Section 1.7. Group Actions 61 Proof. For each a G, define “translation” τa : G G by τa(x) = ax for every x G (if a = 1, then τa is not a homomorphism). For a, b G, (τaτb)(x) = τa(τb(x)) = τa(bx) = a(bx) = (ab)x, by associativity, so that τaτb = τab. It follows that each τa is a bijection, for its inverse is τa−1 : τaτa−1 = τaa−1 = τ1 = 1G = τa−1a, and so τa SG. Define ϕ : G SG by ϕ(a) = τa. Rewriting, ϕ(a)ϕ(b) = τaτb = τab = ϕ(ab), so that ϕ is a homomorphism. Finally, ϕ is an injection. If ϕ(a) = ϕ(b), then τa = τb, and hence τa(x) = τb(x) for all x G in particular, when x = 1, this gives a = b, as desired. The last statement follows from Exercise 1.46 on page 44, which says that if X is a set with |X| = n, then SX = Sn. The reader may note, in the proof of Cayley’s Theorem, that the permutation τa : x ax is just the ath row of the multiplication table of G. To tell the truth, Cayley’s Theorem itself is only mildly interesting, but a generalization having the identical proof is more useful. Theorem 1.96 (Representation on Cosets). If H is a subgroup of finite index n in a group G, then there exists a homomorphism ϕ: G Sn with ker ϕ H. Proof. We denote the family of all the left cosets of H in G by G/H, even though H may not be a normal subgroup. For each a G, define “translation” τa : G/H G/H by τa(xH) = axH for every x G. For a, b G, (τaτb)(xH) = τa(τb(xH)) = τa(bxH) = a(bxH) = (ab)xH, by associativity, so that τaτb = τab. It follows that each τa is a bijection, for its inverse is τa−1 : τaτa−1 = τaa−1 = τ1 = 1G/H = τa−1 τa, and so τa SG/H . Define ϕ : G SG/H by ϕ(a) = τa. Rewriting, ϕ(a)ϕ(b) = τaτb = τab = ϕ(ab), so that ϕ is a homomorphism. Finally, if a ker ϕ, then ϕ(a) = 1G/H , so that τa(xH) = xH for all x G in particular, when x = 1, this gives aH = H, and a H, by Lemma 1.46(i). The result follows from Exercise 1.46 on page 44, for |G/H| = n, and so SG/H = Sn. When H = {1}, this is the Cayley Theorem, for then ker ϕ = {1} and ϕ is an injection. We are now going to classify all groups of order up to 7. By Example 1.59, every group of prime order p is isomorphic to Ip, and so, up to isomorphism, there
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