MATH 113 - WORKSHEET 3 SOLUTIONS (1) Let G be a group, and let a 2 G. Let λa be the function G ! G defined by λa(g) = ag. (a) Show that λa is a bijection G ! G. (b) Show that λa is not a homomorphism in general. (c) Is λa ever a homomorphism? λa is called \translation by a", or simply \left multiplication by a". Solution: (a) To show λa is a bijection, it suffices to find an inverse function. We'll show that −1 −1 λa−1 = (λa) . Let g 2 G. Then λa−1 (λa(g)) = a ag = g, and λa(λa−1 (g)) = aa−1g = g. (b) Note that λa(e) = ae = a. Since homomorphisms preserve the identity, λa is not a homomorphism whenever a 6= e. (c) On the other hand, if a = e, then λa(g) = eg = g, so λa is the identity homo- morphism. (2) For this problem, use the notation and result of problem 1(a). Show that there is a one-to-one homomorphism φ : G,! SG, where SG is the symmetric group on the set G, given by φ(a) = λa. This is called Cayley's theorem, and it shows that every group is isomorphic to a group of permutations (i.e. a subgroup of a permutation group). Solution: By problem 1(a), φ(a) = λa 2 SG, so φ is well-defined. φ is a homomorphism: Let a; b 2 G. Then φ(ab) = λab and φ(a)◦φ(b) = λa ◦λb. So we'd like to show that λab = λa ◦ λb. Indeed, for all g 2 G, λab(g) = abg = λa(λb(g)). φ is one-to-one: Let a; b 2 G with φ(a) = φ(b). Then λa = λb, so in particular a = λa(e) = λb(e) = b. (3) Let G be a group, and let a 2 G. Let τa be the function G ! G given by τa(g) = −1 aga . Show that τa is an automorphism of G. τa is called \conjugation by a". 0 0 −1 0 −1 Solution: τa is a homomorphism: Let g; g 2 G. Then τa(g)τa(g ) = (aga )(ag a ) = 0 −1 0 agg a = τa(gg ). −1 τa has an inverse: Indeed, τa−1 = (τa) . Let g 2 G. Then τa(τa−1 (g)) = −1 −1 −1 −1 −1 −1 −1 −1 −1 −1 a(a g(a ) )a = aa gaa = g, and τa−1 (τa(g)) = a (aga )(a ) = g. (4) For this problem, use the notation and result of problem 3. (a) Show that there is a homomorphism φ : G ! Aut(G), where Aut(G) is the group of automorphisms of G, given by φ(a) = τa. 1 (b) Show that this homomorphism is not one-to-one in general. An automorphism of the form τa is called an \inner automorphism", and range(φ) = fτa j a 2 Gg ≤ Aut(G) is called the \group of inner automorphisms of G". Solution: (a) By problem 3, φ(a) = τa 2 Aut(G), so φ is well-defined. Let a; b 2 G. Then φ(ab) = τab, and φ(a)φ(b) = τa ◦τb. So we'd like to show that −1 −1 −1 τab = τa ◦τb. Indeed, for all g 2 G, τab(g) = (ab)g(ab) = abgb a = τa(τb(g)). (b) For a counterexample, let G be any nontrivial abelian group. Then for any −1 −1 a 2 G, φ(a) = τa = idG, since for g 2 G, τa(g) = aga = aa g = g. So φ is the trivial homomorphism sending all elements of G to the identity of Aut(G). (5) In lecture, we showed that A3 is abelian and cyclic (isomorphic to Z3). Is A4 abelian? Solution: No. (123) = (13)(12) and (124) = (14)(12) are elements of A4 which do not commute: (123)(124) = (13)(24) and (124)(123) = (14)(23). A4 is a nonabelian group of order 4!=2 = 24=2 = 12 which we have not met before..