1. The problem statement, all variables and given/known data Prove that isomorphism is an equivalence relation on groups. 2. Relevant equations Need to prove reflexivity, symmetry, and transitivity for equivalence relationship to be upheld. **We will use ≅ to define isomorphic to** 3. The attempt at a solution Let G, H, and K be groups. We want to show that G≅G, G≅H, and if G≅H H≅K then G≅K. Reflexivity: Define f:G->G by f(x)=x. we show f is an isomorphism. For x,y ∈ G if f(x) = f(y), then by definition of f we see that x=y. f is thus one to one. f(x)=x so f is onto. f(xy) = xy = f(x)f(y). There f is an isomorphism and G≅G. Symmetric: We assume G≅H. Then there is an isomorphism f:G->H, that is a bijection and homomorphism. Claim: f^(-1) from H to G is an isomorphism. Since f is a bijection, so is f^(-1). let h, h' ∈H and g,g' ∈G. f^(-1)(h*h') = f^(-1)(h)f^(-1)(h'). Since f^(-1) is a bijection, there exists g,g'∈G such that f(g) = h and f(g') = h'. f(gg') = f(g)f(g') = hh' => f^(-1)(hh') = gg'. f^(-1)(h)f^(-1)(h') = gg'. Thus, f^(-1)(hh') = f^(-1)(h)f^(-1)(h') as required. Therefore H≅G. Transitive: Assume G≅H and H≅K, then there are isomorphisms g:G->H and f->K. Claim: (g°f):G->K is an isomorphism. The composition of two bijections is also a bijection. For any h,h' ∈G. (g°f)(hh') = g(f(hh')) = g(f(h)f(h')) = g(f(h))g(f(h')) = (g°f)(h)(g°f)(h') Hence, G≅K. Thus, isomorphisms of groups is an Equivalence relationship. Question: I am most concerned about the symmetric part. Did I do this part right? How can I arbitrarily define isomorphisms like f:G->G by f(x)=x? Thank you.