Prove that Aut(Z4) ? Z2. Solution k, suppose you have a group, N. we can form it

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Prove that Aut(Z4) ? Z2.

Explanation / Answer

k, suppose you have a group, N. we can form its automorphism group, Aut(N). now let's say we have another group H, and a homomorphism f:H-->Aut(N). this means for every h in H, we have an automorphism of N, f_h. we can use this to create a new group out of N and H, N x| H. the elements of N x| H are just pairs (n,h) with n in N, h in H. so it is the same SET as the direct product, N x H. what's different is the multiplication: (n1,h1)*(n2,h2) = (n1f_h1(n2), h1h2). in other words, we let h1 act on n2 via the automorphism f_h1. now if the homomorphism f:H-->Aut(N) is trivial, that is, if f(h) is always the identity map on N,f_h(n) = n, for all n in N, and every h, the semi-direct prodict IS the direct product. so the semi-direct product is "more general". it sometimes helps to see an example. suppose we have the cyclic group, Z4. Aut(Z4) has two elements: x-->x, and x-->x^3 (= x^-1). so we clearly have a homomorphism f:Z2-->Aut(Z4) (in fact, it is an isomorphism) given by f(0) = x-->x (the identity map) f(1) = x--> x^-1 (the inversion map). so what does Z4 x| Z2 look like in this case? well, it is of order 8, since Z4 x Z2 has eight pairs. now (a,0)(a',b') = (a + a', b') since for b = 0, we have the identity automorphism of Z2. while (a,1) (a',b') = (a - a', b') since for b = 0, we have the inversion automorphism of Z2. note that (a,0)(a',0) = (a + a', 0), so {(0,0) (1,0), (2,0), (3,0)} is a cyclic subgroup of order 4. also (0,b)(0,b') = (0, b+b'), since for either automorphism of Z2, f(0) = 0. thus {(0,0), (0,1)} is a subgroup of order 2. what is (1,0)(0,1)? it's (1+0, 0+1) = (1,1). what is (0,1)(1,0)? it's (0-1, 1+0) = (-1,1) = (3,1). so Z4 x| Z2 is not abelian. note that (3,0)(0,1) = (3+0, 0+1) = (3,1). so if we set r = (1,0), and s = (0,1), then r^4 = s^2 = e = (0,0), and sr = (0,1)(1,0) = (3,1) = r^3s, we see that Z4 x| Z2 is isomorphic to D4, the dihedral group of order 8. (Z4 x| Z2 is a "twisted product" of Z4 and Z2, rotations are "normal" elements (a,0), while reflections are "twisted elements" (a,1) that "flip" the sign of the first coordinate in the second term in a product when we multiply. note that (a,1)(a,1) = (a - a, 1+1) = (0,0), these are all of order 2. geometrically, these means rotations can be done without leaving the space, but reflections require an "out-of-space flip") note that we have the following series of homomorphisms: Z4--> Z4 x| Z2 ---> Z2 where a --> (a,0) and (a,b) --> b, where the kernel of the second homomorphism is Z4.

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