Homework 1, Solutions

1. Let V be a three-dimensional vector space over ℂ with basis {v_{1}, v_{2}, v_{3}} and let G = S_{3} act on V by permuting the indicies. Let V_{1} = ⟨v_{1} + v_{2} + v_{3}⟩ and

V_{2} = {αv_{1} + βv_{2} + γv_{3} : α + β + γ = 0} = ⟨v_{1} − v_{2}, v_{2} − v_{3}⟩

be ℂG-submodules.

(a) Show that V = V_{1} ⊕ V_{2}.

(b) Show that V_{2} is simple.

(c) Let H = ⟨(123)⟩ ≤ G and let G act on G/H by multiplication. Show that ℂG/H is reducible as a ℂG-module.

(d) Let F be a field of characteristic 3 and let V_{1} and V_{2} be the F G-submodules of V defined above. Do we still have V = V_{1} ⊕ V_{2}?

(a) ( \begin{matrix}1 & 1 & 0 \\ 1 & −1 & 1 \\ 1 & 0 & −1 \end{matrix} ) is invertible iff \{v_{1} + v_{2} + v_{3}, v_{1} − v_{2}, v_{2} − v_{3}\} is a basis for

M iff every element of V can be expressed uniquely as a linear combination of v_{1} + v_{2} + v_{3}, v_{1} − v_{2}, and , v_{2} − v_{3} iff every element of V can be expressed uniquely as u_{1}+u_{2} for u_{1} ∈ V_{1} = ⟨v_{1} + v_{2} + v_{3}⟩ and u_{2} ∈ V_{2} = ⟨v_{1} − v_{2}, v_{2} − v_{3}⟩ iff V = V_{1} ⊕ V_{2}.

(b) If W is a non-zero submodule of V_{2}, then there must be some w = αv_{1} + βv_{2} + γv_{3} ∈ W with w ≠ 0. Also, α + β + γ = 0 as W ⊂ V_{2}. Now since W is a ℂS_{3}- submodule, we have that

w − (12) · w = (α − β) (v_{1} − v_{2})
w − (23) · w = (β − γ) (v_{2} − v_{3})
w − (13) · w = (α − γ) (v_{1} − v_{3})

are all in W. If any two of the scalars α − β, β − γ, and α − γ are both zero, then α = β = γ so that w ∉ V_{2}, a contradiction. We therefore have that two of the above scalars, say α − β and β − γ nonzero. But this means that both v_{1} −v_{2} and v_{2} −v_{3} are in W so that W = V_{2}. This shows that V_{2} is irreducible.

(c) We have that ⟨H + (12)H⟩ is a ℂG-submodule of ℂG/H as g · (H + (12)H) = H + (12)H for all g ∈ G. Since this submodule has dimension 1, ℂG/H is reducible.

(d) If F is a field of characteristic 3, then V_{1} = ⟨v_{1} + v_{2} + v_{3}⟩ is a submodule of V_{2} = {αv_{1} + βv_{2} + γv_{3} : α + β + γ = 0} so that V ≠ V_{1} ⊕ V_{2}. Also note that the matrix given above has determinant 3.

2. Let D be a division ring and let A be the submodule of M_{n}(D) of upper triangular matrices.

(a) Describe J (A).

(b) Describe the decomposition of A/J (A) as a direct sum of simple modules.

(a) Define L_{k} = {A = (a_{i,j}) ∈ A : a_{k,k} = 0}. To show that L_{k} is an ideal, it suffices to show that E_{i,j}A ∈ L_{k} for all i ≤ j. But left multiplying A by E_{i,j} results in a matrix with the jth row of A in the ith row and zeros elsewhere. Also, since i < j, we have, that this operation has the effect of moving rows upward and clobbering the rest so that the k, k entry of E_{i,j}A is 0. To show that L_{k} is a maximal ideal, let B ∈ A\L_{k}. Then (b_{k,k}) ≠ 0. But then E_{k,k}b_{k,k}^{−1}B = E_{k,k} ∈ L_{k} so that (L_{k}, B) = A. We therefore have that

J (A) ⊂ ⋂_{k=1}^{n} L_{k} = {A = (a_{i,j}) ∈ A : a_{k,k} = 0, 1 ≤ k ≤ n}.

However, since the righthand side is a nilpotent ideal, we have the reverse inclusion since J (A) contains all nilpotent ideals. Remark: The final answer is that J is precisely the set of upper triangular matrices with 0 along the main diagonal. This is the right hand side above. Check that this is a nilpotent ideal, by showing that the k-th power of a matrix of that form has the k-th diagonal (i.e. the k − 1-st diagonal to the right of the main diagonal) equal to zero.

(b) We have that A_{1} and A_{2} will be in the same coset of J (A) if and only if they have the same diagonal elements. Therefore, if we define

M_{j} = {α_{j}E_{j,j} + J (A) : α_{j} ∈ D} ,

then M_{j} is irreducible as D is a division ring and

A/ (JA) = ⨁_{j=1}^{n} M_{j}.

Homework 2 Solutions

Added to Marcus’ Homework

Here are some general comments on the decomposition of F [x]-modules, F a field. Let V be a cyclic F [x]-module, i.e. V has a basis {v, xv, … x^{n−1}v} and the minimal polynomial f(x) of x (as a linear transformation of V) has degree equal to the dimension n of V. Then in fact V is isomorphic to F [x]/(f(x)) and the latter can be written as ⨁_{1}^{k} F [x]/(p_{i}(x)^{r_{i}}), where each factor F [x]/(p_{i}(x)^{r_{i}}) is cyclic and the p_{i}(x) are distinct and irreducible. (Note that in general, if we have a decomposition of an F [x]-module according to the ”rational form” as in DF, p.455, we have to apply the above to each cyclic factor.) Now consider a cyclic module V = F [x]/(p(x)^{r}). Then V is indecomposable as F [x]-module and is irreducible if r = 1. To see that such a V is indecomposable, suppose V = V_{1} ⊕ V_{2}. Then on V_{1} (resp. V_{2}), the minimal polynomial of x would be of the form p(x)^{k} (resp. p(x)^{l}) where k, l < r and then the minimal polynomial on V would have degree less than deg(p).r. Similarly if r = 1 and V had a proper submodule V_{1} (not zero) then the minimal polynomial of x on V_{1} would be a factor of p(x) of smaller degree, but p(x) is irreducible. We can apply these arguments to the problems below. In Problem 1 the modules are not cyclic, but we apply the above criteria to each cyclic factor. In Problem 2 the module Q[x]/(x^{n} − 1) is cyclic.

1. Let A = ℚ[x], V an A-module where xv = T v for v ∈ V with T the linear transformation given by

(a) ( \begin{matrix}2 & −2 & 14 \\ 0 & 3 & −7 \\ 0 & 0 & 2 \end{matrix} ) with respect to the basis \{e_{1}, e_{2}, e_{3}\} of V. Show that V is

completely reducible. Describe V as a direct sum of simple modules.

(b) ( \begin{matrix}1 & 2 & −4 & 4 \\ 2 & −1 & 4 & −8 \\ 1 & 0 & 1 & −2 \\ 0 & 1 & −2 & 3 \end{matrix} ) with respect to the basis \{e_{1}, e_{2}, e_{3}, e_{4}\} of V. Show that

V is not completely reducible. Describe V as the direct sum of two indecomposable submodules and give a composition series for V

(a) We have that the Jordan form of T is J = ( \begin{matrix}2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 2 \end{matrix} ) with S = ( \begin{matrix}2 & −2 & 0 \\ 7 & 1 & 7 \\ 1 & 0 & 1 \end{matrix} )

such that J = S^{−1}T S. Then with respect to the basis

{f_{1} = 2e_{1} + 7e_{2} + e_{3}, f_{2} = −2e_{1} + e_{2}, f_{3} = 7e_{2} + e_{3}}

of V, T has matrix J. This means that ⟨f_{1}⟩ , ⟨f_{2}⟩,⟨f_{3}⟩ are A-submodules of V since these are the J-stable ℚ-subspaces of V. Also,

V = ⟨f_{1}⟩ ⊕ ⟨f_{2}⟩ ⊕ ⟨f_{3}⟩

since the f_{j} are the columns of S, which is invertible.

(b) The Jordan form of T is R = ( \begin{matrix}1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{matrix} ). Letting \{f_{1}, f_{2}, f_{3}, f_{4}\} be the

columns of the conjugating matrix as before, we have that V_{1} = ⟨f_{1}, f_{2}⟩ and V_{2} = ⟨f_{3}, f_{4}⟩ are A-submodules of V which have irreducible sub-submodules ⟨f_{1}⟩ and ⟨f_{3}⟩ respectively. Then V_{1} and V_{2} are indecomposable (by the general arguments given above, taking p(x) = x − 1 and noting that V_{1} and V_{2} are isomorphic to Q[x]/(x − 1)^{2} and are cyclic) and V = V_{1} ⊕ V_{2}. A composition series for V is

V = ⟨f_{1}, f_{2}, f_{3}, f_{4}⟩ ≥ ⟨f_{1}, f_{2}, f_{3}⟩ ≥ ⟨f_{1}, f_{2}⟩ ≥ ⟨f_{1}⟩ ≥ 0.

2. Let G = ⟨a : a^{n} = 1⟩ be the cyclic group of order n and let R = ℚG. Describe the Wedderburn decomposition of R and find the number and the degrees of the irreducible representations of G over ℚ. In particular, show that if n = p is a prime, then G has exactly one nontrivial irreducible representation over ℚ and this representation has degree p − 1.

The map ℚ[x] → ℚG defined x^{j} → a^{j} is a surjective ring homomorphism with kernel (1 − x^{n}) so that

ℚ[x]/ (1 − x^{n}) ≅ ℚG.

Next, the polynomial 1 − x^{n} ∈ ℚ[x] factors as

1 − x^{n} = ∏_{d|n} Φ_{d}(x)

where Φ_{d}(x) is the dth cyclotomic polynomial. Then by the Chineese Remainder Theorem, since the (Φ_{d}(x)) are comaximal, Φ_{d}(x) being irreducible, we have

ℚG ≅ ℚ[x]/ (1 − x^{n}) ≅ ∏_{d|n} ℚ[x]/ (Φ_{d}(x))

as rings. Note that as a ℚ[x] module, ℚ[x]/ (Φ_{d}(x)) is irreducible, again by the general arguments. Note also that ℚ[x]/ (Φ_{d}(x)) has dimension φ(d). Therefore,

ℚG ≅ ⨁_{d|n} ℚ[x]/ (Φ_{d}(x))

as modules so that

ℚG ≅ ⨁_{d|n} M_{1} (D_{d})

where D_{d} = ℚ(ζ) where ζ is a primitive d-th root of unity.

Homework 3 Solutions, added to Marcus’

Let G = S_{3} and A = ℚG.

1. Find the central idempotents e_{1}, e_{2} and e_{3} such that A = Ae_{1} ⊕ Ae_{2} ⊕ Ae_{3} where the Ae_{j} are ideals which are simple algebras with dim Ae_{1} = dim Ae_{2} = 1 and dim Ae_{3} = 4.

2. Find orthogonal idempotents (not necessarily central) f_{1}, f_{2} f_{3} and f_{4} such that AA = ⨁_{j=1}^{4} Af_{j} where the Af_{j} are minimal left ideals (simple A-modules).

3. Let p = 2. We consider two modular representations of G. Let R ⊂ ℚ be given by R = \{\frac{a}{b} : a, b ∈ ℤ, 2 |̸b\}. Then R is a local ring and R/(2) ≅ 𝔽_{2}. Let Ã = RG and \overline{A} = 𝔽_{2}G. Using the decomposition in (1), write Ã = ÃE_{1} ⊕ ÃE_{2} where E_{1} and E_{2} are central idempotents. It will not be possible to reduce e_{1}, e_{2}, and e_{3} mod 2 as some of them will not lie in R. So you will have to combine them into two idempotents. Finally, write \overline{A} = \overline{A}\overline{E}_{1} ⊕ \overline{A}\overline{E}_{2} after reducing mod 2.

4. Try doing the same procedure as in (3) for p = 3. Here R = \{\frac{a}{b} : a, b ∈ ℤ, 3 |̸b\}. Can you decompose Ã = RG?

1. Let e_{1} = \frac{1}{6} ∑_{g∈G} g. Then

e_{1}^{2} = (\frac{1}{6} ∑_{g} g) (\frac{1}{6} ∑_{g} g) = \frac{1}{36} ∑_{g} (g ∑_{g} g) = \frac{1}{36} ∑_{g} (∑_{g} g) = \frac{1}{6} (∑_{g} g)

so that e_{1} is idempotent. Also,

(∑_{g} α_{g}g) (\frac{1}{6} ∑_{g} g) = \frac{1}{6} ∑_{g} α_{g}g (∑_{g} g) = \frac{1}{6} ∑_{g} α_{g} (∑_{g} g)  (1)

= \frac{1}{6} ∑_{g} (∑_{g} g) α_{g} = \frac{1}{6} ∑_{g} (∑_{g} g) α_{g}g = \frac{1}{6} (∑_{g} g) (∑_{g} α_{g}g)

so that e_{1} is central. Let e_{2} = \frac{1}{6} ∑_{g} (−1)^{sgn g} g. We similarly compute that e_{2} is a central idempotent. We define e_{3} = 1 − e_{1} − e_{2} = e_{3} and confirm directly that e_{3} is central.

Ae_{1} and Ae_{2} are 1-dimensional, since, as suggested in (1), ae_{1} = αe_{1} for some α ∈ ℚ and similarly for e_{2}. We confirm that Ae_{3} is 4-dimensional by explicitly computing its elements and confirming independence of four of them. Alternately, GAP verifies this claim and everything stated above as follows.

gap> G:=SymmetricGroup(3);A:=GroupRing(Rationals,G);
Sym( [ 1 .. 3 ] )
<algebra-with-one over Rationals, with 2 generators>
gap> e:=CentralIdempotentsOfAlgebra(A);
[ (1/6)*()+(-1/6)*(2,3)+(-1/6)*(1,2)+(1/6)*(1,2,3)+(1/6)*(1,3,2)+(-1/6)*(1,3),
  (1/6)*()+(1/6)*(2,3)+(1/6)*(1,2)+(1/6)*(1,2,3)+(1/6)*(1,3,2)+(1/6)*(1,3),
  (2/3)*()+(-1/3)*(1,2,3)+(-1/3)*(1,3,2) ]
gap> Dimension(A*e[1]);Dimension(A*e[2]);Dimension(A*e[3]);
1
1
4

2. To find f_{3} and f_{4}, we consider the representation ρ of ℚG given by

(12) ↦ ( \begin{matrix}−1 & 1 \\ 0 & 1 \end{matrix} ) and (123) ↦ ( \begin{matrix}0 & −1 \\ 1 & −1 \end{matrix} )  (2)

and take the inverse images of the matrices ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) and ( \begin{matrix}0 & 0 \\ 0 & 1 \end{matrix} ). We begin

with the following. The newlines were added to this document for legibility. A.1 and A.2 give the first and second generators of A which are then printed to the standard output.

gap> s:=A.1;t:=A.2;
(1)*(1,2,3)
(1)*(1,2)
gap> S:=[[0,-1],[1,-1]];T:=[[-1,1],[0,1]];
[ [ 0, -1 ], [ 1, -1 ] ]
[ [ -1, 1 ], [ 0, 1 ] ]

The next command defines f to be the algebra homomorphism given by the images of the generators in 2.

gap> f:=AlgebraHomomorphismByImages(
        A,FullMatrixAlgebra(Rationals,2),[t,s],[T,S]);
[ (1)*(1,2), (1)*(1,2,3) ] ->
[ [ [ -1, 1 ], [ 0, 1 ] ], [ [ 0, -1 ], [ 1, -1 ] ] ]
v:=PreImagesRepresentative(f,[[1,0],[0,0]]);
(1/3)*()+(-1/3)*(2,3)+(-2/3)*(1,2)+(-1/3)*(1,2,3)
gap> Dimension(A*v);
4

It is somewhat disturbing that GAP’s preimage generates a subalgebra of dimension 4, but we consider the possibility that the GAP has returned the sum of the

preimage of ( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix} ) and something in the kernel of ρ (note that e_{1}, e_{2} ∈ ker ρ).

We experiment with e_{1} and e_{2} to find f_{3} and f_{4} as follows.

gap> f3:=v-e[1]+e[2];
(1/3)*()+(-1/3)*(1,2)+(-1/3)*(1,2,3)+(1/3)*(1,3)
gap> Dimension(A*f3);
2
gap> f3^2=f3;
true
gap> f4:=e[3]-f3;
(1/3)*()+(1/3)*(1,2)+(-1/3)*(1,3,2)+(-1/3)*(1,3)
gap> Dimension(A*f4);
2
gap> f4^2=f4;
true

Addendum (B.S) :

Thus, one solution for f_{3} and f_{4} is given by:

f_{3} = (1/3)[1 − (1, 2) + (1, 2, 3) − (1, 3)], f_{4} = e_{3} − f_{3}.

Other solutions for f_{3} might be obtained by observing that the representation ρ of

ℚG is given by (12) ↦ ( \begin{matrix}−1 & 1 \\ 0 & 1 \end{matrix} ) , (123) ↦ ( \begin{matrix}0 & −1 \\ 1 & −1 \end{matrix} ) , (13) ↦ ( \begin{matrix}0 & −1 \\ −1 & 0 \end{matrix} ) , (132) ↦

( \begin{matrix}−1 & 1 \\ −1 & 0 \end{matrix} ) , (23) ↦ ( \begin{matrix}1 & 0 \\ 1 & −1 \end{matrix} ).

Thus by inspection we see easily that −(13) − (123) − 132) ↦ ( \begin{matrix}1 & 1 \\ 1 & 1 \end{matrix} ), and so we

could take f_{3} = −1/2[(13) + (123) + (132)].

3. We have

E_{2} := e_{3} = 1 − e_{1} − e_{2} = \frac{2}{3} − \frac{1}{3} (123) − \frac{1}{3} (132) ∈ R

6 and observe that

E_{1} = e_{1} + e_{2} = \frac{1}{3} + \frac{1}{3} (123) + \frac{1}{3} (132) ∈ R

and that E_{1} is idempotent by FOIL and orthogonality. As in (1), ÃE_{2} is irreducible and of dimension 4, the only difference being that R is a subring of ℚ.

Then Ã = ÃE_{1} ⊕ ÃE_{2}. Reducing both sides by the ideal generated by 2 · 1, which has the effect of reducing all the coefficients modulo the ideal (2), we have \overline{A} = \overline{A}E_{1} ⊕ \overline{A}E_{2}.

However, if we replace 2 by 3 we find that the R-algebra A is not decomposable into a sum of proper two-sided ideals.

Homework 4

Let G = SL (2, 3). You are given that the representatives of the conjugacy classes of

G are \{g_{j} : 1 ≤ j ≤ 7\} where g_{1} = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ), g_{2} = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ), g_{3} = ( \begin{matrix}0 & 1 \\ −1 & 0 \end{matrix} ),

g_{4} = ( \begin{matrix}1 & 0 \\ 1 & 1 \end{matrix} ), g_{5} = g_{4}^{−1}, g_{6} = g_{2}g_{4}, and g_{7} = g_{2}g_{5}. Fill in the character table by the

following steps.

1. (Projective Geometry) G/Z = PSL (2, 3) permutes the 4 points of the projective line over 𝔽_{3} and is isomorphic to A_{4}. Use this to fill in χ_{2}, χ_{3}, and χ_{4}.

2. Find the other character degrees and the values on Z.

3. Of the three characters χ_{5}, χ_{6}, and χ_{7}, at least one has values in ℝ since complex conjugate characters come in pairs. Suppose this is χ_{5} with corresponding representation ρ_{5}. Consider the eigenvalues of ρ_{5} (g_{r}) and use χ_{5} (g_{4}) ∈ ℝ to find χ_{5} (g_{4}). Then find χ_{5} (g_{5}),χ_{5} (g_{6}) and χ_{5} (g_{7}).

4. Tensoring with suitable linear characters gives χ_{6} and χ_{7} on g_{i} for 4 ≤ i ≤ 7.

5. Finally, find χ_{5} (g_{3}),χ_{6} (g_{3}) and χ_{7} (g_{3}) (g_{3}) by orthogonality.

1. Given that

φ : GL (2, 3) /Z = P SL(2, 3) ⟶^{∼} A_{4}

is an isomorphism for some map φ, we have that |Z| = |GL (2, 3)| / |A_{4}| = 2 so that Z = {g_{1}, g_{2}} since g_{1} and g_{2} must be central in SL (2, 3), being central in all of M_{2} (𝔽_{3}).

This means that elements are in the same coset of the quotient if they differ by g_{2}.

Therefore, the conjugacy classes represented by g_{4}Z and g_{6}Z = g_{2}g_{4}Z must collapse under φ to the same conjugacy class and this class must either be (123) or (124) since |φ (g_{4}Z)| must divide |g_{4}Z| = 3 and isn’t 1 because φ is injective. Similarly, g_{5}Z and g_{7}Z must collapse to the conjugacy class represented by the other three cycle. By elimination, we have that the conjugacy class of g_{3}Z corresponds with the conjugacy class of (12) (34) and observe that these classes are in bijection unlike the others.

We therefore lift the characters of A_{4} to characters of G as follows

\begin{matrix}SL(2, 3) & g_{1} & g_{2} & g_{3} & g_{4} & g_{5} & g_{6} & g_{7} \\ χ_{1} & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & 1 & 1 & ω & \overline{ω} & ω & \overline{ω} \\ χ_{3} & 1 & 1 & 1 & \overline{ω} & ω & \overline{ω} & ω \\ χ_{4} & 3 & 3 & −1 & 0 & 0 & 0 & 0 \end{matrix}

2. Let n_{j} = dim χ_{j}. Then since ∑_{j=1}^{7} n_{j}^{2} = 24, we have ∑_{j=5}^{7} n_{j}^{2} = 12. The only squares ≤ 12 are 1, 2, and 9. If any n_{j}^{2} = 9, then the sum of the squares of the other two must be 4, but 4 is not the sum of two squares. Similarly n_{j}^{2} ≠ 1 for any j so that 4 = n_{5}^{2} = n_{6}^{2} = n_{7}^{2} = χ_{5} (g_{1}) = χ_{6} (g_{1}) = χ_{7} (g_{1}).

Since g_{2} ∈ Z, we have that ρ_{j} (g_{2}) must be a diagonal matrix for 5 ≤ j ≤ 7. The diagonal entries of ρ_{j} (g_{2}) are ±1. The trace of such a matrix is therefore −2, 0, or 2.

However, we must have χ_{5} (g_{2}) = χ_{6} (g_{2}) = χ_{7} (g_{2}) = −2 since this is the only way that

0 = ⟨g_{1}, g_{2}⟩ = 1^{2} + 1^{2} + 1^{2} + 3^{2} + 2χ_{5} (g_{2}) + 2χ_{6} (g_{2}) + 2χ_{7} (g_{2})

will be satisfied. We therefore have the following situation.

SL(2, 3) g_{1} g_{2} g_{3} g_{4} g_{5} g_{6} g_{7}
χ_{5}  2 −2
χ_{6}  2 −2
χ_{7}  2 −2

3. Assume χ_{5} is an irreducible real-valued character. We have |ρ_{5} (g_{4})| divides |g_{4}| = 3 so that |ρ_{5} (g_{4})| is 1 or 3. If |ρ_{5} (g_{4})| = 1, then g_{4} ∈ ker ρ_{5} ◁ G, but the only normal subgroup of A_{4} has order 4. This subgroup pulls back under π ∘ φ to a normal subgroup of order 8, but g_{4} cannot be in this subgroup, having order 3.

Therefore |ρ_{5} (g_{4})| = 3. Let ε_{1} and ε_{2} be its eigenvalues with eigenvectors v_{1} and v_{2}. Then v_{j} = (ρ_{5} (g_{4}))^{3} v_{j} = ε_{j}^{3}v_{j} so that ε_{j}^{3} = 1. We have agreed that χ_{5} (g_{4}) = ϵ_{1} + ϵ_{2} ∈ ℝ. But for \{ε_{1}, ε_{2}\} ⊂ \{1, ω, \overline{ω}\}, and ε_{1} + ε_{2} ∈ ℝ, we must have either have ε_{1} = ε_{2} = 1 or \{ε_{1}, ε_{2}\} = \{ω, \overline{ω}\}. But the first case is impossible since we have seen that ρ_{5} (g_{4}) ≠ I. This means we have the latter case and χ_{5} (g_{4}) = −1. Similarly, since |ρ_{5} (g_{5})| = 3, we have χ_{5} (g_{5}) = −1.

We have observed above that ρ_{5} (g_{2}) = ( \begin{matrix}−1 & 0 \\ 0 & −1 \end{matrix} ) (with respect to any basis). This means that ρ_{5} (g_{6}) v_{j} = ρ_{5} (g_{2}) ρ_{5} (g_{4}) v_{j} = ρ_{5} (g_{2}) ϵ_{j}v_{j} = −ϵ_{j}v_{j} so that χ_{5} (g_{6}) = 1 and similarly for g_{7}.

SL(2, 3) g_{1} g_{2} g_{3} g_{4} g_{5} g_{6} g_{7}
χ_{5}  2 −2  −1 −1 1 1

4. We provisionally put χ_{6} = χ_{5} ⊗ χ_{2} and χ_{7} = χ_{5} ⊗ χ_{3}.

SL(2, 3) g_{1} g_{2} g_{3} g_{4} g_{5} g_{6} g_{7}
χ_{1}  1 1 1 1 1 1 1
χ_{2}  1 1 1 ω \overline{ω} ω \overline{ω}
χ_{3}  1 1 1 \overline{ω} ω \overline{ω} ω
χ_{4}  3 3 −1 0 0 0 0
χ_{5}  2 −2  −1 −1 1 1
χ_{6}  2 −2  −ω −\overline{ω} ω \overline{ω}
χ_{7}  2 −2  −\overline{ω} −ω \overline{ω} ω

5. To use the first orthogonality relation, we need to compute the class sizes. We have |C_{1}| = |C_{2}| = 1 since g_{1} and g_{2} are central. We saw that C_{3} corresponds under the isomorphism φ to the conjugacy class of A_{4} containing (12) (34) which has 3 members. Therefore, |C_{3}| = 6. Finally, we have |C_{j}| = 4 for 4 ≤ j ≤ 7 by the second orthogonality relation. For example,

|C_{5}| = 1 + 1 + 1 + 0 + 1 − 1 − 1 = 4.

Since we have assumed χ_{5} to be an irreducible character, we can use the first orthogonality relation to compute its remaining element

4 + 4 + 6χ_{5} (g_{3}) \overline{χ_{5} (g_{3})} + 4 + 4 + 4 + 4 = |G| = 24

giving χ_{5} (g_{3}) = 0. Finally, χ_{6} (g_{3}) = χ_{5} (g_{2}) χ_{2} (g_{2}) = 0 and similarly for χ_{7} (g_{3}) giving the final character table.

SL(2, 3) g_{1} g_{2} g_{3} g_{4} g_{5} g_{6} g_{7}
1 1 6 4 4 4 4
χ_{1}  1 1 1 1 1 1 1
χ_{2}  1 1 1 ω \overline{ω} ω \overline{ω}
χ_{3}  1 1 1 \overline{ω} ω \overline{ω} ω
χ_{4}  3 3 −1 0 0 0 0
χ_{5}  2 −2 0 −1 −1 1 1
χ_{6}  2 −2 0 −ω −\overline{ω} ω \overline{ω}
χ_{7}  2 −2 0 −\overline{ω} −ω \overline{ω} ω

It remains to confirm that χ_{6} and χ_{7} are irreducible by checking that ⟨χ_{6}, χ_{6}⟩ = ⟨χ_{7}, χ_{7}⟩ = 1.

Homework 5

0.1 Introduction

We compute the conjugacy classes of G by enumerating all the permissible rational canonical forms, we compute the orders of the representatives, and we match the representatives with the columns of the given table by order. We have the following table.

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} )
pseudonym  g_{1}  g_{2}  g_{3}  g_{4}  g_{5}  g_{6}
order of element  1  2  3  4  7  7
class size  1  21  56  42  24  24
χ_{1}  1  1  1  1  1  1
χ_{2}  6  2  0  0  −1  −1
χ_{3}  7  −1  1  −1  0  0
χ_{4}  8  0  −1  0  1  1
χ_{5}  3  −1  0  1  \frac{−1+i\sqrt{7}}{2} \frac{−1−i\sqrt{7}}{2}
χ_{6}  3  −1  0  1  \frac{−1−i\sqrt{7}}{2} \frac{−1+i\sqrt{7}}{2}

0.2 Stabilizer of Lines

Let V be a vector space over 𝔽_{2} with basis {v_{1}, v_{2}, v_{3}}. Then G acts transitively on the set of one-dimensional subspaces of V. We count that V has 7 one-dimensional subspaces {⟨αv_{1} + βv_{2} + γv_{3}⟩ : α, β, γ ∈ 𝔽_{2}}.

With respect to this basis, the subgroup P = ( \begin{matrix}♣ & ♣ & ♣ \\ 0 & ♣ & ♣ \\ 0 & ♣ & ♣ \end{matrix} ) ≤ G is the stabilizer of the line ⟨v_{1}⟩. Therefore, elements in the same coset of G/P act equivalently on ⟨v_{1}⟩. Since G acts transitively on the set of lines, there should be one coset for each line. \frac{|G|}{|P |} = 7 confirms this claim.

Let {x_{j} : 1 ≤ j ≤ 7} be coset representatives of G/P and let 1 denote the trivial character of G/P. We will use the following observation to compute

Ind_{P}^{G}1 (g) = ∑_{j=1}^{7} 1 (x_{j}^{−1}gx_{j}) .

We have x_{j}^{−1}gx_{j} ∈ P iff gx_{j} ∈ x_{j}P iff gx_{j}P = x_{j}P. Therefore, Ind_{P}^{G}1 (g) counts 1 for each coset x_{j}P that g fixes. But since the elements of the coset x_{j}P act equivalently on ⟨v_{1}⟩, we can regard x_{j}P as the well defined line x_{j} ⟨v_{1}⟩ so that Ind_{P}^{G}1 (g) is the number of lines g fixes.

We directly compute that g_{2} fixes the lines ⟨v_{1}⟩, ⟨v_{2} + v_{3}⟩, and ⟨v_{1} + v_{2} + v_{3}⟩, g_{3} fixes the line ⟨v_{1} + v_{2} + v_{3}⟩, and g_{4} fixes the line ⟨v_{1} + v_{3}⟩. We therefore have the following character.

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} )
Ind_{P}^{G} (1)  7  3  1  1  0  0

Taking inner products, we have

Ind_{P}^{G} (1) = χ_{1} + χ_{2}

giving us χ_{2}.

0.3 Stabilizer of Flags

G also acts transitively on the set of complete flags. Let B = \{( \begin{matrix}♣ & ♣ & ♣ \\ 0 & ♣ & ♣ \\ 0 & 0 & ♣ \end{matrix} )\} be the stabilizer of the complete flag ⟨v_{1}⟩ ≤ ⟨v_{1}, v_{2}⟩. Then by the same argument used in section 0.2, Ind_{B}^{G} (1) (g) is the number of flags fixed by g.

For each one-dimensional subspace of V enumerated in section 0.2, there are three two-dimensional subspaces of V containing it, making a total of 21 complete flags. Thus, Ind_{B}^{G} (1) (g_{1}) = 21.

To count flags fixed by g_{2} = ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ), we recall from section 0.2 that the only onedimensional subspaces of V fixed by g_{2} are ⟨v_{1}⟩, ⟨v_{2} + v_{3}⟩, and ⟨v_{1} + v_{2} + v_{3}⟩. By enumerating the two-dimensional subspaces containing these, we see that g_{2} fixes the following flags

⟨v_{1}⟩ ≤ ⟨v_{1}, v_{2} + v_{3}⟩
⟨v_{2} + v_{3}⟩ ≤ ⟨v_{2}, v_{3}⟩
⟨v_{2} + v_{3}⟩ ≤ ⟨v_{1}, v_{2} + v_{3}⟩
⟨v_{2} + v_{3}⟩ ≤ ⟨v_{1} + v_{2}, v_{2} + v_{3}⟩
⟨v_{1} + v_{2} + v_{3}⟩ ≤ ⟨v_{1}, v_{1} + v_{2} + v_{3}⟩

and g_{4} fixes the flag

⟨v_{1} + v_{3}⟩ ≤ ⟨v_{1} + v_{3}, v_{2} + v_{3}⟩

so that we have the following character.

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} )
Ind_{B}^{G} (1) 21  5  0  1  0  0

Taking inner products, we see that

Ind_{B}^{G} (1) = χ_{1} + 2χ_{2} + χ_{4}

giving us χ_{4}.

0.4 The Parabolic Subgroup

Consider the subgroup

P = \{( \begin{matrix}♣ & ♣ & ♣ \\ ♣ & ♣ & ♣ \\ 0 & 0 & 1 \end{matrix} )\} ⊃ \{( \begin{matrix}♣ & ♣ & 0 \\ ♣ & ♣ & 0 \\ 0 & 0 & 1 \end{matrix} )\} := L ≅ GL_{2} (𝔽_{2}) × GL_{1} (𝔽_{2}) ≅ S_{3}.

Since P = UL for U = \{( \begin{matrix}1 & 0 & ♣ \\ 0 & 1 & ♣ \\ 0 & 0 & 1 \end{matrix} )\} ◁ P, we will lift a character from L ≅ P/U up to P and induce from this a character of G.

Again, using the rational form, we compute representatives of the conjugacy classes of GL_{2} (𝔽_{2}) to be ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ), ( \begin{matrix}0 & 1 \\ 1 & 0 \end{matrix} ), and ( \begin{matrix}0 & 1 \\ 1 & 1 \end{matrix} ) so that representatives of the conjugacy classes of L are ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ), ( \begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} ), ( \begin{matrix}0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ). The classes in L represented by these three elements do not merge in P since these elements have different orders.

We take the following character from the character table for S_{3}.

\begin{matrix}L & ( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) & ( \begin{matrix}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} ) & ( \begin{matrix}0 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) \\ ε & 1 & −1 & 1 \end{matrix}

We will compute

Ind_{P}^{G}ε (g_{2}) = |C_{G} (g_{2})| (∑_{j=1}^{s} \frac{1}{|C_{P} (h_{j})|}ε (h_{j}))

where h_{j} are the representatives of the conjugacy classes in P whose union is the conjugacy class of g_{2} in G.

The GAP program given in the appendix gives the following conjugacy classes of P.

gap> CCC;
[ [ [ [ 0, 1, 0 ], [ 1, 0, 0 ], [ 0, 0, 1 ] ],
[ [ 0, 1, 1 ], [ 1, 0, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 0, 0 ], [ 1, 1, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 0, 0 ], [ 1, 1, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 0 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 1 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ] ],
[ [ [ 0, 1, 0 ], [ 1, 0, 1 ], [ 0, 0, 1 ] ],
[ [ 0, 1, 1 ], [ 1, 0, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 0, 1 ], [ 1, 1, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 0, 1 ], [ 1, 1, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 0 ], [ 0, 1, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 1 ], [ 0, 1, 1 ], [ 0, 0, 1 ] ] ],
[ [ [ 0, 1, 0 ], [ 1, 1, 0 ], [ 0, 0, 1 ] ],
[ [ 0, 1, 0 ], [ 1, 1, 1 ], [ 0, 0, 1 ] ],
[ [ 0, 1, 1 ], [ 1, 1, 0 ], [ 0, 0, 1 ] ],
[ [ 0, 1, 1 ], [ 1, 1, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 0 ], [ 1, 0, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 0 ], [ 1, 0, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 1 ], [ 1, 0, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 1, 1 ], [ 1, 0, 1 ], [ 0, 0, 1 ] ] ],
[ [ [ 1, 0, 0 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ] ],
[ [ [ 1, 0, 0 ], [ 0, 1, 1 ], [ 0, 0, 1 ] ],
[ [ 1, 0, 1 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ],
[ [ 1, 0, 1 ], [ 0, 1, 1 ], [ 0, 0, 1 ] ] ] ]

Since the factorization ( \begin{matrix}a & b & c \\ d & e & f \\ 0 & 0 & 1 \end{matrix} ) = ( \begin{matrix}1 & 0 & c \\ 0 & 1 & f \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}a & b & 0 \\ d & e & 0 \\ 0 & 0 & 1 \end{matrix} ) ∈ UL is transparent, we can easily see which elements are in which coset of G/P.

We observe that the first and the last conjugacy class listed contain elements of order 2. Since only one class of G contains elements of order 2, namely the class containing g_{2} at which we’re trying to compute the character, we must have that the class of g_{2} splits in P into the first and last classes above. The first class contains a transposition, so ε must be −1 on that class, and the last class is the identity coset, so ε must be 1 on that class. Therefore,

Ind_{P}^{G}ε (g_{2}) = \frac{1}{8} (\frac{1}{4} (−1) + \frac{1}{8} (1)) = −1.

12 Similarly for the other values giving the following character.

( \begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} ) ( \begin{matrix}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix} ) ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} )
Ind_{P}^{G} (1)  7  −1  1  −1  0  0

We confirm that this character is irreducible by taking inner products, or alternately, we notice that this character is χ_{3} from the given table.

0.5 The 7-Subgroup

Next, consider the Sylow 7-subgroup S = ⟨g_{5}⟩. We will induce the following characters.

\begin{matrix}S & g_{1} & g_{5} & g_{5}^{2} & g_{5}^{3} & g_{5}^{4} & g_{5}^{5} & g_{5}^{6} \\ ψ_{1} & 1 & ω & ω^{2} & ω^{3} & ω^{4} & ω^{5} & ω^{6} \\ ψ_{2} & 1 & ω^{3} & ω^{6} & ω^{2} & ω^{5} & ω & ω^{4} \end{matrix}

We determine which conjugacy classes of the powers of g_{5} lie in by computing their rational forms. Gap gives the characteristic polynomials for the powers of g_{5} below.

gap> g5:=[[0,0,1],[1,0,0],[0,1,1]];
[ [ 0, 0, 1 ], [ 1, 0, 0 ], [ 0, 1, 1 ] ]
gap> for j in [0..6] do
Print(j,": ",CharacteristicPolynomial(g5^j) mod 2,"\n")$
0: 1+x_1+x_1^2+x_1^3
1: 1+x_1^2+x_1^3
2: 1+x_1^2+x_1^3
3: 1+x_1+x_1^3
4: 1+x_1^2+x_1^3
5: 1+x_1+x_1^3
6: 1+x_1+x_1^3

We check that both 1 + x^{2} + x^{3} and 1 + x + x^{3} are irreducible over 𝔽_{2} so that the rational forms of A^{j} have one block containing the companion matrices ( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 1 \end{matrix} ) and

( \begin{matrix}0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} ) respectively. We conclude that

g_{5}^{3} ∼ g_{5}^{5} ∼ g_{5}^{6} ∼ g_{6}

and

g_{5} ∼ g_{5}^{2} ∼ g_{5}^{4}.

in G. Of course, in S, none of these elements are conjugate. This means that the conjugacy class of g_{5} in G splits into three conjugacy classes of H and similarly for g_{6}. We compute

Ind_{S}^{G}ψ_{1} (g_{5}) = 7 · \frac{1}{7} (ω + ω^{2} + ω^{4})

and similarly for the remaining values giving the following table.

g_{1} g_{2} g_{3} g_{4}  g_{5}  g_{6}
Ind_{S}^{G} (ψ_{1}) 24 0 0 0 ω + ω^{2} + ω^{4} ω^{3} + ω^{5} + ω^{6}
Ind_{S}^{G} (ψ_{2}) 24 0 0 0 ω^{3} + ω^{5} + ω^{6} ω + ω^{2} + ω^{4} 13

Taking inner products, we have

Ind_{S}^{G}ψ_{1} = χ_{2} + χ_{3} + χ_{4} + χ_{6}
Ind_{S}^{G}ψ_{2} = χ_{2} + χ_{3} + χ_{4} + χ_{5}

giving us χ_{5} and χ_{6}.

0.6 Appendix: The Conjugacy Classes of P

The following GAP program was used to compute the conjugacy classes of P. Our method is extremely simple minded. We simply compute all the conjugates of all the elements of P by all the elements of P. While this may not be such a good algorithm for larger groups, it works quite well for P. Since P = UL, we produce P by first producing the sets U and L.

UU:=[];LL:=[];
for i in [0..1] do for j in [0..1] do
for k in [0..1] do for l in [0..1] do
B:=[[i,j,0],[k,l,0],[0,0,1]];
if not Determinant(B) = 0 then
UniteSet(LL,[B]);
fi;od;od;od;od;
for i in [0..1] do for j in [0..1] do
B:=[[1,0,i],[0,1,j],[0,0,1]];
UniteSet(UU,[B]);
od;od;

Then we take all conjugates. The following routine makes use of the idempotency of GAP’s UniteSet function.

CCC:=[];
for i in [1..4] do for j in [1..6] do
DD:=[];
for U in UU do for L in LL do
B:=(U*L)^(-1)*UU[i]*LL[j]*U*L mod 2;
UniteSet(DD,[B]);
od;od;
UniteSet(CCC,[DD]);
od;od;

Then the set CCC contains all the conjugacy classes of P.

Homework 6

0.7 The Sylow 7-Subgroup

We recall that G has the following character table.

G  g_{1} g_{2} g_{3} g_{4}  g_{5}  g_{6}
order of element 1 2 3 4  7  7
class size  1 21 56 42 24  24
χ_{1}  1 1 1 1  1  1
χ_{2}  6 2 0 0  −1  −1
χ_{3}  7 −1 1 −1  0  0
χ_{4}  8 0 −1 0  1  1
χ_{5}  3 −1 0 1 \frac{−1+i\sqrt{7}}{2} \frac{−1−i\sqrt{7}}{2}
χ_{6}  3 −1 0 1 \frac{−1−i\sqrt{7}}{2} \frac{−1+i\sqrt{7}}{2}

Let P_{7} ∈ Syl_{7} (G). We compute the conjugacy classes of N_{G} (P_{7}) and its character table by restricting characters of G.

We have from DF that |N_{G} (P_{7})| = 21. Since P_{7} ◁ N_{G} (P_{7}), we have that N_{G} (P_{7}) has one Sylow 7-subgroup so that N_{G} (P_{7}) has 6 elements of order 7. By our computations in Homework 5, we have that these 6 elements form two distinct conjugacy classes. The remaining elements of N_{G} (P_{7}) must be the identity and 14 elements of order 3. The elements of order 3 lie in two conjugacy classes of N_{G} (P_{7}) as follows.

Let Q ∈ Syl_{3} (N_{G} (P_{7})). There are 7 Sylow 3-subgroups of N_{P_{7}} (P_{3}) since this is the only way we can account for the the 14 elements of order 3. Therefore,

[N_{G} (P_{7}) : N_{N_{G}(P_{7})} (Q)] = 7.

Then |N_{N_{G}(P_{7})} (Q)| = 3 so that N_{N_{G}(P_{7})} (Q) = Q. Let q ∈ Q. Then since Q is cyclic, we have C_{N_{G}(P_{7})} (q) = C_{N_{G}(P_{7})} (Q) and this set at least contains Q. But

Q ≤ C_{N_{G}(P_{7})} (q) = C_{N_{G}(P_{7})} (Q) ≤ N_{N_{G}(P_{7})} (Q) = Q

so that |C_{N_{G}(P_{7})} (q)| = 3. Therefore the size of the conjugacy class of q in N_{G} (P_{7}) is 7. This means that N_{G} (P_{7}) has two conjugacy classes of elements of order 3.

We begin with the restrictions of χ_{1}, χ_{5} and χ_{6} to N_{G} (P_{7}) and the characters induced from the nontrivial characters ε_{2} and ε_{3} of Q ≤ N_{G} (P_{7}).

order of element 1 3 3  7  7
class size  1 7 7  3  3
χ_{1}  1 1 1  1  1
χ_{5}  3 0 0 ω + ω^{2} + ω^{4} ω^{3} + ω^{5} + ω^{6}
χ_{6}  3 0 0 ω^{3} + ω^{5} + ω^{6} ω + ω^{5} + ω^{6}
ε_{2}  7 ζ ζ^{2}  0  0
ε_{3}  7 ζ^{2} ζ  0  0

where ω is a primitive 7th root of unity and ζ is a primitive 3rd root of unity. We find ψ_{1}, ψ_{5} and ψ_{6} to be irreducible by taking inner products, we compute

⟨ε_{2}, χ_{5}⟩ = ⟨ε_{3}, χ_{5}⟩ = ⟨ε_{2}, χ_{6}⟩ = ⟨ε_{3}, χ_{6}⟩ = 1

and we find ε_{2} − χ_{5} − χ_{6} and ε_{3} − χ_{5} − χ_{6} to be irreducible giving the following character table.

order of element 1 3 3  7  7
class size  1 7 7  3  3
ψ_{1}  1 1 1  1  1
ψ_{2}  1 ζ ζ^{2}  1  1
ψ_{3}  1 ζ^{2} ζ  1  1
ψ_{5}  3 0 0 ω + ω^{2} + ω^{4} ω^{3} + ω^{5} + ω^{6}
ψ_{6}  3 0 0 ω^{3} + ω^{5} + ω^{6} ω + ω^{5} + ω^{6}

0.8 The Sylow 2- and 3-Subgroups

Next, let P_{2} ∈ Syl_{2} (G) and P_{3} ∈ Syl_{3} (G) We have from DF that the number of Sylow 2- and 3-subgroups of G are 21 and 28^{1} so that |N_{G} (P_{2})| = 8 and |N_{G} (P_{3})| = 6. We therefore have N_{G} (P_{2}) = P_{2} ≅ D_{8} by DF and N_{G} (P_{3}) ≅ S_{3} by elimination.

To compute the character table for D_{8} = ⟨r, s : r^{4} = s^{2} = 1, rs = sr^{3}⟩, we lift the three nontrivial characters from the quotient D_{8}/ ⟨r^{2}⟩ giving χ_{2}, χ_{3}, and χ_{4}, and induce a nontrivial character from ⟨r⟩ giving χ_{5}. We thus have the following table.

\begin{matrix}D_{8} & 1 & r & s & sr & r^{2} \\ χ_{1} & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & −1 & −1 & 1 & 1 \\ χ_{3} & 1 & −1 & 1 & −1 & 1 \\ χ_{4} & 1 & 1 & −1 & −1 & 1 \\ χ_{5} & 2 & 0 & 0 & 0 & −2 \end{matrix}

The character table for S_{3} was computed previously and is reprinted below.

\begin{matrix}S_{3} \\ χ_{1} & 1 & 1 & 1 \\ χ_{2} & 1 & −1 & 1 \\ χ_{1} & 2 & 0 & 1 \end{matrix}

0.9 The Issaacs-Navarro Conjecture for G

When p = 2, we have

M_{1} (G) = 4 = M_{1} (N (P_{2})) ,

for p = 3 we have

M_{1} (G) = 3 = M_{1} (N (P_{3})) ,

and for p = 7 we have

M_{1} (G) = 3 = M_{1} (N (P_{5})) ,
M_{2} (G) = 0 = M_{2} (N (P_{5})) ,
M_{3} (G) = 2 = M_{3} (N (P_{5}))

1There is a typographic error in the statement of Proposition 14 in DF, which has n_{3} = 7. The proof, however, has n_{3} = 28.

satisfying the Issaacs-Navarro Conjecture.

0.10 The Issaacs-Navarro Conjecture for A_{5}

We compute the character degrees with GAP as follows.

1 gap> G:=AlternatingGroup(5);
2 Alt( [ 1 .. 5 ] )
3 gap> P:=Normalizer(G,SylowSubgroup(G,2));
4 Group([ (1,3)(2,4), (1,2)(3,4), (2,3,4) ])
5 gap> Q:=Normalizer(G,SylowSubgroup(G,3));
6 Group([ (1,2,3), (2,3)(4,5) ])
7 gap> R:=Normalizer(G,SylowSubgroup(G,5));
8 Group([ (1,2,3,4,5), (2,5)(3,4) ])
9 gap> CharacterDegrees(G);
10 [ [ 1, 1 ], [ 3, 2 ], [ 4, 1 ], [ 5, 1 ] ]
11 gap> CharacterDegrees(P);
12 [ [ 1, 3 ], [ 3, 1 ] ]
13 gap> CharacterDegrees(Q);
14 [ [ 1, 2 ], [ 2, 1 ] ]
15 gap> CharacterDegrees(R);
16 [ [ 1, 2 ], [ 2, 2 ] ]

For example, the output on line 12 tells us that the Sylow 2-subgroup has three characters of degree 1 and one character of degree 3. For p = 2, we have

M_{1} (G) = 4 = M_{1} (P) ,

for p = 3 we have

M_{1} (G) = 3 = M_{1} (Q) ,

and for p = 5 we have

M_{1} (G) = 2 = M_{1} (Q) ,
M_{2} (G) = 2 = M_{2} (Q)

satisfying the Issaacs-Navarro Conjecture.

Homework 8

1. Find the p-blocks of G for p = 2, 3, 5.

2. Construct the isotypies between B_{3} (G) and B_{3} (N_{G} (P_{3})) and between B_{7} (G) and B_{7} (N_{G} (P_{7})), where P_{3} and P_{7} are Sylow 3- and 7-subgroups.

3. In constructing isptypies, a sign is attached to an irreducible character in B_{p} (G). Is there any connection with the sign which appears in Issaacs-Navarro?

4. Let p = 3. Find (a) the Brauer characters

(b) the decomposition numbers

(c) the Cartan invariants

(d) the dimension of the PIMs. and repeat for p = 7.

We have the following table for GL (3, 2).

order of element 1 2 3 4  7  7
class size  1 21 56 42 24  24
χ_{1}  1 1 1 1  1  1
χ_{2}  6 2 0 0  −1  −1
χ_{3}  7 −1 1 −1  0  0
χ_{4}  8 0 −1 0  1  1
χ_{5}  3 −1 0 1 \frac{−1+i\sqrt{7}}{2} \frac{−1−i\sqrt{7}}{2}
χ_{6}  3 −1 0 1 \frac{−1−i\sqrt{7}}{2} \frac{−1+i\sqrt{7}}{2}

We therefore have the following table for Z (\overline{A})

class size 1 21 56 42  24  24
ω_{1}  1 21 56 42  24  24
ω_{2}  1 7 0 0  −4  −4
ω_{3}  1 −3 8 −6  0  0
ω_{4}  1 0 −7 0  3  3
ω_{5}  1 −7 0 14 −4 + 4i\sqrt{7} −4 − 4i\sqrt{7}
ω_{6}  1 −7 0 14 −4 − 4i\sqrt{7} −4 + 4i\sqrt{7}

Reducing this mod 2R, 3R, and 7R, we have the following.

class size 1 21 56 42 24 24
ω_{1}  1 1 0 0 0 0
ω_{2}  1 1 0 0 0 0
ω_{3}  1 1 0 0 0 0
ω_{4}  1 0 1 0 1 1
ω_{5}  1 1 0 0 0 0
ω_{6}  1 1 0 0 0 0
ω_{1}  1 0 2 0 0 0
ω_{2}  1 1 0 0 2 2
ω_{3}  1 0 2 0 0 0
ω_{4}  1 0 2 0 0 0
ω_{5}  1 2 0 2 ♣ ♣
ω_{6}  1 2 0 2 ♣ ♣
ω_{1}  1 0 0 0 3 3
ω_{2}  1 0 0 0 3 3
ω_{3}  1 4 1 1 0 0
ω_{4}  1 0 0 0 3 3
ω_{5}  1 0 0 0 ♣ ♣
ω_{6}  1 0 0 0 ♣ ♣

From the table mod 2, we see that the 2-blocks of G are {χ_{1}, χ_{2}, χ_{3}, χ_{5}, χ_{6}} and {χ_{4}}. Some of the elements of the table mod 3 have not been reduced. From the other elements, however, we can see that {χ_{1}, χ_{3}, χ_{4}} and {χ_{2}} form 3-blocks. Since 3 divides χ_{5} (1) = χ_{6} (1) = 3, we have that χ_{5} and χ_{6} mod 3 are the only irreducible Brauer characters in their blocks. This means that {χ_{5}} and {χ_{6}} are the remaining 3-blocks.

Next, we compute the 7-blocks. Since 7 ∈ 𝔭 by assumption, we have that

−7 = \sqrt{−7}\sqrt{−7} ∈ 𝔭,

but since 𝔭 is prime, this means that \sqrt{−7} ∈ 𝔭. Thus,

24 − (−4 + 4\sqrt{−7}) = 28 − 4\sqrt{−7} ∈ 𝔭

which shows that χ_{1} and χ_{5} are in the same 7-block. Repeating this argument, we have that {χ_{1}, χ_{2}, χ_{4}, χ_{6}, χ_{7}} is a 7-block. This means that {χ_{3}} is the remaining 7-block.

We next look at the character tables for N (P_{3}) and N (P_{7}) where P_{3} and P_{7} are Sylow 3- and 7- subgroups.

N (P_{3}) = S_{3}  Table  ω_{i} (C_{j})

order of element 1 2 3

class size  1 3 2
ψ_{1}  1 1 1 1 3 2
ψ_{2}  1 −1 1 1 −3 2
ψ_{3}  2 0 −1 1 0 −1

N (P_{7})  Character Table  ω_{i} (C_{j})
order of element 1 3 3  7  7
class size  1 7 7  3  3
ψ_{1}  1 1 1  1  1  1 7 7  3  3
ψ_{2}  1 ζ ζ^{2}  1  1  1 7ζ 7ζ^{2}  3  3
ψ_{3}  1 ζ^{2} ζ  1  1  1 7ζ^{2} 7ζ  3  3
ψ_{4}  3 0 0 \frac{−1+i\sqrt{7}}{2} \frac{−1−i\sqrt{7}}{2} 1 0 0 \frac{−1+i\sqrt{7}}{2} \frac{−1−i\sqrt{7}}{2}
ψ_{5}  3 0 0 \frac{−1−i\sqrt{7}}{2} \frac{−1+i\sqrt{7}}{2} 1 0 0 \frac{−1−i\sqrt{7}}{2} \frac{−1+i\sqrt{7}}{2}

where ω is a primitive 7th root of unity and ζ is a primitive 3rd root of unity,

Reducing ω_{i} (C_{j}) from the first table mod 3R, we see that B_{3} (N_{G} (P_{3})) = {ψ_{1}, ψ_{2}, ψ_{3}}. For ω_{i} (C_{j}) from the second table, we observe that

\frac{7}{2} · 2 = 7 ∈ 𝔭
\frac{\sqrt{−7}}{2} · 2 = \sqrt{−7} ∈ 𝔭,

but since 2 ∉ 𝔭 else 7 − 2 − 2 − 2 = 1 ∈ 𝔭, we must have that \frac{7}{2} ∈ 𝔭 and \frac{\sqrt{−7}}{2} ∈ 𝔭. Thus,

\frac{−1 + \sqrt{−7}}{2} − 3 = \frac{−7 + \sqrt{−7}}{2} ∈ 𝔭

which shows that χ_{1} and χ_{4} are in the same 7-block. Repeating this argument, we have B_{7} (N_{G} (P_{7})) = {ψ_{1}, ψ_{2}, ψ_{3}, ψ_{4}, ψ_{5}}.

Next, we see that the maps

I_{3} : \{ \begin{matrix}χ_{1} \\ χ_{3} \\ χ_{4} \end{matrix} \} → \{ \begin{matrix}ψ_{1} \\ ψ_{2} \\ ψ_{3} \end{matrix} \}

and

I_{7} : \{ \begin{matrix}χ_{1} \\ −χ_{2} \\ χ_{4} \\ χ_{5} \\ χ_{6} \end{matrix} \} → \{ \begin{matrix}ψ_{1} \\ ψ_{2} \\ ψ_{3} \\ χ_{4} \\ χ_{5} \end{matrix} \}

preserve the character degrees mod 3 and 7 and preserve the values of the characters on the 3- and 7-elements. Then I_{3} and I_{7} are the isotypies.

For p = 3, we have that the trivial character χ_{1} is a Brauer character and that χ_{2}, χ_{5} and χ_{6} are also Brauer characters since 3 divides χ_{2} (1), χ_{5} (1), and χ_{6} (1). We then have the decomposition matrix

φ_{1} φ_{2} φ_{3} φ_{4} φ_{5}
deg 1  · 6 3 3
χ_{1} 1 1 ⋯
χ_{3} 7  m n ⋯
χ_{4} 8 m + 1 n ⋯
χ_{2} 6  ·  · 1 · ·
χ_{5} 3  ·  · · 1 ·
χ_{6} 3  ·  ⋯ 1 20

Let m and n be the multiplicities of φ_{1} and φ_{2} in χ_{3} as in the table. Then

0 = χ_{3} (g) = mφ_{1} (g) + nφ_{2} (g) = m + nφ_{2} (g) = Φ (ω)  (3)

for g ∈ G an element of order 7. The right hand side of (3) is a sum of 7^{th} roots of unity and is therefore some polynomial Φ in ω . Since the minimal polynomial for ω is 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}, we must have either Φ (x) = 0 or Φ (x) = 1 + x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6}. Thus, we either have that m = 0, in which case χ_{3} is an irreducible Brauer character, or else m = 1, (1 is the only integer in the sum) in which case χ_{3} = φ_{1} + φ_{2}.

In the latter case,

so that

C =^{t} DD = (\begin{matrix}2 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{matrix}) .

Final Thoughts on GL (3, 2)

Let G = GL (3, 2), the simple group of order 168. We begin by computing the conjugacy classes of G. We do this by enumerating all the possible rational forms. There are 6 conjugacy classes.

Next, we construct the characters of G. We begin with χ_{1}, the trivial character. Next, we define the subgroup P = \{( \begin{matrix}♣ & ♣ & ♣ \\ 0 & ♣ & ♣ \\ 0 & ♣ & ♣ \end{matrix} )\} ≤ G and compute Ind_{P}^{G} (1) where 1 is the trivial character of P. To do this, we observe that P is the stabilizer of the line ⟨v_{1}⟩ in the vector space V = ⟨v_{1}, v_{2}, v_{3}⟩ over 𝔽_{2} so that Ind_{P}^{G} (1) (g) is the number of lines fixed by g ∈ G. This gives us χ_{2}.

Similarly, B = \{( \begin{matrix}♣ & ♣ & ♣ \\ 0 & ♣ & ♣ \\ 0 & 0 & ♣ \end{matrix} )\} ≤ G is the stabilizer of the flag ⟨v_{1}⟩ ≤ ⟨v_{1}, v_{2}⟩ in V so

that Ind_{B}^{G} (1) (g) is number of flags fixed by g. This gives us χ_{4}.

We next construct the character of degree 7 using Harish-Chandra induction. We take

P = \{( \begin{matrix}♣ & ♣ & ♣ \\ ♣ & ♣ & ♣ \\ 0 & 0 & ♣ \end{matrix} )\} ⊃ L = \{( \begin{matrix}♣ & ♣ & 0 \\ ♣ & ♣ & 0 \\ 0 & 0 & ♣ \end{matrix} )\} ≅ S_{3}

We lift the sign character of S_{3} ≅ L to a character of P and induce from this a character of G. This is not extremely easy, most of the complication arising from determining how the conjugacy classes of G split in P, a computation required in the induced character formula. This can at least be accomplished by explicit computation, which I’m sorry to say is what I did. This gives us an irreducible character χ_{3} of G.

We compute the remaining two characters for G by inducing two characters from a 7-Sylow subgroup S ≤ G. Again, the difficulty is in determining how the conjugacy classes of G split in S, but again, this can at least be accomplished with an explicit computation. This produces the characters χ_{5} and χ_{6} of G.

The characters computed in the discussion are presented in the character table at the head of this document.

Next, we determine the Sylow 2-, 3-, and 7-subgroups P_{2}, P_{3}, and P_{7} of G and their normalizers. By order considerations and other algebraic manipulations, we find that P_{2} = N_{G} (P_{2}) ≅ B and P_{3} ◁ N_{G} (P_{3}) ≅ L ≅ S_{3}. For P_{7}, we use the construction of the characters for groups of order pg to compute the character table for N_{G} (P_{7}) which has order 3 · 7. The character tables for N_{G} (P_{3}) and N_{G} (P_{7}) are presented on pages 19 and 20 of this document.

Having the character tables for N_{G} (P_{2}) N_{G} (P_{3}), and N_{G} (P_{7}), we were able to verify the Issaacs-Navarro conjecture for G.

Finally, in this document, we computed various p-blocks, isotypies, Brauer characters and Cartan invariants for G.