Math 503: Abstract Algebra

Homework 4
Lawrence Tyler Rush
<me@tylerlogic.com> In Collaboration With
Keaton Naff
February 22, 2014
http://coursework.tylerlogic.com/courses/upenn/math503/homework04

1

Let G be a finite group and S be a finite set on which G acts, and the action is given by μ : G × S → S. Let V be the space of ℂ valued functions on S. Let ρ : G → GL(V ) be the linear representation of G defined by
(ρg(f))(s) = f (μ(g-1,s)) ∀g ∈ G, s ∈ S,f ∈ ℂS

Let χρ be the character of (V,ρ).

Let the elements of S be {s1,⋅⋅⋅,sm}, and denote by B = {fsi}⊂ V the basis of functions such that fsi(sj) = δij for each sj ∈ S.

(a) Prove that (χρ|1G) is the number of conjugacey class in G

Fix an s ∈ S, and define f to be the vector of V by
f = ∑ f s∈Gs si i

Then for any g ∈ G and sj ∈ S

{ ∑ ∑ ( -1 ) 1 μ(g- 1,sj) ∈ Gs ρg(f)(sj) = s ∈Gsρg(fsi) = s∈Gsfsi μ(g ,sj) = 0 otherwise i i

However, if g-1sj ∈ Gs, then there exists a g′∈ G with g-1sj = g′s, which implies sj = gg′s, meaning that sj is also in Gs. Hence the above equation boils down to

{ 1 sj ∈ Gs ∑ ρg(f)(sj) = 0 otherwise = fsi(sj) = f(sj) si∈Gs

Thus the 1-dimensional subspace Ws = span{f} is G-stable. Since there is one such Ws corresponding to each conjugacey class Gs, then (χρ|1G) is the number of conjugacey classes. 1

(b)

Now let G operate transitively on S, and define U ⊂ V be the set of functions f for which ∑ s∈Sf(s) = 0. Then for any f ∈ U, f = a1fs1 + ⋅⋅⋅ + amfsm
∑ ∑ f(s) = a1fs1(s)+ ⋅⋅⋅+ amfsm(s) = a1 + ⋅⋅⋅+ am = 0 s∈S s∈S

implying that a1 + ⋅⋅⋅ + am-1 = -am, which further implies that U has dimension dimV - 1.

Now because G is transitive, then for any g ∈ G

∑ ∑ ∑ ρg(f)(s) = f(μ(g-1,s)) = f(s) = 0 s∈S s∈S s∈S

implying that ρg(f) ∈ U, i.e. U is G-stable. Therefore V can be decomposed into V = U ⊕W for some G-stable subspace W, by Maschke’s theorem. However, since U has dimension dimV - 1, then W has dimension 1, and since G is acts transitively on S, then (χρ|1G) = 1, implying that W must be the single trivial representation in the decomposition of V . Thus, denoting the character of U by χU,

χU = χρ - 1

(c)

(d)

2

(a)

(b)

(c)

(d) Extra Credit

3

(a)

(b)

(c)

4

(a)

(b) Extra Credit

References

[Sho]   Clayton Shonkwiler. Algebra hw 2. http://www.math.uga.edu/~clayton/courses/503/503_3.pdf.