MAT 150C, Spring 2017
Solutions to HW1
1.1. (25 points) Show that the image of a representation of dimension 1 of a finite
group is a cyclic group.
Solution: The image of a representation of dimension 1 is a finite subgroup of
GL(1, C) = C∗ . So we need to prove that every finite subgroup H of C∗ is cyclic. Indeed,
by Lagrange theorem the order of every element x ∈ H divides H, so x|H| = 1. The
equation x|H| = 1 has exactly |H| solutions, so H is a cyclic group with the generator
2πi
e |H| .
2.2. (25 points) Consider the standard two-dimensional representation of the dihedral
group Dn . For which n this is an irreducible complex representation?
Solution: The group Dn is generated by the rotation by angle 2π
and a reflection in
n
a line l passing through the origin. It is easy to see that for n = 1 and n = 2 this line l is
invariant under the rotation by π and by 2π, and hence it an invariant subspace for the
action of Dn , and the representation is reducible. Let us prove that it is irreducible for
n ≥ 3.
Indeed, the only invariant subspaces for the reflection in l are l and l⊥ . On the other
hand, for n ≥ 3 the matrix of rotation has no real eigenvectors, so neither l nor l⊥ are
invariant under rotation. Therefore there are no one-dimensional invariant subspaces, and
the representation of Dn is irreducible.
2.3. (25 points) Let V be a representation of S3 .
a) Show that V contains a nonzero invariant subspace of dimension at most 2.
Solution: Let v ∈ V be a nonzero vector. To save space, we will denote ρ(g)v by gv
for g ∈ S3 . Consider the vector
s = v + (1 2)v + (1 3)v + (2 3)v + (1 2 3)v + (1 3 2)v.
Clearly, gs = s for all g ∈ S3 . We have the following cases:
(1) If s 6= 0 then it spans a one-dimensional invariant subspace isomorphic to the
trivial representation of S3 .
(2) If s 6= 0, consider the vector u = v + (1 2 3)v + (1 3 2)v. We have
(1 2 3)u = (1 3 2)u = u, (1 2)u = (1 3)u = (2 3)u = (1 2)v + (1 3)v + (2 3)v = s − u = −u
If u 6= 0, then it spans a one-dimensional invariant subspace isomorphic to the sign
representation of S3 .
(3) If u = 0, consider the vector w = v +(1 2)v. If w 6= 0 then w+(1 2 3)w+(1 3 2)w =
s = 0, so w and (1 2 3)w span one- or two-dimensional representation of S3 .
(4) Finally, if w = u = s = 0 then (1 2)v = −v, so v and (1 2 3)v span one- or
two-dimensional representation of S3 .
b) Prove that all two-dimensional representations of S3 are isomorphic, and determine
all irreducible representations of S3 .
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Solution: If V is irreducible then it should coincide with the invariant subspace
constructed above. In (1) and (2) we identified is with the trivial and sign representation,
respectively. If in (3) we get a one-dimensional subspace, then
(1 2)w = w, (1 2 3)w = λw ⇒ λw = (1 2)(1 2 3)w = (2 3w) = (1 3 2)(1 2)w = λ2 w,
so λ = λ2 and λ = 1, so we get a trivial representation. If in (4) we get a one-dimensional
subspace, then (1 2)v = −v, (1 2 3)v = λv and similarly λ = 1, so we get a sign
representation.
Finally, we need to prove that two-dimensional representations in cases (3) and (4)
are isomorphic. This follows from the fact that both are isomorphic to the standard
representation of D3 , where we choose w to be parallel to the axis of one of the reflections
in (3), and we choose v to be perpendicular to this axis in (4).
A. (25 points) Consider the cyclic group G = hx|xn = 1i.
a) Describe all one-dimensional complex representations of G
Solution: A one-dimensional representation assigns to x a complex number z, and
2πim
to xk the number z k . Since xn = 1, we get z n = 1, so z = e n , m = 0, 1, . . . , n − 1.
Therefore G has exactly n different one-dimensional representations:
ρm (xk ) = e
2πimk
n
, m = 0, 1, . . . , n − 1.
b) Prove that every complex representation of G has a one-dimensional invariant subspace.
Solution: In a complex representation x corresponds to a complex matrix ρ(x). It
has an eigenvector v such that ρ(g) · v = λv for some λ. Then ρ(g k )v = ρ(g)k v = λk v, so
the one-dimensional subspaces spanned by v is invariant under the action of ρ(G).
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