Problem : Let $S$ be the relation G~H iff G is isomorphic to H.
Show reflexive, transitivity and symmetric.
First show G is automorphism, which will imply G~G. So the identity mapping gives us this automorphism. So G~G. My question is, is this the right process for showing reflexiveness?
Now we show symmetry. Assume G is isomorphic to H. Thus there exists a mapping $\phi:G ->H$. We must show H is isomorphic to G. Since $\phi$ is this mapping, consider $\phi^{-1}.$ This is a bijection since $\phi$ is a bijection. How do I show it is homomorphic?
Finally show its transitive. This I have done in a previous problem.
This is different than the duplicate as suggested because previously I had done it incorrectly. I know now I must do it this way.
If $\phi$ is an isomorphism, we have $\phi(a)\phi(b)=\phi(ab)$. Therefore, taking $a=\phi^{-1}(u)$ and $b=\phi^{-1}(v)$ for any $u$ and $v$, we deduce that $uv=\phi(\phi^{-1}(u)\phi^{-1}(v))$, or $\phi^{-1}(uv)=\phi^{-1}(u)\phi^{-1}(v)$. In other words, $\phi^{-1}$ is an isomorhism.