Isomorphism between Abelian groups

Question

Let G, H be abelian groups. Prove that if $\phi$ : G $\rightarrow$ H is an isomorphism, then|g| = |$\phi$(g)|for all g in G, that is, the order of g is the same as the order of the image of g under $\phi$ .

Answer 1

Like Kyle said in the comment, if you have a question, you need to tell us what you've tried and where you're stuck.

I'll give you a hint.

if $\phi(x)$ is an isomorphims, then $\phi(xy) = \phi(x)\phi(y)$. Also $\phi(e_{G}) = e_{H}$