#6) Prove that isomorphism is an equivalence relation. That is for any groups G,
ID: 1892756 • Letter: #
Question
#6) Prove that isomorphism is an equivalence relation. That is for any groups G, H, and K, (G is isomorphic to G), (G is isomorphic to H) implies (H is isomorphic to G), and (G is isomorphic to H) and (H is isomorphic to K) implies (G is isomorphic to K).Explanation / Answer
a) 1) Reflexivity: Any group is isomorphic to itself via the identity map from G to G. (This is clear!) 2) Symmetry: Suppose that G is isomorphic to H. Then, there exists f:G --> H, that is a bijection and a homomorphism. Claim: f^(-1) from H to G is also an isomorphism. Since f is a bijection, so is f^(-1). To show f^(-1)(hh') = f^(-1)(h) f^(-1)(h): Since f^(-1) is a bijection, there exist g,g' in G such that f(g) = h and f(g') = h'. (i.e., f^(-1)(h) = g and f^(-1)(h') = g'.) On one hand, f(gg') = f(g)f(g') = hh' ==> f^(-1)(hh') = gg'. On the other hand, f^(-1)(h) f^(-1)(h') = g'g'. Thus, f^(-1)(hh') = f^(-1)(h) f^(-1)(h), as required. Therefore, H is isomorphic to G. 3) Suppose that G is isomorphic to H and H is isomorphic to K. Then, we have isomorphism maps f: G--> H and g:H-->K Claim: (g composed f): G --> K is also an isomorphism. The composition of two bijections is also a bijection and the homomorphism condition follows from that of g and h: (I'll write the composition as g*f.) For any h,h' in G, (g*f)(hh') = g(f(hh')) = g(f(h) f(h')), f is a homom. = g(f(h)) g(f(h')), g is a homom. = (g*f)(h) (g*f)(h'). Hence, G is isomorphic to K. Thus, isomorphism of groups is an equivalence relation. (b) This is false; |S_4| = 24, while |D_4| = 8. On the other hand, this is true if you replace the 4 with a 3. This map can be given explicitly element-wise, because there are only six elements. For example, define f: S_3 --> D_3 via f(123) = r and f(12) = f, and extend multiplicatively. I hope that helps!
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