HW ALGEBRA 4
CLAY SHONKWILER
1 Prove that the number of non-isomorphic one-dimensional representations of a nite group G is |G|/|[G, G]|, where [G, G] denotes the commutator subgroup. Proof. Suppose : G C is a one-dimensional representation of G. Then we know that factors through the quotient G/[G, G], which is to say that, if : G G/[G, G] is the natural map, then there exists a homomorphism : G/[G, G] C such that = . In other words, each one-dimension representation of G corresponds to a one-dimensional representation of G/[G, G]. Furthermore, this correspondance is bijective. Surjectivity is clear; we see that this correspondance is injective since, if , are nonisomorphic one-dimensional representations of G, then there exists g G such that (g) (g); then, if (g) = g, (g) = (g). Now, since G/[G, G] is abelian, it has |G/[G, G]| = |G|/|[G, G]| conjugacy classes, meaning it has |G|/|[G, G]| irreducible representations, each of dimension 1. Since each corresponds to a one-dimensional representation of G, we see that the number of one-dimensional representations of G is |G|/|[G, G]|. 2 Let Wi be an irreducible representation of G, V an arbitrary representation, and Vi the direct sum of all subrepresentations of V isomorphic to Wi . Let Hi be the space of linear maps f : Wi V such that V (g)f = f (Wi )(g) for all g G. (a) Show that the dimension of Hi is equal to the number of times that Wi occurs in V . Proof. Suppose Hi . Then : Wi V . Now, V = Uj , so dene pj : V Uj to be the projection onto the jth coordinate. pj is a homomorphism, so j = pj : Wi Uj is a homomorphism of C[G]-modules as well. Now, by Schurs Lemma, if Wi Uj , then j is an isomorphism given by multiplication by a scalar;
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CLAY SHONKWILER
otherwise j is the trivial map. Hence, Hom(Wi , Uj ) is isomorphic either to C or to {0}. Now Hi = Hom(Wi , V ) = Hom(Wi , Uj ) = Hom(Wi , Uj ) = Wi
Uj C.
From this, then, we conclude that the dimension of Hi is equal to the number of times that Wi occurs in V . (b) Let G act on Hi Wi through the tensor product of the trivial representation on Hi and the given representation on Wi . Show that the map: F : Hi W i Vi dened by F( T w ) = T (w ) is an isomorphism of Hi Wi onto Vi . Proof. We must show that F commutes with the action of G on Hi Wi and on Vi . That is, we must show the following diagram commutes: Hi W i Vi |
Wi
F
=1Hi
Vi
Hi W i Vi .
F
If f Hi , w Wi , g G then (F (1H Wi )(g))(f w) = F (f Wi (g)(w)) = f Wi (g)(w) = V (g)f (w) = |Vi (g)f (w) = (|Vi F )(f w), by the properties of f , so the diagram above does indeed commmute. Now, suppose v Vi . Then each v lies in an isomorphic copy of Wi in V , there exists an isomorphism T and an element w Wi such that T (w ) = v . Hence, F( T w ) = T (w ) = v .
Therefore, F is surjective. Now, to see that F is injective, and therefore an isomorphism, we need only count dimensions. dim(Hi Wi ) = dim(Hi ) dim(Wi ). On the other hand, as we showed in part (a), Vi is the direct sum of dim(Hi ) copies of Wi , each of dimension dim(Wi ), so dim(Vi ) = dim(Hi ) dim(Wi ). Since F is a surjective linear map between vector spaces of the same dimension, must it be injective as well.
ALGEBRA HW 4
3
3 Let be the character of 2-dimensional representation of G, and let x be an element of order 2 in G. Prove that (x) = 2, 0, 2. Generalize this to n-dimensional representations. Proof. Let : B V be the 2-dimensional representation. Note that we can think of V as being simply C2 . Now, suppose (x) = a b c d .
Then 2 (x) = (x2 ) = (1) = Id2 . Also, note that (x) = Tr((x)) = a + d. Squaring the matrix for (x), then, we see that the following system of equations must hold: a2 + bc = 1 ab + bd = 0 ac + cd = 0 bc + d2 = 1. We note that the bottom equation implies bc = 1 d2 , which, replacing for bc in the top equation yields 1 = a2 + 1 d2 0 = a2 d2 = (a + d)(a d). If a + d = 0, then (x) = Tr((x)) = a + d = 0. On the other hand, if a d = 0, then a = d. Substituting into the second equation in the above system, we see that 0 = ab + ba = 2ab. If a = 0, then d = a = 0, so (x) = a + d = 0. If b = 0, then, from the rst equation, we see that a2 = 1 a = 1. Hence d = a = 1, so (x) = a + d = 2. Since weve covered all the possibilities, we see that, indeed, (x) = 2, 0, 2. 4 Let be an irreducible character of G. Prove that for every element g in Z(G), the center of G, we have (g) = (1), where is a root of unity. Proof. Let g Z(G). Let : G GL(V ) be the irreducible representation with character . Then, for any x G, (g) (x) = (gx) = (xg) = (x) (g). Since is irreducible and (g) is certainly a linear map from V to itself, we can use Schurs Lemma to conclude that (g) is simply a scalar multiple of the identity; (g) = Id. Hence (g) = Tr((g)) = Tr(Id) = Tr(Id) = (1).
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CLAY SHONKWILER
Now, we note that, since G is nite, there exists n N such that g n = 1. Hence Id = (g n ) = ((g))n = (Id)n = n Id, so n = 1. Therefore, is a root of unity. 5 Let be the character of some representation of G. Prove that (a) If (g) = (1) then g Ker(). Proof. Le...