Do the isomorphism's of groups form an equivalence relation on the class of all groups? An isomorphism is simply a bijective homomorphism. How would one show that isomorphism's are symmetric, reflexive, and transitive?
 A: I think you are a little confused about what the relation is. The set is the class of all groups, and two groups $G_1$ and $G_2$ are isomorphic (denoted $G_1 \cong G_2$) if there exists an isomorphism $f:G_1 \to G_2$. You want to show that being isomorphic is an equivalence relation.


*

*Reflexivity: show that any group isomorphic to itself

*Symmetry: show that $G_1 \cong G_2$ implies $G_2 \cong G_1$.

*Transitivity: show that $G_1 \cong G_2$ and $G_2 \cong G_3$ together imply $G_1 \cong G_3$.

A: The relation being considered is: if $G,H$ are groups, then $G\sim H$ if there is an isomorphism from $G$ to $H$.
Symmetry
This amounts to showing that if there is an isomorphism $f: G\to H$, then there is also an isomorphism $g: H\to G$. To construct $g$, we note that since $f$ is bijective, we can consider the inverse function $f^{-1}$, which is also bijective. It remains to show that $f^{-1}$ is a homomorphism from $H$ to $G$.
Suppose we have two elements $h_1,h_2\in H$. Since $f$ is a surjection, we can write $h_1=f(g_1)$ and $h_2=f(g_2)$ for some elements $g_1,g_2\in G$. Since $f$ is a homomorphism, we then have that
$$f^{-1}(h_1h_2)=f^{-1}(f(g_1)f(g_2))=f^{-1}(f(g_1g_2))=g_1g_2=f^{-1}(h_1)f^{-1}(h_2)$$
so $g:=f^{-1}$ works perfectly well as an isomorphism from $H$ to $G$.
Reflexivity
We need to show that there is an isomorphism from any group to itself. This is easy: just take the identity mapping. It is very clearly bijective, and the homomorphism property is very easy to verify since elements are simply mapped to themselves.
Transitivity
This amounts to showing that if there are isomorphisms $f: G\to H$ and $g: H\to K$, then there is also an isomorphism $h: G\to K$. The easiest thing to try (which indeed works) is to take $h:= g\circ f$. If you already know that the composition of bijections is a bijection, then it is clear that this definition gives a bijection from $G$ to $K$. You just need to show that this definition of $h$ gives a homomorphism. This follows from the fact that $f$ and $g$ are homomorphisms:
$$h(g_1g_2)=g(f(g_1g_2))=g(f(g_1)f(g_2))=g(f(g_1))g(f(g_2))=h(g_1)h(g_2)$$
