Exercise 23.1 . . . 106 Exercise 23.2 . . . 106 Exercise 23.3 . . . 106 Exercise 23.4 . . . 107 Exercise 23.5 . . . 108 Exercise 23.6 . . . 110 Exercise 23.7 . . . 111 Exercise 23.8 . . . 112 Exercise 23.9 . . . 113 Exercise 23.10 . . . 114 Exercise 23.1. Let G be a group of odd order. If x ∈ G is real then x = x−1 ⇒ x2 = 1, so if x 6= 1 this means G contains an element of order 2. However this would imply 2 | |G|
but this is not possible as G has odd order, so x = 1.
Exercise 23.2. Let G ∼= Cn1× · · · × Cnr be a finite abelian group. Let g1, . . . , gr be gener-ators for the respectively cyclic components of G and λ1, . . . , λn be respectively n1, . . . , nr primitive roots of unity. Then every irreducible character of G is linear and equivalent to one of
χ(g1i1, . . . , grir) = (λi11. . . λirr).
Now assume χ is a real irreducible character of G. Recall that any linear character of G is also a homomorphism of G. Therefore we have
χ(g) = χ(g)⇔ χ(g) = χ(g−1)⇔ χ(g2) = χ(1)⇔ χ(g)2= 1⇔ χ(g) = ±1.
In other words we have λi11. . . λirr = ±1 but this happens if and only if λijj = ±1, which happens if and only if 2 | nj. If 2 | nj then the only values which take ±1 are λ0j and λnjj/2. If ` is the number of nj which are even then there are 2` such irreducible characters.
Conversely, how many involutions are there? If g2 = 1 then either we have ij = 0 or 2| nj and we have ij = nj/2. Therefore there are 2` such elements and we’re done.
Exercise 23.3. We have the character tables of D2n, for n odd and even, to be as in tables23.1 and 23.2.
106
Chapter 23 107
gi 1 ar
16r 6n−12
b
|CG(gi)| 2n n 2
χ1(g) 1 1 1
χ2(g) 1 1 −1
ψj(g)
16j6n−12
2 εj r + ε−j r 0
Table 23.1: The character table of D2n, for n odd
gi 1 am ar
16r 6m−1 b ab
|CG(gi)| 2n 2n n 4 4
χ1(g) 1 1 1 1 1
χ2(g) 1 1 1 −1 −1
χ3(g) 1 (−1)m (−1)r 1 −1 χ4(g) 1 (−1)m (−1)r −1 1
ψj(g)
16j6m−1
2 2(−1)j εj r + ε−j r 0 0
Table 23.2: The character table of D2n, for n even
Clearly the χi are real characters in both tables but what about the ψj? Well for any complex number z we have z + z = 2 Re(z). Also it’s clear that ε = e−2πi /n = ε−1 and so ψj is also a real character of D2n. Therefore all the conjugacy classes of D2n are real conjugacy classes. This means that every element of D2n is real and so there are 2n real elements. Clearly every irreducible character is real, which gives us ιχ = 1 for all irreducible characters χ.
Exercise 23.4. Let V be the 2-dimensional irreducible CG module associated to the repre-sentation ρ. Let B = {v1, v2} be the basis of V then we have B0 ={v1⊗ v2− v2⊗ v1} is a basis of A(V ⊗ V ). Let the matrix of g with respect to B be
gρB =
"
a b c d
# . Therefore we have v1g = av1+ bv2 and v2g = c v1+ d v2.
We now consider the action of g on the antisymmetric tensor product module A(V ⊗V ).
Chapter 23 108
We have
(v1⊗ v2− v2⊗ v1)g = v1g ⊗ v2g− v2g⊗ v1g,
= (av1+ bv2)⊗ (cv1+ d v2)− (cv1+ d v2)⊗ (av1+ bv2),
= ad (v1⊗ v2) + bc (v2⊗ v1)− cb(v1⊗ v2)− ad (v2⊗ v1),
= (ad− bc)(v1⊗ v2− v2⊗ v1).
Therefore χA(g) = ad − bc = det(gρ). Now ιχ = −1 if and only if 1G is a constituent of χA but this happens if and only if χA = 1G, (as χA is a linear character and hence irreducible). In other words det(g) = 1 for all g ∈ G.
Exercise 23.5. We let G = T4n = ha, b | a2n = 1, an = b2, b−1ab = a−1i and ε 6= ±1 a 2nth root of unity.
(a) Let ρ : G → GL(2, C) be the irreducible matrix represenation associated to V . It’s clear that for a and b we have
aρ =
"
ε 0
0 ε−1
#
bρ =
"
0 1 εn 0
#
Now clearly det(aρ) = εε−1 = 1 and det(bρ) = −εn. Recall the determinant is multiplicative and every element in T4n can be written in the form aibj with 1 6 i 6 2n and 0 6 j 6 1. Now ιχ = −1 if and only if det(xρ) = 1 for all x ∈ T4n. Therefore this only happens if −εn = 1⇒ εn=−1. So we have ιχ = 1 if εn= 1 and ιχ = −1 if εn =−1.
(b) We check the action of a on the bilinear form
β(v1a, v1a) = β(εv1, εv1) = ε2β(v1, v1) = 0 = β(v1, v1), β(v2a, v2a) = β(ε−1v2, ε−1v2) = ε−2β(v2, v2) = 0 = β(v2, v2), β(v1a, v2a) = β(εv1, ε−1v2) = εε−1β(v1, v2) = β(v1, v2), β(v2a, v1a) = β(ε−1v2, εv1) = ε−1εβ(v2, v1) = β(v2, v1).
Therefore β is invariant under the action of a. Now considering the action of b on the bilinear form
β(v1b, v1b) = β(v2, v2) = 0 = β(v1, v1),
β(v2b, v2b) = β(εnv1, εnv1) = ε2nβ(v1, v1) = 0 = β(v2, v2), β(v1b, v2b) = β(v2, εnv1) = εnβ(v2, v1) = ε2n = 1 = β(v1, v2),
Chapter 23 109
β(v2b, v1b) = β(εnv1, v2) = εnβ(v1, v2) = εn = β(v2, v1).
Therefore β is invariant under the action of b, hence invariant under the action of all elements x ∈ T4n.
Using Theorem 23.16 we have ιχ = 1 if there exists a symmetric bilinear form on V, which happens if β(v2, v1) = β(v1, v2) = 1, in other words εn = 1. Alternatively the theorem tells us that ιχ = −1 if there exists a skew-symmetric bilinear form on V, which happens if β(v2, v1) =−β(v1, v2), in other words εn=−1.
(c) Every element in T4n has the form aibj for some 1 6 i 6 2n and 0 6 j 6 1. Now assume j = 0 then (ai)2 = a2i = 1 ⇔ i = nor 2n, however if i = 2n then ai = 1, which is not an element of order 2. Assume j = 1 then (aib)2 = aibaib = aia−ib2= an 6= 1. Therefore the only element of order 2 in T4n is an.
(d) Now the character table for T4n with n even is as in table 23.3.
g 1 an ar
16r 6n−1 b ba
g2 1 1 a2r an an
|CG(g)| 4n 4n 2n 4 4
1G 1 1 1 1 1
φ1 1 1 1 −1 −1
φ2 1 1 (−1)r 1 −1
φ3 1 1 (−1)r −1 1
χk(g)
06k6n−1
2 2(−1)k εk r + ε−kr 0 0
χ2k(g) 4 4 2 + ε2k r + ε−2kr 0 0 (χk)S(g) 3 3 1 + ε2k r + ε−2kr (−1)k (−1)k
(χk)A(g) 1 1 1 (−1)k +1 (−1)k +1
Table 23.3: The character table of T4n, with n even
It’s obvious that ι1G = ιφ1 = ιφ2 = ιφ3 = 1 because 12G = φ21 = φ22 = φ23 = 1G and (1G)S = 1G, (1G)A= 0. Now all that’s left to consider is the indicator of the χk. It’s clear that χk is real because εk r+ ε−kr = εk r + εk r = 2 Re(εk r).
Now in the table above we have considered χ2k and decomposed it into its sym-metric and alternating parts. If k is odd then k + 1 is even and (χk)A = 1G, which gives us ιχk =−1. If k is even then k + 1 is odd and (χk)A= φ1. Therefore 1G must
Chapter 23 110
be a constituent of (χk)S and so (ιχk) = 1. So we have the Frobenius-Schur Count of Involutions is
X
χ
(ιχ)χ(1) = 1 + 1 + 1 + 1 +
n−1
X
k =1
2(−1)k = 4− 2 = 2, because n is even so the sum equals −2.
If n is odd then we have the character table of T4n to be as in table 23.4.
g 1 an ar
16r 6n−1 b ba
g2 1 1 a2r an an
|CG(g)| 4n 4n 2n 4 4
1G 1 1 1 1 1
φ1 1 −1 (−1)r i −i
φ2 1 1 1 −1 −1
φ3 1 −1 (−1)r −i i
χk(g)
16k6n−1
2 2(−1)k εk r + ε−kr 0 0
χ2k(g) 4 4 2 + ε2k r + ε−2kr 0 0 (χk)S(g) 3 3 1 + ε2k r + ε−2kr (−1)k (−1)k
(χk)A(g) 1 1 1 (−1)k +1 (−1)k +1
Table 23.4: The character table of T4n, with n odd
It’s clear that 12G = φ22= 1Gso ι1G = ιφ2= 1and φ1, φ3aren’t real so ιφ1= ιφ3= 0. Now all that’s left to consider is the indicator of the χk. However the situation here is clearly identical to the situation for when n is even. Therefore the Frobenius-Schur Count of Involutions is
X
χ
(ιχ)χ(1) = 1 + 0 + 1 + 0 +
n−1
X
k =1
2(−1)k = 2 + 0 = 2,
because n is odd so the sum equals 0. This confirms that there is only one involution in T4n.
Exercise 23.6. Let χ be an irreducible character of G such that ιχ = −1. Let V be the irreducible CG module which affords χ as a character. By Theorem 23.16 we have ιχ = −1
Chapter 23 111
if and only if there exists a non-zero G-invariant skew-symmetric bilinear form on V . Let’s call this bilinear form β. Now let v1, . . . , vm be a basis for the CG module V . We consider the vector subspace
U ={u ∈ V | β(u, v ) = 0 for all v ∈ V }.
It’s clear that this is a vector subspace but it is also a CG module. Recall that V G = V and so given u ∈ U, v ∈ V and g ∈ G we have β(ug, vg) = β(u, v) = 0 and so UG = U.
Now V is an irreducible CG submodule and β is non-zero so we must have U = {0}. Now let β(vi, vj) = xi j for all 1 6 i 6 j 6 m then we can construct a matrix X with
X =
0 x1,2 . . . x1,m−1 x1,m
−x1,2 ... x2,m
... ... ...
−x1,m−1 ... xm−1,m
−x1,m −x2,m . . . −xm−1,m 0
.
We can see that the matrix X is skew-symmetric, i.e. XT = −X, because it’s coming from a skew-symmetric bilinear form. By the fact that U = {0} we know that non of the off diagonal entries in X are 0, hence the determinant of X is non-zero. We consider the determinant of XT in two ways. We can see that we can get from XT to X by applying m row and column swaps and multiplying by −1. Every time we make a row or column swap we multiply the determinant by −1 and so det(XT) = (−1)m+1det(X). Alterantively det(XT) = det(−X) = − det(X), so this gives us det(X) = (−1)mdet(X). Therefore m is even and we’re done.
Exercise 23.7. Let V be a vector space over R and let {v1, . . . , vn} be any basis of V . Recall that given a vector subspace U ⊆ V we define U⊥ with respect to β1 by
U⊥ ={v ∈ V | β1(u, v ) = 0for all u ∈ U}.
Also, as vector spaces, we have V = U ⊕U⊥. We now wish to construct a basis {e1, . . . , en} of V which is orthonormal with respect to β1.
We start by letting f10 = v1. Consider the vector subspace U1 = span{v1} and the standard projection map π1 : V → U1. Define f20 = v2− π1(v2) ∈ U1⊥. Note that f20 6= 0 because if f20 = 0 then v2 = π1(v2)⇒ v2 ∈ span{v1}. However this cannot happen because {v1, . . . , vn} is a basis for V .
In a similar fashion we consider the vector subspace U2 = span{v1, v2}and the standard
Chapter 23 112
projection map π2: V → U2. Define f30 = v3− π2(v3) ∈ U2⊥. Again we have f30 6= 0 because if f30 = 0 then v3 = π2(v3) ⇒ v3 ∈ span{v1, v2}. However this cannot happen because {v1, . . . , vn} is a basis for V .
Carrying on with this process gives us a set {f10, . . . , fn0} such that β(fi0, fj0) = 0 for all i 6= j. Now we define fi = fi0/pβ1(fi0, fi0), then {e1, . . . , en} is an orthonormal basis of V with respect to β1. Note that pβ(fi0, fi0)∈ R because β(w , w ) > 0 for all non-zero w ∈ V . Note that in general this process is known as the Gram Schmidt algorithm.
We now want to show that {f1, . . . , fn} can be transformed into a basis which is also orthogonal on β. Let X be an n × n matrix such that Xi j = β(fi, fj) then X is a real symmetric matrix. By general matrix theory there exists a real orthogonal matrix Q, (i.e.
QT = Q−1), such that QXQ−1 = D where D is a diagonal matrix. Note that this a special case of the usual diagonalisation process for complex valued matrices.
The above process corresponds to a change of basis for V . Let Q = (qi j)then we define a new basis
ei =
n
X
j =1
qi jfj.
Now because QXQ−1 is diagonal we have β(ei, ej) = 0 for any i 6= j. Also the ma-trix representing β1(fi, fj) is the identity so conjugating by Q does nothing. This means β1(ei, ej) = δi j as required.
Exercise 23.8. Let V and W be RG-modules.
(a) Let ϑ : V → W be an RG-homomorphism. Now im(ϑ) is an RG submodule of W and ker(ϑ) is an RG submodule of V . We have V and W are irreducible RG modules so either Im(ϑ) = W and ker(ϑ) = {0} or Im(ϑ) = {0} and ker(ϑ) = V . This gives you either ϑ is an RG isomorphism or ϑ is the zero map.
(b) Now if ϑ : V → V is an RG isomorphism then ϑ is also a CG isomorphism. If V is an irreducible CG module then by Schur’s Lemma there exists λ ∈ C such that ϑ = λ1V. However ϑ is also an RG homomorphism so we must have λ ∈ R otherwise im(ϑ) = V is not an RG-module.
(c) Consider G = C3 =hx | x3 = 1i. We have a 2-dimensional irreducible RG submodule V = span{1 − a, 1 − a2}. Let v1 = 1− a and v2= 1− a2 then we have the action of a on these basis elements to be
v1a = (1− a)a = a − a2 = v2− v1 v2a = (1− a2)a = a− 1 = −v1. We can define a map ϑ : V → V by vϑ = av. Note that multiplication on the left
Chapter 23 113
is well defined as V is a submodule of the group algebra. Also, as G is commutative, we will have the action of the group on the left to be the same as the action of the group on the right.
So, with respect to the basis B = {v1, v2} we have
[a]B =
"
−1 −1
1 0
# .
Now considering the eigenvalues of this matrix we have the characteristic polynomial to be
det([a]B− x I2) =
−1 − x −1
1 −x
=−x (−1 − x ) + 1 = x2+ x + 1.
This polynomial has eigenvalues −1±3i2 6∈ R. Therefore the map ϑ cannot be a real scalar multiple of the identity map.
Exercise 23.9. Let Ω = {Hx1, . . . , Hxn}. The map ρg : Ω→ Ω is a permutation of Ω if it is a bijection. It’s clearly 1-1 because
(Hxi)ρg = (Hxj)ρg⇒ Hxig = Hxjg ⇒ Hxi = Hxj ⇒ xi = xj
because the right cosets of H partition the group G. It is also surjective as given Hxi ∈ Ω we have Hxig−1 ∈ Ω and (Hxig−1)ρg = Hxig−1g = Hxi. Therefore ρg is a permutation of Ω.
We need to show that for all g, h ∈ G we have (gh)ρ = (gρ)(hρ) or in other words ρgh = ρgρh. For any Hxi ∈ Ω we have
(Hxi)ρgh = Hxigh = (Hxig)ρh= (Hxi)ρgρh.
Now our choice of Hxi was arbitrary and so ρgh= ρgρh, which gives us ρ is a homomorphism from G to Sym(Ω).
Now we have ker(ρ) = {g ∈ G | ρg = idΩ}. If ρg is such a map then for all Hxi ∈ Ωwe have
(Hxi)ρg = (Hxi) idΩ⇔ Hxig = Hxi,
⇔ H = Hxigxi−1,
⇔ xigxi−1∈ H,
Chapter 23 114
⇔ g ∈ xi−1Hxi. Therefore ker(ρ) = ∩x∈Gx−1Hx as required.
We have a bijection ϕ : Ω → {1, . . . , n} given by (Hxi)ϕ = i. We have a map from ψ : Sym(Ω) → Sn given by σ 7→ ϕ−1σϕ. Now this is in fact an isomorphism, we can see it’s a homomorphism as for any σ, τ ∈ Sym(Ω) we have
(στ )ψ = ϕ−1(στ )ϕ = (ϕ−1σϕ)(ϕ−1τ ϕ) = (σψ)(τ ψ).
Also we can define an inverse map ψ−1: Sn→ Sym(Ω) by πψ−1 = ϕπϕ−1. Note that the composition of homomorphisms is again a homomorphism. Therefore if H is a subgroup of index n in G then we have a homomorphism ρ : G → Sym(Ω), hence a homomorphism ρψ : G → Sn. It’s clear that ker(ρψ) = ker(ρ) = ∩x∈Gx−1Hx ⊆ H.
Exercise 23.10. Let G be a finite group with an involution t ∈ G such that CG(t) ∼= C2. Now let χ1, . . . , χk be the irreducible characters of G and ψ1, ψ2be the irreducible characters of C2. Recall that all the irreducible characters of C2 are linear.
As t2 = 1 we have t−1 ∈ tG, which means χi(t) is an integer for each 1 6 i 6 k.
Furthermore we know that χi(t) ≡ χi(1) (mod 2) for each 1 6 i 6 k. From the column orthogonality relations we have
k
X
i =1
χi(1)2=|G|
k
X
i =1
χi(t)2= 2.
As the χi(t) are integers we must have χi(t) = ±1 for two i, j ∈ {1, . . . , k} with i 6= j and χi(t) = 0 for all other irreducible characters. As χi(1) ≡ χi(t) (mod 2) for every irreducible character this means that only two of the character degrees are odd and the rest are even.
We now consider the restriction of the irreducible characters to the centraliser CG(t). We know that χi ↓ CG(t) = d1ψ1 + d2ψ2 for each 1 6 i 6 k and that d12+ d22 6 [G : CG(t)] = 2. Therefore the only possibilities are that d1, d2 are 0 or 1. We have that
χi(1) = (χi ↓ CG(t))(1) = d1ψ1(1) + d2ψ2(1) = d1+ d2.
So the degree of every irreducible character of G is either 1 or 2. However we know there are two characters of odd degree and all the other characters have even degree. This means G has two linear characters and all other irreducible characters are of degree 2. In other words [G : G0] = 2.
Chapter 23 115
Assume G is a simple group. We have G0C G is a normal subgroup of G and clearly G0 6= G so G0 ={1}. This means G is abelian, which means there are only two characters and they’re both linear. So G has the character table of CG(t) ∼= C2 which means G ∼= C2.