Prove that if $f : G \rightarrow H$ is a group isomorphism, then for any element $x \in G$ one has $o(x) = o(f(x))$, where $o(a)$ denotes the order of the element $a$.
I feel that this proof should be online somewhere but haven't been able to find anything so would be helpful if anyone could point me in the right direction.
Here's a way through:
Suppose $f:G\to H$ is a homomorphism with $f(g)=h$
If the order of $g$ is $n$ so that $g^n=1_G$ (with $1_G$ the group identity in $G$) and the order of $h$ is $m$, do you know how to prove that $n$ is an integer multiple of $m$ ie $m|n$ or $n=km$ for some integer $k$? That is straightforward.
Then for an isomorphism, you have homomorphisms in both directions. Hence both $m|n$ and $n|m$ and for positive integers this implies $n=m$.