1 \\ χ_{V} & 3 & 1 & −1 & 0 & −1 \end{matrix}

Let g_{1} = 1, g_{2} = (15), g_{3} = (25), g_{4} = (35), and g_{5} = (45) be representatives of the cosets of S_{4} in S_{5}. Then the character of S_{5} induced by the character χ_{V} of S_{4} evaluated at (12) is

Ind_{S_{4}}^{S_{5}}χ_{V} (12) = ∑_{j} χ_{V} (g_{j}^{−1}gg_{j})
= χ_{V} (1(12)1) + χ_{V} (15)(12)(15) + χ_{V} (25)(12)(25) + χ_{V} (35)(12)(35) + χ_{V} (45)(12)(45)
= χ_{V} (12) + χ_{V} (25) + χ_{V} (15) + χ_{V} (12) + χ_{V} (12)
= 1 + 0 + 0 + 1 + 1
= 3.

Similarly for the other elements. Then the induced character together with the character for ⋀^{2} V are

1 10  15  20  20  30  24
S_{5}  1 (12) (12)(34) (123) (123)(45) (1234) (12345)
χ_{⋀^{2} V} 6 0  −2  0  0  0  1
χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} 15 3  −1  0  0  −1  0.

Then

⟨χ_{⋀^{2} V} , χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} ⟩ = \frac{1}{120} ∑_{j} χ_{⋀^{2} V} (g_{j})\overline{χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} (g_{j})} = 1

where here the g_{j} are the elements S_{5}. Similarly,

⟨χ_{U}, χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} ⟩ = 0
⟨χ_{U′}, χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} ⟩ = 0
⟨χ_{V} , χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} ⟩ = 1
⟨χ_{V′}, χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} ⟩ = 0

Now let m and n be the degrees of the two remaining representations. Then

120 = ∑_{j} (χ_{j}(1))^{2} = 70 + m^{2} + n^{2}

and since m and n both divide 120, trial and error gives that m = n = 5. Hence, from the inner product argument above, we have

χ_{Ind}_{S_{4}}^{S_{5}} = χ_{V} + χ_{⋀^{2} V} + χ_{W}

where W is one of the remaining modules of degree 5. Then subtraction gives

\begin{matrix}S_{5} & 1 & (12) & (12)(34) & (123) & (123)(45) & (1234) & (12345) \\ χ_{V} & 4 & 2 & 0 & 1 & −1 & 0 & −1 \\ χ_{⋀^{2} V} & 6 & 0 & −2 & 0 & 0 & 0 & 1 \\ χ_{Ind_{S_{4}}^{S_{5}}χ_{V}} & 15 & 3 & −1 & 0 & 0 & −1 & 0 \\ χ_{W} & 5 & 1 & 1 & −1 & 1 & −1 & 0 \end{matrix}

so that the second orthogonality relation gives

1 10  15  20  20  30  24
S_{5} 1 (12) (12)(34) (123) (123)(45) (1234) (12345)
χ_{U} 1 1  1  1  1  1  1
χ_{U′} 1 −1  1  1  −1  −1  1
χ_{V} 4 2  0  1  −1  0  −1
χ_{V′} 4 −2  0  1  1  0  −1
χ_{⋀^{2} V} 6 0  −2  0  0  0  1
χ_{W} 5 1  1  −1  1  −1  0
χ_{W′} 5 −1  1  −1  −1  1  0

2. We first note that if V is one-dimensional vector space over ℂ and ρ : G ⟶ GL(V) ≅ ℂ^{∗} is a representation of G over V, then since G/ ker ρ ≅ imρ ≤ ℂ is abelian, we must have that G′ ≤ ker ρ. We’ll keep in mind during the following discussion that this means that ρ maps all the elements of each coset of G′ to the same number.

Now since G/G′ is abelian, all of the representations of G/G′ are one-dimensional and there are exactly [G : G′] of them since there are [G : G′] elements in G/G′, all of which are singleton conjugacy classes. They all lift to different one-dimensional characters of G so that G has at least [G : G′] one-dimensional characters.

Now if ρ is a one-dimensional character of G, then as we observed above, ρ is constant on the cosets of G′ so that ρ̃: gG′ ↦ ρ(g) well-defined homomorphism ρ̃: G/G′ ⟶ ℂ^{∗} and hence is a one-dimensional representation of G/G′. Hence, there can be no more than [G : G′] one-dimensional representations of G. This shows that there are exactly [G : G′] one-dimensional representations of G.

Now since G′ is normal, we have

G′ = ⋂ {ker ρ_{j}|G′ ≤ ker ρ_{j}}

so that G′ can be determined from the character table by taking those elements which have character 1 for all one-dimensional characters. Of course, the other elements in those conjugacy classes are also in G′ as G′ is normal.

3. We compute that

( \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} )^{−1} = ( \begin{matrix}1 & −a & ac − b \\ 0 & 1 & −c \\ 0 & 0 & 1 \end{matrix} )

and that

( \begin{matrix}1 & −x & xz − y \\ 0 & 1 & −z \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & a & b − az − xc \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} )

so that conjugation changes only the (1, 3) entry. Moreover, whenever a ≠ 0 or c ≠ 0,

we see that the conjugacy class of ( \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} ) contains three elements since we can

select the conjugating matrix ( \begin{matrix}1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{matrix} ) with x and z such that b − az − xc =

0, 1, or 2. This gives the following conjugacy classes and sizes, where \begin{aligned}a & ∗ \\ & c \end{aligned} represents

the conjugacy class containing ( \begin{matrix}1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{matrix} ).

1  1  1  3  3  3  3  3  3  3  3
G 0 0 0 0 1 0 0 2 0 \begin{aligned}1 & ∗ \\ & 0 \end{aligned} \begin{aligned}2 & ∗ \\ & 0 \end{aligned} \begin{aligned}0 & ∗ \\ & 1 \end{aligned} \begin{aligned}1 & ∗ \\ & 1 \end{aligned} \begin{aligned}2 & ∗ \\ & 1 \end{aligned} \begin{aligned}0 & ∗ \\ & 2 \end{aligned} \begin{aligned}1 & ∗ \\ & 2 \end{aligned} \begin{aligned}2 & ∗ \\ & 2 \end{aligned}

The conjugacy classes of G happen to be the same sets as the cosets of G′ in G with the exception that the three central elements comprise the trivial coset G′ but are in singleton conjugacy classes.

We next show that G/G′ ≅ ℤ_{3} × ℤ_{3}. Define φ : ℤ_{3} × ℤ_{3} ⟶ G/G′ by

φ(x, y) = ( \begin{matrix}1 & x & ∗ \\ 0 & 1 & y \\ 0 & 0 & 1 \end{matrix} ) G′.

Then φ is easily seen to be a well-defined, surjective homomorphism so that G/G′ ≅ ℤ_{3} × ℤ_{3}.

The computation of the character table for ℤ_{3} ×ℤ_{3} is based on the following procedure. If ρ_{1}, ρ_{2} are representations of G on V, then ρ_{1} × ρ_{2} defined

(ρ_{1} × ρ_{2})(g_{1}, g_{2}) : v_{1} ⊗ v_{2} ↦ (ρ_{1}g_{1})v_{1} ⊗ (ρ_{2}g_{2})v_{2}

is a representation of G × G on V ⊗ V, where v_{j} is a basis for V. Then (ρ_{1} × ρ_{2})(g_{1}, g_{2}) has matrix

( a_{1,1}B a_{1,2}B ⋯; a_{2,1}B a_{2,2}B ⋯; ⋮ )

with respect to v_{j} where A is the matrix for ρ_{1}g_{1} and B is the matrix for ρ_{2}g_{2}. Thus, the character for ρ_{1} × ρ_{2} at (g_{1}, g_{2}) is χ_{1}g_{1}χ_{2}g_{2}. The details of all the above assertions are available upon request.

We recall the character table for ℤ_{3}:

\begin{matrix}ℤ_{3} & 0 & 1 & 2 \\ χ_{0} & 1 & 1 & 1 \\ χ_{1} & 1 & ω & ω^{2} \\ χ_{2} & 1 & ω^{2} & ω \end{matrix}

Using these characters, we construct the characters for ℤ_{3} × ℤ_{3}. For example, the character for χ_{1} × χ_{0} at (1, 0) is χ_{1}(1)χ_{0}(0) = ω · 1 = ω and similarly,

\begin{matrix}ℤ_{3} × ℤ_{3} & (0, 0) & (1, 0) & (2, 0) & (0, 1) & (1, 1) & (2, 1) & (0, 2) & (1, 2) & (2, 2) \\ χ_{1} × χ_{0} & 1 & ω & ω^{2} & 1 & ω & ω^{2} & 1 & ω & ω^{2} \end{matrix}

This produces 9 one-dimensional characters for ℤ_{3} × ℤ_{3} ≅ G/G′ which we lift to characters for G, keeping in mind the trivial coset G′, which corresponds under the isomorphism above with (0, 0), splits into three conjugacy classes, each with the same values for each character. Writing χ_{i,j} for the character lifted from χ_{i} × χ_{j}, we have

1  1  1  3  3  3  3  3  3  3  3
G 0 0 0 0 1 0 0 2 0 \begin{aligned}1 & ∗ \\ & 0 \end{aligned} \begin{aligned}2 & ∗ \\ & 0 \end{aligned} \begin{aligned}0 & ∗ \\ & 1 \end{aligned} \begin{aligned}1 & ∗ \\ & 1 \end{aligned} \begin{aligned}2 & ∗ \\ & 1 \end{aligned} \begin{aligned}0 & ∗ \\ & 2 \end{aligned} \begin{aligned}1 & ∗ \\ & 2 \end{aligned} \begin{aligned}2 & ∗ \\ & 2 \end{aligned}
χ_{0,0}  1  1  1  1  1  1  1  1  1  1  1
χ_{1,0}  1  1  1  ω  ω^{2}  1  ω  ω^{2}  1  ω  ω^{2}
χ_{2,0}  1  1  1  ω^{2}  ω  1  ω^{2}  ω  1  ω^{2}  ω
χ_{0,1}  1  1  1  1  1  ω  ω  ω  ω^{2}  ω^{2}  ω^{2}
χ_{1,1}  1  1  1  ω  ω^{2}  ω  ω^{2}  1  ω^{2}  1  ω
χ_{2,1}  1  1  1  ω^{2}  ω  ω  1  ω^{2}  ω^{2}  ω  1
χ_{0,2}  1  1  1  1  1  ω^{2}  ω^{2}  ω^{2}  ω  ω  ω
χ_{1,2}  1  1  1  ω  ω^{2}  ω^{2}  1  ω  ω  ω^{2}  1
χ_{2,2}  1  1  1  ω^{2}  ω  ω^{2}  ω  1  ω  1  ω^{2}

Now consider the subgroup

H := ⟨( \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )⟩ = \{( \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) , ( \begin{matrix}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) , ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )\} ≤ G

We compute the character Ind_{H}^{G}(ψ) of G where ψ is the character of H given by

\begin{matrix}H & ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) & ( \begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) & ( \begin{matrix}1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) \\ ψ & 1 & ω & ω^{2} \end{matrix}

For example,

Ind_{H}^{G}(ψ) ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) = |C_{G} ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} )| ∑_{j} \frac{ψ(g_{j})}{|C_{H}g_{j}|}

= 27\frac{1}{3}
= 9

where C is the conjugacy class of G containing ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) and g_{j} are representa-

tives of the conjugacy classes D_{j} of H with ⋃_{j} D_{j} = C ∩ H. In this case, C ∩ H =

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) = g_{1}. Similarly for the other elements. We then have that

\begin{matrix}G & 0 0 0 & 0 1 0 & 0 2 0 & \begin{aligned}1 & ∗ \\ & 0 \end{aligned} & \begin{aligned}2 & ∗ \\ & 0 \end{aligned} & \begin{aligned}0 & ∗ \\ & 1 \end{aligned} & \begin{aligned}1 & ∗ \\ & 1 \end{aligned} & \begin{aligned}2 & ∗ \\ & 1 \end{aligned} & \begin{aligned}0 & ∗ \\ & 2 \end{aligned} & \begin{aligned}1 & ∗ \\ & 2 \end{aligned} & \begin{aligned}2 & ∗ \\ & 2 \end{aligned} \\ Ind_{H}^{G}ψ & 9 & 0 & 0 & 9ω & 9ω^{2} & 0 & 0 & 0 & 0 & 0 & 0 \end{matrix}

We compute that

9 = ⟨Ind_{H}^{G}ψ, Ind_{H}^{G}ψ⟩ = ∑_{j}⟨Ind_{H}^{G}, χ_{j}⟩^{2}

where χ_{j} are the irreducible characters of G. This is only possible if ⟨Ind_{H}^{G}, χ_{j}⟩ = 3 for some j or if Ind_{H}^{G}, χ_{j}⟩ = 1 for 9 different j’s. In the latter case, we must have ⟨Ind_{H}^{G}, χ_{j}⟩ = 1 for the 9 characters χ_{j} already computed as the remaining two have degree 3. We eliminate this possibility since Ind_{H}^{G}ψ ≠ ∑_{j=1}^{9} χ_{j}. Hence, Ind_{H}^{G}ψ = 3χ_{j} for some new character χ_{j}. The conjugate of this character is also an irreducible character so that we have

1  1  1  3  3  3  3  3  3  3  3
G  0 0 0 0 1 0 0 2 0 \begin{aligned}1 & ∗ \\ & 0 \end{aligned} \begin{aligned}2 & ∗ \\ & 0 \end{aligned} \begin{aligned}0 & ∗ \\ & 1 \end{aligned} \begin{aligned}1 & ∗ \\ & 1 \end{aligned} \begin{aligned}2 & ∗ \\ & 1 \end{aligned} \begin{aligned}0 & ∗ \\ & 2 \end{aligned} \begin{aligned}1 & ∗ \\ & 2 \end{aligned} \begin{aligned}2 & ∗ \\ & 2 \end{aligned}
χ_{0,0}  1  1  1  1  1  1  1  1  1  1  1
χ_{1,0}  1  1  1  ω  ω^{2}  1  ω  ω^{2}  1  ω  ω^{2}
χ_{2,0}  1  1  1  ω^{2}  ω  1  ω^{2}  ω  1  ω^{2}  ω
χ_{0,1}  1  1  1  1  1  ω  ω  ω  ω^{2}  ω^{2}  ω^{2}
χ_{1,1}  1  1  1  ω  ω^{2}  ω  ω^{2}  1  ω^{2}  1  ω
χ_{2,1}  1  1  1  ω^{2}  ω  ω  1  ω^{2}  ω^{2}  ω  1
χ_{0,2}  1  1  1  1  1  ω^{2}  ω^{2}  ω^{2}  ω  ω  ω
χ_{1,2}  1  1  1  ω  ω^{2}  ω^{2}  1  ω  ω  ω^{2}  1
χ_{2,2}  1  1  1  ω^{2}  ω  ω^{2}  ω  1  ω  1  ω^{2}
Ind_{H}^{G}ψ  3  0  0  3ω  3ω^{2}  0  0  0  0  0  0
Ind_{H}^{G}ψ  3  0  0  3ω^{2}  3ω  0  0  0  0  0  0

The Mackey Problem

(a) Note that

⟨ψ_{1}^{(12)}, ψ_{1}^{(12)}⟩ = ∑_{g} ψ_{1}((12)g(12))\overline{ψ_{1}((12)g(12))}
= ∑_{g} ψ_{1}(g)\overline{ψ_{1}(g)}
= ⟨ψ_{1}, ψ_{1}⟩
= 1

so that ψ_{1}^{(12)} is an irreducible character of H. We have that G = H1H ∪ H(12)H so that D = {1, (12)} may be taken as a set of double coset representatives. Then

⟨Indψ_{1}Indψ_{2}⟩ = ⟨ψ_{1}^{1}|_{1H1∩H}, ψ_{2}|_{1H1∩H}⟩ + ⟨ψ_{1}^{(12)}|_{(12)H(12)∩H}, ψ_{2}|_{(12)H(12)∩H}⟩
= ⟨ψ_{1}, ψ_{2}⟩ + ⟨ψ_{1}^{(12)}, ψ_{2}⟩

Suppose now that ψ_{1} = ψ_{1}^{(12)}. Then

⟨Indψ_{1}Indψ_{2}⟩ = ⟨ψ_{1}, ψ_{2}⟩ + ⟨ψ_{1}, ψ_{2}⟩ = \{ \begin{matrix}2 & if ψ_{1} = ψ_{2} \\ 0 & if ψ_{1} ≠ ψ_{2} \end{matrix}

and if ψ_{1} ≠ ψ_{1}^{(12)}

⟨Indψ_{1}, Indψ_{2}⟩ = ⟨ψ_{1}^{(12)}, ψ_{2}⟩ + ⟨ψ_{1}, ψ_{2}⟩ = \{ \begin{matrix}1 & if ψ_{1}^{(12)} = ψ_{2} or ψ_{1} = ψ_{2} \\ 0 & if ψ_{1}^{(12)} ≠ ψ_{2} and ψ_{1} ≠ ψ_{2} \end{matrix} .

(b) Let χ be an irreducible character of G and consider Res χ. We have that no irreducible character ψ of H can appear in Res U more than 2 times since if ⟨Res χ, ψ⟩ > 2, then by Frobenius, ⟨χ, Ind ψ⟩ > 2, that is, χ appears more than 2 times in Ind ψ. This means that {number of constituents of Ind ψ} = ⟨Ind ψ, Ind ψ⟩ > 2, which contradicts part (a). Now suppose that ψ is a constituent of Res χ. Then, of course, we have deg ψ ≤ deg Res χ = deg χ. However, in this situation, we also have that χ is a constituent of Ind ψ so that

deg χ ≤ deg Ind ψ = [G : H] deg ψ = 2 deg ψ.

Compiling this information, we have

deg ψ ≤ deg χ ≤ 2 deg ψ

and this inequality is only satisfied when deg ψ = deg χ or when deg ψ = \frac{1}{2} deg χ. This shows that Res χ can have no more than two different consituents.

(c) Restricting the characters of G and taking inner products, we have that the following characters are irreducible.

1 20  15  12  12
A_{5} 1 (123) (12)(34) (12345) (21345)
Res U 1 1  1  1  1
Res V 4 1  0  −1  −1
Res W 5 −1  1  0  0

Also, Res U′ = Res U, Res V′ = Res V, and Res W′ = Res W provide no new information. We have, however, that the two remaining characters both have degree 3 since ∑_{j}(χ_{j}(1))^{2} = 60.

Restricting ⋀^{2} V,

\begin{matrix}A_{5} & 1 & (123) & (12)(34) & (12345) & (21345) \\ Res ⋀^{2} V & 6 & 0 & −2 & 1 & 1 \end{matrix}

we find that ⟨Res ⋀^{2} V, Res ⋀^{2} V, ⟩ = 2 so that by part (b), Res ⋀^{2} V is the sum of two irreducible characters Y and Z. We observe that these two characters must be different since if Res ⋀^{2} V = 2Y, then

\begin{matrix}A_{5} & 1 & (123) & (12)(34) & (12345) & (21345) \\ Y & 3 & 0 & −1 & \frac{1}{2} & \frac{1}{2} \end{matrix}

and ⟨Y, Y ⟩ = \frac{1}{2}, which contradicts the assumption that Y is irreducible. We next observe that ⟨Res ⋀^{2} V, Res U⟩ = 0, ⟨Res ⋀^{2} V, Res V ⟩ = 0, and ⟨Res ⋀^{2} V, Res U⟩ = 0. These lead to the system

20y_{1} + 15y_{2} + 12y_{3} + 12y_{4} = 3
20y_{1} − 12y_{3} − 12y_{4} = −12
−20y_{1} + 15y_{2} = −15

where y_{j} are the remaining values of Y:

1 20  15  12  12
A_{5} 1 (123) (12)(34) (12345) (21345)
Res U 1 1  1  1  1
Res V 4 1  0  −1  −1
Res W 5 −1  1  0  0
Y  3 y_{1}  y_{2}  y_{3}  y_{4}

This system reduces to

y_{1} = 0
y_{2} = −1
y_{3} + y_{4} = 1

Thus, Y so far is the following.

1 20  15  12  12
A_{5} 1 (123) (12)(34) (12345) (21345)
Y 3 0  −1  y_{3}  1 − y_{3}

Now since 1 = ⟨Y, Y ⟩ = \frac{1}{60} (9 + 15 + 12y_{3}^{2} + 12(1 − y_{3})^{2}), we have that 0 = y_{3}^{2} −

y_{3} − 1 = 0. The quadratic formula gives y_{3} = \frac{1 ± \sqrt{5}}{2}. We can take either value for y_{3}. Thus we have

1 20  15  12  12
A_{5} 1 (123) (12)(34) (12345) (21345)
Res U 1 1  1  1  1
Res V 4 1  0  −1  −1
Res W 5 −1  1  0  0
Y  3 0  −1  \frac{1+\sqrt{5}}{2}  \frac{1−\sqrt{5}}{2}
Z 3 0  −1  \frac{1−\sqrt{5}}{2}  \frac{1+\sqrt{5}}{2}

The Prelim Problem.

Let P be a non-abelian group of order p^{3}. Forevermore I will write g_{j} for the representatives of the non-central conjugacy classes of whatever group is under discussion.

We determine the size of Z(P). |Z(P)| ≠ p^{3} since P is non-abelian.

The class equation gives

p^{3} = |Z(P)| + ∑_{j} \frac{p^{3}}{|C_{G}(g_{j})|}.

Now C_{G}(g_{j}) contains at least g_{j} and the identity element, and these are different, so |C_{G}(g)| ≥ p. But as g_{j} is non-central, we have |C_{G}(g_{j})| ≠ p^{3} so that |C_{G}(g_{j})| = p or p^{2}. This means that p divides ∑_{j} \frac{p^{3}}{|C_{G}(g_{j})|}. We then have that p divides |Z(P)| =

p^{3} − ∑_{j} \frac{p^{3}}{|C_{G}(g_{j})|}. This eliminates the possibility that |Z(P)| = 1. (The same argument shows that we have |Z(G)| ≠ 1 for any group G of order p^{n}. We use this observation later.)

Now if |Z(P)| = p^{2}, then C_{G}(g_{j}) contains at least g_{j} and all p^{2} central elements so that |C_{G}(g_{j})| = p^{3}, a contradiction since g_{j} ∉ Z(P).

We conclude then that |Z(P)| = p. Moreover, |C_{G}(g_{j})| includes at least g_{j} and the p central elements, so |C_{G}(g_{j})| > p, but as g_{j} ∉ Z(P), we have |C_{G}(g_{j})| ≠ p^{3} so that |C_{G}(g_{j})| = p^{2}. Thus, each non-central conjugacy class contains exactly p elements. From the class equation, we have

p^{3} = |Z(P)| + ∑_{j} \frac{p^{3}}{C_{G}(g_{j})} = p + rp

yielding r = p^{2} − 1 non-central conjugacy classes.

Next, we deduce that any group G of order p^{2} is abelian. If |Z(G)| = p then |C_{G}(g_{j})| = p^{2}, since it contains at least g_{j} and all p central elements. This contradicts g_{j} ∉ Z(G). As noted above, |Z(G)| ≠ 1. Hence Z(G) = G and G is abelian.

This means that P/Z(P) is abelian as |P/Z(P)| = p^{2}. We then have that P′ ≤ Z(P). But as P′ ≠ 1 since P is non-abelian, we have that P′ = Z(P).

The Dummit and Foote Problem

(a) i. By the prelim discussion above, we have that the commutator subgroup of P has order p so that by the last homework, P has [P : P′] = p^{2} characters of degree 1.

ii. The degrees of the characters of P divide p^{3}. G can have no irreducible character χ_{j_{0}} of degree p^{2} as then

p^{3} = ∑_{j}(χ_{j}(1))^{2} > (χ_{j_{0}}(1))^{2} = p^{4},

where χ_{j} are the irreducible characters of P, and this is a contradiction. Similarly, P can have no irreducible character of degree p^{3}. Hence, all the characters of P have degree 1 or p. Let r be the number of characters of degree p. Then

p^{3} = ∑_{j}(χ_{j}(1))^{2} = p^{2} · 1^{2} + r · p^{2} = p^{2}(1 + r)

so that P has r = p − 1 characters of degree p. iii. The conjugacy classes were determined above. Now suppose that g_{1} ≠ 1 and

that g_{1} = hg_{2}h^{−1}. Then g_{1}g_{2}^{−1} = hg_{2}h^{−1}g_{2}^{−1} ∈ Z(P). This means that

g_{1} = (g_{1}g_{2}^{−1})g_{2} = g_{2}(g_{1}g_{2}^{−1}) ∈ g_{2}Z(P).

This shows that whenever g_{1} and g_{2} are non-central, conjugate elements, they are in the same coset of Z(P). Conversely, if g_{1} and g_{2} are in the same coset…

iv. Let ρ be an irreducible representation of degree p and consider ker ρ. Of course, | ker ρ| ≠ p^{3} else ρ = pρ_{U} is not irreducible. If | ker ρ| = p or p^{2}, then |G/ ker ρ| = p^{2} or p, so that G/ ker ρ is abelian. Now

ρ̃: G/ ker ρ ⟶ GL(V) defined ρ̃(g ker ρ) = ρ(g)

is a representation of degree p on G/ ker ρ. Now,

1 = ⟨ρ, ρ⟩ = \frac{1}{|G|} ∑_{g∈G} ρ(g)\overline{ρ(g)} = \frac{1}{|G|} ∑_{g ker ρ∈G/ ker ρ} | ker ρ|ρ(g)\overline{ρ(g)}

= \frac{1}{|G/ ker ρ|} ∑_{g ker ρ∈G/ ker ρ} ρ̃(g ker ρ)\overline{ρ̃(g ker ρ)} = ⟨ρ̃, ρ̃⟩

so that ρ̃ is irreducible. However, since G/ ker ρ is abelian, it can have no irreducible representations of degree p, a contradiction. This shows that | ker ρ| = 1 so that ρ is faithful.

v. Let z ∈ Z(P) and let V be an irreducible ℂP module of degree p. Then the map φ_{z} : V ⟶ V defined φ_{z}(v) = z · v is a ℂP homomorphism. Let λ be such that det (φ_{z} − λI) = 0. Then φ_{z} − λI is also a ℂP homomorphism and ker (φ − λI) ≠ 0. But as V is irreducible, we have that ker (φ − λI) = V so that z · v = λv for all v ∈ V. Moreover, since z^{p} = 1, we have v = z^{p} · v = λ^{p}v so that λ is a pth root of unity. Also, it follows that χ_{V} (z) = pλ. Note also that if z ∈ Z(P) but z ≠ 1, then z generates Z(P) and we have z^{j} · v = λ^{j}v so that χ_{V} (z^{j}) = pλ^{j}, and so X_{V} is determined for all the elements of Z(P). Now if g ∈ P \Z(P), we have that the conjugacy class containing g is

{g, gz, gz^{2}, ⋯, gz^{p−1}}

by the observation above that the non-identity cosets of Z(P) are conjugacy classes of P. In particular, this means that χ_{V} (gz^{j}) = χ_{V} (g) for all j since χ_{V} is constant on conjugacy classes. However, since z^{j} acts as scalar multiplication by λ^{j}, we have that χ_{V} (gz^{j}) = λ^{j}χ_{V} (g) for all j. Thus,

0 = (∑_{j} λ^{j}) χ_{V} (g) = ∑_{j} χ_{V} (gz^{j}) = pχ_{V} (g)

so that χ_{V} (g) = 0. This determines the character for V vi. Let ρ_{1} and ρ_{2} be irreducible representations of P of degree p. As indicated

above, if ρ_{1}(z) = ρ_{2}(z) for any z ∈ Z(P) other than 1, then ρ_{1} and ρ_{2} agree on all of Z(P), and since ρ_{j} vanishes outside of Z(P), we have that ρ_{1} and ρ_{2} agree on all of P. Hence, if ρ_{1} ≠ ρ_{2}, then ρ_{1}(z) ≠ ρ_{2}(z) for all z ∈ Z(P)\{1}. Hence, the p − 1 representations of degree p can be produced by assigning one of the pth roots of unity other than 1 to some z ∈ Z(P) other than 1.

The Group of order 42.

(a) The character table for G is the following.

1 6 7 7 7 7 7
G 1 a b c d e f
χ_{1} 1 1 1 1 1 1 1
χ_{2} 1 1 1 ζ \overline{ζ} ζ \overline{ζ}
χ_{3} 1 1 1 \overline{ζ} ζ \overline{ζ} ζ
χ_{4} 1 1 −1 −ζ −\overline{ζ} ζ \overline{ζ}
χ_{5} 1 1 −1 −\overline{ζ} −ζ \overline{ζ} ζ
χ_{6} 1 −1 −1 −1 1 1 1
χ_{7} 6 −1 0 0 0 0 0

We see from the table that ker ρ_{1} ∩ ker ρ_{2} ∩ ker ρ_{3} = O_{1} ∪ O_{a} ∪ O_{b}. Since ker ρ_{j} is a normal subgroup of G and the intersection of three normal subgroups is normal, we have that N := O_{1} ∪ O_{a} ∪ O_{b} is a normal subgroup of order 14. Moreover, since |G/N| = 3, we have that G/N ≅ ℤ_{3} and that the cosets of N are exactly O_{c} ∪ O_{e} and O_{d} ∪ O_{f} as only this combination causes χ_{1}, χ_{2}, and χ_{3} to agree with the characters of ℤ_{3}. (?) N is not ℤ_{14}, so we must have N ≅ D_{14}.

(b) We briefly compute the conjugacy classes of D_{14} = ⟨r, s|r^{7} = s^{2} = 1, rs = sr^{6}⟩. Since r^{j}, j = 1 … 6, commutes only with r^{k}, k = 0 … 6, we have |C_{D_{14}}(r^{j})| = 7, j = 1 … 6 so that |O_{r^{j}}| = 2. We explicitly compute that O_{r} = {r, r^{6}}, O_{r^{2}} = {r^{2}, r^{5}}, and O_{r^{3}} = {r^{3}, r^{4}}. By order considerations, we must have that the G orbit of a is the union of these three N orbits. Similarly, sr^{j}, j = 0 … 6 commutes only with itself and with 1 so that |C_{D_{14}}(sr^{j})| = 2 and |O_{sr^{j}}| = 7. We must have then that O_{s} = {sr^{j}}_{j=0}^{6} and of course O_{1} = {1}. The distinct restrictions then are the following.

1 2 2 2 7
N 1 r r^{2} r^{3} s
Resχ_{1} 1 1 1 1 1
Resχ_{4} 1 1 1 1 −1
Resχ_{7} 6 −1 −1 −1 0

We compute ⟨Res χ_{1}, Res χ_{1}⟩ = ⟨Res χ_{4}, Res χ_{4}⟩ = 1 so that Res χ_{1} and Res χ_{4} are irreducible. Unfortunately, Res χ_{1} and Res χ_{4} are not constituents of Res χ_{7} as ⟨Res χ_{1}, Res χ_{7}⟩ = 0 and ⟨Res χ_{4}, Res χ_{7}⟩ = 0. However, the remaining three characters of N must each have degree 2 since ∑ χ_{j}(1)^{2} = 14. Now since ⟨Res χ_{7}, Res χ_{7}⟩ = 3 and this number is the sum of the squares of the multiplicities of the constituents, we must have that these multiplicities are all 1 so that Res χ_{7} is the sum of the three remaining characters.

(c) n_{7} = 7j + 1 and divides 6. This forces n_{7} = 1. n_{3} = 3j + 1 and divides 14. Then n_{3} = 1 or 7. Write G/N = {N, g_{1}N, g_{2}N} for some g_{1}, g_{2}. Now (g_{1}N)^{3} = g_{1}^{3}N = N so that g_{1} ∈ N. Now since |N| = 14, we have that (g_{1}^{3})^{14} = (g_{1}^{14})^{3} = 1. Also, we have that g_{1}^{14} ≠ 1 else (g_{1}N)^{14} = g_{1}^{14}N = N = (g_{1}N)^{3} which forces g_{1}N = N since 3 and 14 are relatively prime, a contradiction. Incidentally, I stole this argument from Herstein’s proof of the Cauchy theorem, which I enjoyed immensely. This shows that |g_{1}^{14}| = 3. Similarly, |g_{2}^{14}| = 3 and g_{1}^{14} ≠ g_{2}^{14} since g_{1} ≠ g_{2}, being representatives of different cosets. Thus there are at least 2 elements of order 3 so that n_{3} ≠ 1. This forces n_{3} = 7. Finally, we have n_{2} = 2j + 1 and divides 21. Now since N itself has 7 elements of order 2, namely sr^{j} for j = 0 … 6, we have that G has at least 7 such elements, and these are the elements in O_{b} as adduced above. There can be no other elements of order 2 since χ_{2} is ζ or \overline{ζ} on elements of O_{c}, O_{d}, O_{e}, and O_{f} and this is not possible if any of these elements has order 2.

(d) As indicated above, a and b have orders 7 and 2 respectively. χ_{2}(c) = χ_{3}(d) = ζ so that 3 divides the orders of these elements. However, χ_{6}(c) = χ_{6}(d) = −1, a second root of unity, so 2 divides the orders of c and d also. Obviously, these elements don’t have order 42, so we are left with 6 as the only possible order for c and d. By elimination, e and f must have order 3 to account for the 14 elements of order 3 in the 7 Sylow 3-subgroups.

Next, let P be a Sylow 7 subgroup and Q_{j}, j = 1 … 7 the Sylow 3 subgroups. Then since P ⊴ G, we have that P Q_{j} ≤ G with |P Q_{j}| = 21. P Q_{j} has index 2 and so is normal in G. We have that Q_{j} ≤ P Q_{j}, but we also have that since the Q_{j} are conjugate,

Q_{k} = gQ_{j}g^{−1} ≤ gP Q_{j}g^{−1} = P Q_{j}

so that all the Q_{k} lie in P Q_{j}. Also, the 14 elements of order 3 in the 7 Sylow 3-subgroups together with the 6 elements of order 7 from the 7-subgroup and the identity constitute all the elements of P Q_{j}.

(a) Define φ(aa_{λ}b_{λ}) = aa_{λ}b_{λ}a_{λ}. Then φ : Aa_{λ}b_{λ} ⟶ Ab_{λ}a_{λ} as aa_{λ} ∈ A. Moreover, φ is a A-module homomorphism as

φ(a_{1}a_{λ}b_{λ} + a_{2}a_{λ}b_{λ}) = φ((a_{1} + a_{2})a_{λ}b_{λ})
= (a_{1} + a_{2})a_{λ}b_{λ}a_{λ}
= a_{1}a_{λ}b_{λ}a_{λ} + a_{2}a_{λ}b_{λ}a_{λ}
= φ(a_{1}a_{λ}b_{λ}) + φ(a_{2}a_{λ}b_{λ})

and

φ(a_{1}a_{2}a_{λ}b_{λ}) = a_{1}a_{2}a_{λ}b_{λ}a_{λ} = a_{1}φ(a_{2}a_{λ}b_{λ}).

Similarly, the map ψ : Ab_{λ}a_{λ} ⟶ Aa_{λ}b_{λ} defined ψ(ab_{λ}a_{λ}) = ab_{λ}a_{λ}b_{λ} is an Amodule homomorphism. Moreover, φ and ψ are almost two sided inverses as

ψ ∘ φ(aa_{λ}b_{λ}) = aa_{λ}b_{λ}a_{λ}b_{λ} = ac_{λ}^{2} = adc_{λ} = daa_{λ}b_{λ}

and

φ ∘ ψ(ab_{λ}a_{λ}) = ab_{λ}a_{λ}b_{λ}a_{λ} = ac_{λ}^{2} = adc_{λ} = dab_{λ}a_{λ}

where d ∈ ℂ is such that c_{λ}^{2} = dc_{λ}. Incidentally, the same d works for b_{λ}a_{λ} as

a_{λ}b_{λ}a_{λ}b_{λ}a_{λ} = (a_{λ}b_{λ}a_{λ}b_{λ})a_{λ} = (da_{λ}b_{λ})a_{λ} = a_{λ}(db_{λ}a_{λ})

so that by by right cancelation, b_{λ}a_{λ}b_{λ}a_{λ} = db_{λ}a_{λ}. We then have that φ is surjective since for any a ∈ A we have φ(d^{−1}ψ(a)) = a and φ is injective since whenever φ(a_{1}) = φ(a_{2}) we have ψ(φ(a_{1})) = ψ(φ(a_{2})) so that a_{1} = a_{2}. Thus, φ is an isomorphism and Aa_{λ}b_{λ} ≅ Ab_{λ}a_{λ}

(b) The map φ : Aa_{λ} ⟶ Aa_{λ}b_{λ} given by φ(aa_{λ}) = aa_{λ}b_{λ} is clearly surjective. Hence, A_{λ} is the image of Aa_{λ} under φ.

(c) We have

Ab_{λ′} ≅ Ind_{Q_{λ′}}^{G} Ũ′
= Ind_{P_{λ}}^{G}Ũ′
= Ind_{P_{λ}}^{G} (Ũ′ ⊗ U)
= Ind_{P_{λ}}^{G} (Res_{P_{λ}}^{G}U′ ⊗ U)
= U′ ⊗ Ind_{P_{λ}}^{G}U
≅ U′ ⊗ Aa_{λ}.

where Ũ′ is the alternating module for Q_{λ′} = P_{λ}, U′ is the alternating module for G, and U is the trivial module for P_{λ}. By a parallel argument, we have that Aa_{λ′} ≅ U′ ⊗ Ab_{λ}. Now by part 2, we have that the image of Aa_{λ′} under right multiplication by b_{λ′} is V_{λ′}. We want to apply this same map to Ab_{λ} ⊗ U′, but first we note that

(u ⊗ ab_{λ})b_{λ′} = (u ⊗ ab_{λ}) ∑_{j}(sgn p_{j})p_{j}
= ∑_{j}(u ⊗ ab_{λ})(sgn p_{j})p_{j}
= ∑_{j} u(sgn p_{j})p_{j} ⊗ ab_{λ}(sgn p_{j})p_{j}
= ∑_{j} up_{j} ⊗ ab_{λ}p_{j}
= ∑_{j}(u ⊗ ab_{λ})p_{j}
= (u ⊗ ab_{λ}) ∑_{j} p_{j}
= (u ⊗ ab_{λ})a_{λ}

so that in U′ ⊗ Ab_{λ}, right mulfftiplication by b_{λ′} is the same as right multiplication by a_{λ}. Again, by part 2, we have that the image of U′ ⊗ Ab_{λ} under right multiplication by a_{λ} is U′ ⊗ V_{λ}. Now since U′ ⊗ Ab_{λ} and Aa_{λ′} are isomorphic, we have that their images under right multiplication by b_{λ′} are isomorphic. Hence, we have that V λ′ ≅ V_{λ} ⊗ U′.

Homework 8

15. Prove that ℚ(φ) is a finite extension

Let B be a fixed basis of some vector space V of dimension m over F. Now ℚ(φ) = ℚ(a_{1,1}^{g_{1}}, a_{1,2}^{g_{1}} …) where the a_{i,j}^{g_{k}} are the i, j entries of the matrix φ(g_{k}) with respect to B. Then (φ)/ℚ is a finite extension of since there are no more than m^{2}|G| different a_{i,j}^{g_{k}}, each of them algebraic over ℚ, being in F.

16. Prove that φ^{σ} is a representation and that the character of φ^{σ} is ψ^{σ} := σ(ψ(g)) where ψ is the character of φ.

Let A = (a_{i,j}) and B = (b_{i,j}) be the matricies φ(g) and φ(h) with respect to some fixed basis. Then φ^{σ}(g) = (σ(a_{i,j})) and φ^{σ}(h) = (σ(b_{i,j})). Now

i, j entry of φ^{σ}(g)φ^{σ}(h) = ∑_{k} σ(a_{i,k})σ(b_{k,j})

= σ (∑_{k} a_{i,k}b_{k,j})

= σ(i, j entry of φ(g)φ(h))
= σ(i, j entry of φ(gh))
= i, j entry of φ^{σ}(gh).

This shows that φ^{σ} is a homomorphism. Next, we compute the character of φ^{σ}.

tr(φ^{σ}(g)) = ∑_{j} σ(a_{j,j})

= σ (∑_{k}(a_{j,j}))
= σ (tr(φ(g)))
= σ (ψ(g))

17. Show that φ is irreducible if and only if φ^{σ} is irreducible.

We note that

\overline{σ(φ(g))} = \overline{σ(a + bi)}
= \overline{σ(a) + σ(b)i}
= σ(a) − σ(b)i
= σ(a − bi)
= σ(\overline{a + bi})
= σ(\overline{φ(g)})

where \overline{·} denotes complex conjugation. Then

⟨φ^{σ}, φ^{σ}⟩ = \frac{1}{|G|} ∑_{g} φ^{σ}(g)\overline{φ^{σ}(g)}

= \frac{1}{|G|} ∑_{g} σ(φ(g))\overline{σ(φ(g))}

= \frac{1}{|G|} ∑_{g} σ(φ(g))σ(\overline{φ(g)})

= σ (\frac{1}{|G|} ∑_{g} φ(g)\overline{φ(g)})
= σ (⟨φ, φ⟩)

and we have that ⟨φ^{σ}, φ^{σ}⟩ = 1 if and only if ⟨φ, φ⟩ = 1 since automorphisms map the element 1 to itself.

18. Prove that ℚ(ψ) ⊂ ℚ(ϵ) where ℚ(ψ) is the extension of ℚ generated by ψ(g), g ∈ G, and ϵ is an nth root of unity where n = |G|. Deduce that ℚ(ψ) is a Galois extension of ℚ with abelian Galois group.

ψ(g) is a sum of eigenvalues of φ(g), which are all nth root of unity. This means that

ψ(g) ∈ ℚ(ϵ)

for all g so that

ℚ(ψ) = ℚ(ψ(g_{1}), ψ(g_{2}), … ψ(g_{n})) ⊆ ℚ(ϵ).

Since the Galois group for ℚ(ϵ)/ℚ is (ℤ/nℤ)^{×}, we see by the Galois correspondence that ℚ(φ), being a subfield of ℚ(ϵ) containing ℚ, corresponds with some subgroup H of (ℤ/nℤ)^{×}. Now H is normal since (ℤ/nℤ)^{×} is abelian so that ℚ(ψ)/ℚ is Galois.

19. Let σ_{a} ∈ Gal(ℚ(ϵ)/ℚ) be defined σ_{a}(ϵ) = ϵ^{a} and show that ψ^{σ_{a}}(g) = ψ(g^{a}) for all g ∈ G.

Let φ be a representation of G on some vector space V of dimension m. Let φ have character ψ and let B be a fixed basis of V with respect to which φ(g) = (b_{i,j}) with b_{i,j} = 0 for i ≠ j and b_{j,j} a kth root of unity where k = |g|. Write b_{j,j} = ϵ^{c}. Then

σ_{a}(b_{j,j}) = σ_{a}(ϵ^{c}) = (σ_{a}(ϵ))^{c} = (ϵ^{a})^{c} = ϵ^{ca} = (b_{j,j})^{a}.

Also note that with respect to B, we have that φ(g^{a}) = (b_{i,j}^{a}) since φ is diagonal. Then

ψ^{σ_{a}}(g) = ∑_{j=1}^{m} σ_{a}(b_{j,j}) = ∑_{j=1}^{m}(b_{j,j}^{a}) = ψ(g^{a})

20. If g ∈ G is conjugate with g^{a} for all a with (a, n) = 1, then ψ(g) ∈ ℚ for all characters ψ of G.

We have

ψ(g) = ψ(g^{a}) = ψ^{σ_{a}}(g) = ∑_{j=1}^{m} σ_{a}(b_{j,j}) = σ_{a} (∑_{j=1}^{m} b_{j,j}) = σ_{a}(ψ(g))

for all characters ψ and all σ_{a} ∈ Gal(ℚ(ϵ)/ℚ) = {σ_{a}|(a, n) = 1}. By definition of the Galois group, since ψ(g) ∈ ℚ(ϵ) is fixed by all σ_{a} ∈ Gal(ℚ(ϵ)/ℚ), we have that ψ(g) ∈ ℚ for all ψ.

21. For g ∈ G fixed, g is conjugate with g^{a} for all (a, |G|) = 1 iff g is conjugate with g^{a} for all (a, |g|) = 1.

Suppose g is conjugate with g^{a} for all (a, |g|) = 1 and let a be such that (a, |G|) = 1. Then we certainly have that (a, |g|) = 1 so that g is conjugate with g^{a} by assumption.

Conversely, suppose g is conjugate with g^{a} for all (a, |G|) = 1 and let a be such that (a, m) = 1 where m = |g|. Let l be such that n = lm and let k be such that l = dk where d = (l, m). We then have that n = dkm and (m, k) = 1. Write 1 = rm + sk for some r, s ∈ ℤ and define x = ask + rm. Then

x ≡_{m} ask + rm ≡_{m} ask ≡_{m} a

since sk ≡_{m} rm + sk ≡_{m} 1.

This equivalence implies that m divides x − a so that my = x − a for some y ∈ ℤ. Then

g^{x}g^{−a} = g^{x−a} = g^{my} = 1

so that g^{a} = g^{x}. We similarly have that

x ≡_{k} ask + rm ≡_{k} rm ≡_{k} 1

since rm ≡_{k} rm + sk ≡_{k} 1.

This equivalence imples that k|x − 1. Now if some prime p divides k, then p divides x − 1 so that p cannot divide x. This shows that (k, x) = 1.

Finally, we want to show that (x, n) = 1. Indeed, if some prime p were a common factor of both x and n = dkm, then by the preceeding paragraph, p would not be a factor of k. Now since x ≡_{m} a, we have that m divides x − a. If p were a factor of m, then since p is a factor of x, we would have that p would be a factor of a, a contradiction since (a, m) = 1. Finally, p cannot be a factor of d = (l, m) as then it would have to be a factor of m. Hence, n and x can have no common factors, that is, (n, x) = 1. By assumption, we then have that g is congruent with g^{x} = g^{a}.

22. Every character of S_{n} is rational valued.

Let g ∈ S_{n} and let g = τ_{1}τ_{2} ⋯ τ_{m} where τ_{j} are disjoint n_{j}-cycles. Then |g| = lcm n_{j} and if a is such that (a, lcm n_{j}) = 1, then we must have that (a, n_{j}) = 1 for all j, for if (a, n_{j_{0}}) ≠ 1 for some j_{0}, then we have

(a, n_{j_{0}}) | n_{j_{0}} | lcm n_{j},

but we have also that (a, n_{j_{0}}) | a so that (a, n_{j_{0}}) is a common divisor of both lcm n_{j} and a. This means that (a, lcm n_{j}) > 1, a contradiction. We must have that (a, n_{j}) = 1 for all j. Then the ath power of each n_{j}-cycle is also an n_{j}-cycle so that g^{a} has the same cycle type as g. Thus, g^{a} is conjugate with g. Thus, we’ve shown that g is conjugate with g^{a} for all (a, |g|) = 1 and for all g ∈ G. Thus, by 21, we have that g^{a} is conjugate with g for all (a, |G|) = 1. Thus, by 20, we have that ψ(g) ∈ ℚ for all g ∈ G and all characters ψ of G.

Homework 9

(a) Show that |G| = q(q − 1)(q^{2} − 1)).

We count the number of bases \{(\begin{matrix}a \\ b \end{matrix}) , (\begin{matrix}c \\ d \end{matrix})\} of 𝔽_{q}^{2} since these are in bijection

with the elements of GL(2, 𝔽_{q}).

Now there are q choices for a and q choices for b except that (\begin{matrix}a \\ b \end{matrix}) should not be

(\begin{matrix}0 \\ 0 \end{matrix}) so that we have q^{2} − 1 ways to select (\begin{matrix}a \\ b \end{matrix}). (\begin{matrix}c \\ d \end{matrix}) can be anything not in

the span of (\begin{matrix}a \\ b \end{matrix}). The span of (\begin{matrix}a \\ b \end{matrix}) contains q different vectors, including the

forbidden (\begin{matrix}0 \\ 0 \end{matrix}) so we have q^{2} − q choices for (\begin{matrix}c \\ d \end{matrix}). This gives a grand total of

(q^{2} − 1)(q^{2} − q) different bases for 𝔽_{q}^{2}.

(b) Let B = ( \begin{matrix}⋆ & ⋆ \\ 0 & ⋆ \end{matrix} ) ≤ G, T = ( \begin{matrix}⋆ & 0 \\ 0 & ⋆ \end{matrix} ) ≤ G, and U = ( \begin{matrix}1 & ⋆ \\ 0 & 1 \end{matrix} ) ≤ G.

Show that B is the semidirect product B = T ⋉ U.

We have that B = T U since

(\begin{matrix}a & b \\ 0 & c \end{matrix}) = (\begin{matrix}a & 0 \\ 0 & c \end{matrix}) (\begin{matrix}1 & ba^{−1} \\ 0 & 1 \end{matrix}) ∈ T U.

Next, we see that T ≤ N_{B}(U) since

(\begin{matrix}a & 0 \\ 0 & b \end{matrix}) (\begin{matrix}1 & c \\ 0 & 1 \end{matrix}) (\begin{matrix}a^{−1} & 0 \\ 0 & b^{−1} \end{matrix}) = (\begin{matrix}1 & abc \\ 0 & 0 \end{matrix}) ∈ U.

Of course, we have that U ≤ N_{B}(U) so that B = T U ≤ N_{B}(U) so that U ⊴ B.

We also have that T ∩ U = (\begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}) so that B = T U = T ⋉ U.

Now B/U = T U/U ≅ T/(T ∩ U) ≅ T/{1} ≅ T by the isomorphism theorems. T is abelian so that T has |T | = (q − 1)^{2} characters, all of which are of degree 1. These all lift to distinct characters of of B so that B has at least (q − 1)^{2} characters of degree one. To show that B has exactly (q − 1)^{2} characters of degree one, we will show that U = B′, the commutator subgroup of B, so that we will have exactly [B : B′] = [B : U] = (q − 1)^{2} characters of degree one. Indeed, we have B′ ≤ U since B/U ≅ T is abelian. Imagine now that B′ is strictly contained in U so that B/B′ is isomorphic with a subgroup T̃ of B containing T which has order strictly larger than |T |. Let

a := (\begin{matrix}x & y \\ 0 & z \end{matrix}) ∈ T̃\T.

Then x and z are non-zero else a is not invertible, and y is non-zero else a ∈ T. Then

(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix}) = (\begin{matrix}zy^{−1} & 0 \\ 0 & 1 \end{matrix}) (\begin{matrix}x & y \\ 0 & z \end{matrix}) (\begin{matrix}x^{−1}yz^{−1} & 0 \\ 0 & z^{−1} \end{matrix}) ∈ ⟨(\begin{matrix}⋆ & 0 \\ 0 & ⋆ \end{matrix}) , (\begin{matrix}x & y \\ 0 & z \end{matrix})⟩ ,

the subgroup of B generated by a and the elements of T, but this means that

U = ⟨(\begin{matrix}1 & 1 \\ 0 & 1 \end{matrix})⟩ ≤ ⟨(\begin{matrix}⋆ & 0 \\ 0 & ⋆ \end{matrix}) , (\begin{matrix}x & y \\ 0 & z \end{matrix})⟩ ,

so that

B = T U ≤ ⟨(\begin{matrix}⋆ & 0 \\ 0 & ⋆ \end{matrix}) , (\begin{matrix}x & y \\ 0 & z \end{matrix})⟩ ≤ T̃

so that T̃ = B. But this is impossible since B is non-abelian. Thus, we must have that T = T̃ and U = B′. Now to compute the characters of T, we note that T ≅ 𝔽_{q}^{×} × 𝔽_{q}^{×} ≅ ℤ_{q−1} × ℤ_{q−1}. The characters of ℤ_{q−1} correspond to the assignment of a (q − 1)st root of unity to a generator of ℤ_{q−1}, and the characters of ℤ_{q−1} × ℤ_{q−1} correspond with the product of two characters of ℤ_{q−1}. Then we can index the characters of T as follows

ψ_{i,j} = \{ \begin{matrix}(a, 1) & ↦ & ζ^{i} \\ (1, b) & ↦ & ζ^{j} \end{matrix}

where 0 is the identity element of ℤ_{q−1}, a and b are generators of ℤ_{q−1}, and ζ is a primative (q − 1)st root of unity. Or more compactly, taking a = b = 1 we have

ψ_{i,j}(x, y) = ζ^{xi+yj}

(c) Show that B\G/B has representatives B and BwB where w = (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}).

Let (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ G. If c = 0, then (\begin{matrix}a & b \\ c & d \end{matrix}) ∈ B. If c ≠ 0, then

(\begin{matrix}a & b \\ c & d \end{matrix}) = \underbrace{(\begin{matrix}a & b \\ c & d \end{matrix}) (\begin{matrix}1 & c^{−1}d \\ 0 & 1 \end{matrix}) (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix})}_{∈B} (\begin{matrix}0 & 1 \\ 1 & 0 \end{matrix}) \underbrace{(\begin{matrix}1 & −c^{−1}d \\ 0 & 1 \end{matrix})}_{∈B} ∈ BwB.

(d) Let ψ be a character of B of degree 1 constructed as in (2). Analyze ⟨Ind_{B}^{G}(ψ), Ind_{B}^{G}(ψ)⟩ by Mackey’s Theorem and describe when Ind_{B}^{G}(ψ) is irreducible. How many irreducible characters of G do you get in this way? We have that w^{−1}Bw = wBw is the subgroup of lower-triangular matricies since the w on the left swaps the rows and the w on the right swaps the columns. Thus, x^{−1}Bw ∩ B is exactly T, the subgroup of diagonal matricies. Also, we have ψ^{w}(b) = ψ^{w}(w^{−1}wbw^{−1}w) = ψ(wbw^{−1}). Then by Mackey,

⟨Ind_{B}^{G}(ψ), Ind_{B}^{G}(ψ)⟩ = ∑_{x∈1,w} ⟨ψ^{x}|_{x^{−1}Bx∩B}, ψ|_{x^{−1}Bx∩B}⟩_{x^{−1}Bx∩B}

= ⟨ψ, ψ⟩_{B} + ⟨ψ^{w}|_{T} , ψ|_{T} ⟩_{T}

Now ⟨ψ^{w}|_{T} , ψ|_{T} ⟩_{T} = ⟨ψ^{w}, ψ⟩_{B} since ψ is lifted from some character of T. We see then that ⟨ψ, ψ⟩_{B} + ⟨ψ^{w}, ψ⟩_{B} = 1 when ⟨ψ^{w}, ψ⟩_{B} = 0, that is, when ψ and ψ^{w} are different characters of T. Using the notation of (2), we have that ψ_{i,j} = ψ_{i,j}^{w} = ψ_{j,i} when i = j. Hence, we can produce (q − 1)^{2} − (q − 1) different characters of G in this manner.