I heard this statement in group isomorphism:
A group $G$ is isomorphic to $G'$ if $\exists\ \phi : G \to G'$ such that $\forall\ g \in G, |g|=|\phi(g)|$.
I don't have the proof of this property and I am not sure if it is correct. I think the property may be incorrect for groups of infinite order but I am not sure. For finite groups I think that saying that the order of $g$ is equal to the order of $\phi(g)$ is not enough. I think we also need to say that order of $G$ is equal to order of $G'$. Please help me understand this property and prove it if it is valid.
Take $I:G\rightarrow G\times G$ defined by $I(g)=(g,e)$. It is not an isomorphism, but $g$ and $I(g)$ have the same order.
But it is true if you assume that $\phi$ is also onto. In this case, let $x\in Ker(\phi)$, $\phi(x)=e$, implies that the order of $x$ is $1$, thus $x=e$. This implies that $\phi$ is injective, since it is supposed surjective, it is bijective and hence an isomorphism.