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Math 902
Solutions to Homework # 1
1. Let ρ : G → GLn (C) be a complex representation of a finite group G and χ its
associated character. Prove that for g ∈ G, |χ(g)| = χ(1) if and only if ρ(g) = λI for
some root of unity λ.
Solution: Since g has finite order, ρ(g)s = I for some s > 0. Hence, the minimal
polynomial over of ρ(g) divides xs − 1, which has distinct roots in C. Thus, ρ(g) is
diagonalizable. Let λ1 , · · · , λn be the eigenvalues of ρ(g). Then, as ρ(g)s = I, each λi
is an sth root of unity. In particular, |λi | = 1 for all i. Recall the triangle inequality
law for complex numbers: for w, z ∈ C, |w + z| ≤ |w| + |z| with equality if and only if
w = 0 or z = rw for some nonnegative real number r. Hence, we have |λ1 + λ2 | ≤ 2
with equality if and only if λ1 = λ2 . By induction, |λ1 + · · · λn | ≤ n with equality if
and only if λi = λj for all i, j. Thus, |χ(g)| = |λ1 + · · · + λn | ≤ n = χ(1) with equality
if and only if λi = λj for all i, j. Note that the last condition is equivalent to the
condition that ρ(g) is similar to λI for some root of unity λ. Since λI is in the center
of GLn (C), ρ(g) is similar to λI if and only if ρ(g) = λI.
2. Let G 6= 1 be a finite group. Prove that G is simple if and only if for all nontrivial
irreducible complex characters χ of G and for all g 6= 1 one has χ(g) 6= χ(1).
Solution: Suppose G is not simple. Let H be a nontrivial normal subgroup of G.
Then G/H has a nontrivial irreducible character χ. Let χ be the irreducible character
of G induced by χ; i.e., χ(g) = X(g) for all g ∈ G. Let g ∈ H, g 6= 1. Then
χ(g) = χ(1) = χ(1). Conversely, suppose there exists g ∈ G, g 6= 1 and a nontrivial
irreducible complex character χ of G such that χ(g) = χ(1). By problem #1, ρ(g) = λI
for some λ ∈ C, where ρ : G → GLn (C) is a representation of G whose character is χ.
Then χ(g) = nλ = n = χ(1), so λ = 1. Thus, ρ(g) = I. Hence, ker ρ 6= {1}. As ρ is not
the trivial representation, ker ρ is a nontrivial normal subgroup of G. Consequently, G
is not simple.
3. Let G be a finite group and G0 its commutator subgroup. Prove that the number of
complex degree one characters of G is [G : G0 ].
Solution: Let t = [G : G0 ] = |G/G0 |. Since G/G0 is abelian, it has t distinct degree
one characters, each of which induces a degree one character on G. Now let χ be a
degree one character of G and ρ : G → GL1 (k) = k ∗ a representation of G whose
character is χ. Since G/ ker ρ is isomorphic to a subgroup of k ∗ , G/ ker ρ is abelian.
Hence, ker ρ ⊇ G0 and there is an induced representation ρ : G/G0 → k ∗ . Then χ is
induced by the character associated to ρ. Thus, every degree one character of G is
induced from a degree one character of G/G0 .
4. Let k be a field of characteristic p > 0 and R a finite dimensional k-algebra. Let M be
a left R-module such that p does not divide dimk M . Prove that χM 6= 0.
Solution: Note that χM (1) = dimk M 6= 0, as p does not divide dimk M .
5. Let ρ : G → GLk (V ) be an k-linear representation of G.
(a) Prove that if ρ(G) spans Endk (V ) as a k-vector space then ρ is irreducible.
Solution: Consider V as a a left k[G]-module. Recall that left multiplication by
an element of k[G] is an k-endomorphism of V . Hence, there is a natural ring
homomorphism from φ : k[G] → Endk (V ). Now, suppose ρ(G) spans Endk (V ).
Since φ(g) = ρ(g) for all g ∈ G, this implies that φ(F [G]) = Endk V . Let
u ∈ V , u 6= 0. We claim that k[G]u = V . Let v ∈ V . Since u is part of
a k-basis for V , there exists f ∈ Endk V such that f (u) = v. Let w ∈ k[G]
such that φ(w) = f . Since φ(w) is just left multiplication by w, this means that
wu = φ(w)(u) = f (u) = v. Hence, k[G]u = V and V is an simple k[G]-module.
Hence, ρ is an irreducible representation of G.
(b) Prove that the converse to part (a) holds if k is algebraically closed and V is finite
dimensional.
Solution: Let φ : k[G] → Endk V be as in the solution to part (a). Since the
k-span of ρ(G) is φ(k[G]), it suffices to prove that φ is surjective. Let R = k[G].
As ρ is irreducible, V is a simple k[G]-module and hence D = EndR V is a division
ring. Since k is algebraically closed, k ⊆ Z(D), and dimk D ≤ dimk Endk V =
(dimk V )2 < ∞, we have that D is algebraic over F . Hence, D = k. By the
Jacobson density theorem, the natural map R → Endk V is surjective. But this
map is exactly the map φ defined above.
(c) Let G = C4 and k = R. Show that G has an irreducible R-representation ρ of
degree 2 and the span of ρ(G) is a proper subspace of M2 (R).
Solution: First, note that R[C4 ] ∼
= R[x]/(x4 − 1). As R[C4 ] is commutative,
the simple R[C4 ]-modules correspond to R[C4 ]/m where m is a maximal ideal
of R[C4 ]. Consider the simple R[C4 ]-module V = R[x]/(x2 + 1). (Note that
(x2 + 1)R[x]/(x4 − 1) is a maximal ideal of R[x]/(x4 − 1).) As dimR V = 2, it
corresponds to an irreducible degree 2 R-representation ρ of C4 . With respect to
the R-basis {1, x}, ρ : C4 → GL2 (R) is given by
0 −1
ρ(g) =
,
1 0
where C4 = hgi. (Note that ρ(g) corresponds to multiplication by x on V .) It is
easily seen that ρ(G) does not span M2 (R).
6. Find the character table (over the complexes) for D8 , the dihedral group of order 8.
Solution: Let D8 = hr, s | r4 = s2 = 1, rs = sr3 i. It is easily seen that D8 has
five conjugacy classes: {1}, {r2 }, {r, r3 }, {s, sr2 }, and {sr, sr3 }. Thus, D8 has five
irreducible representations. We will show that the character table for D8 is
χ1
χ2
χ3
χ4
χ5
1 r2
1 1
1 1
1 1
1 1
2 -2
r s sr
1 1 1
1 -1 -1
-1 1 -1
-1 -1 1
0 0 0
Note that H = {1, r2 } is a normal subgroup of D8 and that D8 /H ∼
= C2 × C2 . Hence,
we get four degree 1 characters for D8 by composing thePfour degree one characters
of C2 × C2 with the canonical map D8 → D8 /H. Since 5i=1 χi (1)2 = |D8 | = 8, we
see that the remaining irreducible character has degree 2. The last row can now be
determined by using that the last four columns must each be orthogonal to the first
column.
7. Let G be a finite group.
(a) For g, h ∈ G prove that g and h are in the same conjugacy class of G if and only
if χ(g) = χ(h) for every irreducible complex character χ of G.
Solution: Since characters are class functions, one direction is clear. Suppose
χ(g) = χ(h) for every irreducible complex character P
χ of G. Let C1 , . . . , Ct be the
distinct conjugacy classes of G, mi = |Ci |, and zi = x∈Ci x. If g ∈ Ci then from
a formula proved in class
t
X
χj (g)
ej ,
zi = mi
χ (1)
j=1 j
where χ1 , . . . , χt are the irreducible characters of G and e1 , . . . , et are the central
idempotents giving the decomposition of C[G] into simple Artinian rings. If h ∈
C` , then a similar formula holds for z` with h in place of g. But since χj (g) = χj (h)
for all j, this implies zi and z` are linearly dependent. Since z1 , . . . , zt are linearly
independent over C, we conclude i = ` and thus g and h are in the same conjugacy
class.
(b) Prove that g is conjugate to g −1 if and only if χ(g) ∈ R for every irreducible
complex character χ of G.
Solution: Recall we proved in class that χ(g −1 ) = χ(g) for all characters χ of G.
Thus, using part (a), we have that g is conjugate to g −1 if and only if χ(g) = χ(g)
for every irreducible character χ of G, which holds if and only if χ(g) ∈ R for
every irreducible character χ of G.
8. Let ρ be a C-representation of a finite group G and χ its associated character. Suppose
1 X
|χ(g)|2 = 3.
|G| g∈G
Prove that ρ is the direct sum of three distinct irreducible representations.
Solution: Let χ1 , . . . , χt be the irreducible complex characters ofPG. Since every
character is uniquely a sum of irreducible characters, we have χ = j mj χj for some
nonnegative integers m1 , . . . , mt . Recall that the χ1 , . . . , χt is form an orthonormal
basis for the C-vector space of class functions FC (G) on G under the inner product
hφ, ψi =
1 X
φ(g)ψ(g).
|G| g∈G
Thus, we have
1 X
|χ(g)|2
|G| g∈G
1 X
=
χ(g)χ(g)
|G| g∈G
3=
= hχ, χi
X
X
=h
mj χj ,
mj χj i
j
=
X
j
m2j .
j
Since the mj ’s are integers, the only possibility is that three of the mj ’s are one and
the rest are zero. Hence, χ is the sum of three distinct irreducible characters and ρ is
the sum of three non-isomorphic irreducible representations.