Ma5c HW 7, Spring 2016
Problem 1. Prove that degree one representations of G are in bijective correspondence with degree 1 representations
of the abelian group G/G0 (where G0 is the commutator subgroup of G).
Proof. Let ρ be a 1-dimensional representation of G. Thus ρ is a group homomorphism G → F× for some field F.
Note that the latter is abelian. Hence, if a, b ∈ G are any two elements, we have ρ([a, b]) = ρ(a)ρ(b)ρ(a)−1 ρ(b)−1 =
1. Thus the commutator subgroup G0 is in the kernel of ρ, and we thus have an induced representation of
G/G0 . Conversely, any such representation ρ0 gives a representation ρ of G: given g ∈ G, define ρ(g) := ρ(g
(mod G)0 ).
Problem 2. Let F = Fpn and G a finite p-group. Show that any irreducible representation of G over F is trivial.
Proof. Let V be such an irreducible representation. Let G act on the nonzero vectors in V (there’s pdim V − 1 of
them). All G-orbits have a power of p size (since they divide the order of G) and add up to pdim V − 1. This
means that some orbits must have size 1, i.e. some nonzero vector is fixed by all of G. This vector generates
an invariant 1-dimensional subspace of V , contradicting the irreducibility (unless dim V = 1, and a generating
vector is fixed by all of G, i.e. V is the trivial representation).
Problem 3. Show that the trace ideal and augmentation ideal (see Page 846, example 11) are in fact 2-sided ideals of
FG.
Proof. We recall that the trace ideal is
N=
nX
o
αg g : αg = αh , g, h ∈ G
Let
x=
X
βg g
Then, if y ∈ N , then
X βg g
αh
X X
=
α
βgh−1 g
xy =
X
h
g
X
=(
βg )g.
Similarly,
X βg g
αh
X X
=
α
βg−1 h gh
yx =
X
h
g
X
=(
βg )h
and similarly for I.
Problem 4. Show that GL2 (R) has no subgroup isomorphic to Q8 . [This may be done by direct computation using the
generators and relations for Q8. Simplify these relations by putting one generator in canonical form.]
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Ma5c HW 7, Spring 2016
4
4
2
2
−1
Proof. Consider the presentation hi,
 j|i =j = 1, i = j = −1, ji = ij i. Sending i and j to ±1 gives 4
i 0
, j 7→ 0−110 gives an irreducible 2 dimensional representation over
1-dimensional representations. i 7→ 
0 −i
C.
This representation cannot be realized over R, hence GL2 (R) has no subgroup isomorphic to Q8 .
Details:
Suppose v is an eigenvalue of ϕ(i), then ϕ(i)4 v = λ4 v = v, hence, the eigenvalues of v satisfy (λ2 +1)(λ2 −1) =
0. This tells us the trace is 0 and the determinant is ±1, giving a canonical form of ϕ(i) by


x
y


ϕ(i) =  ±1 − x2

−x
y
which takes the canonical form.


−1
 = A± .
0
0
ϕ(i) = 
±1
Let us treat these two cases separately. If A = A− and φ(j) = B, we determine
AB + BA = 0,
A2 − B 2 ,
BAB − A = 0.


a b
, we find from the first equation that d = −a and c = −b, leaving two simultaneous
By letting B = 
c d
equations in the non-zero entries of the second two equations
a2 − b2 − 1 = 0,
a 2 − b2 + 1 = 0
which are impossible to solve. Suppose A = A+ then the same set of equations gives d = −a and c = b, however,
the last equation to solve is a2 + b2 + 1 = 0, which is impossible over the reals.
Problem 5. Let G be a finite group and let V be the vector space of functions from G to C. For g ∈ G and f ∈ V let
R(g)(f ) be the function
(R(g)(f ))(x) = f (xg).
Show that this defines a representation of G on V .
Proof. The identity g = 1 fixes V pointwise: (R(id)(f ))(x) = f (x · id) = f (x). If g, h ∈ G, then (R(gh)(f ))(x) =
f (xgh) = (R(h)(f ))(xg) = (R(g)(R(h))(f )(x). Thus R defines a representation of G on V .
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