Symmetry And Group

## Finite Groups and their Representations by Baker A.J.

Posted On March 23, 2017 at 7:36 am by / Comments Off on Finite Groups and their Representations by Baker A.J.

By Baker A.J.

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The contents of this publication were used in classes given by way of the writer. the 1st was once a one-semester path for seniors on the college of British Columbia; it used to be transparent that solid undergraduates have been completely able to dealing with simple workforce conception and its program to easy quantum chemical difficulties.

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58 5. 10. 8. Then gHg −1 = {e, (12), (13), (23)} g∈S3 and N = {e, (123), (132)} is a Frobenius kernel. CHAPTER 6 Automorphisms and extensions In this chapter we will study how a finite group G with a normal subgroup N G is ‘built up’ from N and the quotient group Q = G/N . We will also study actions of one group on another by automorphisms. 1. Automorphisms Let G be a group. A homomorphism ϕ : G −→ G is called an endomorphism of G, and an automorphism if it is invertible. The set of all automorphisms of G is denoted Aut(G), and forms a group under composition of functions.

Vn }. Then the functions δx,j : X −→ V (x ∈ X, 1 j n) given by δx,j (y) = vj if y = x, 0 otherwise, for y ∈ X, form a basis for Map(X, V ). Hence, dim| Map(X, V ) = |X| dim| V. Proof. Let f : X −→ V . Then for any y ∈ X, n fj (y)vj , f (y) = j=1 where fj : X −→ k is a function. It suffices now to show that any function h : X −→ k has a unique expression as h= h x δx x∈X where hx ∈ k and δx : X −→ k is given by δx (y) = 1 if y = x, 0 otherwise. But for y ∈ X, h(y) = hx δx (y) x∈X ⇐⇒ h(y) = hy .

Let G = g0 ∼ = Z/n be cyclic of order n. Let ζn = e2πi/n , the ‘standard’ primitive nth root of unity. Then for each k = 0, 1, . . , (n − 1) we may define a 1-dimensional representation ρk : G −→ C× by ρk (g0r ) = ζnrk . The character of ρk is χk given by χk (g0r ) = ζnrk . Clearly these are all irreducible and non-isomorphic. Let us consider the orthogonality relations for these characters. We have (χk |χk ) = = = For 0 k< 1 n 1 n 1 n n−1 χk (g0r )χk (g0r ) r=0 n−1 ζnkr ζnkr r=0 n−1 1= r=0 n = 1.