Two abelian groups $|G| =n $ and $|H|=n$ are isomorphic if and only if they have same number of elements of order $m$,$1\le m\le n$

Question

Claim : Two abelian groups $|G| =n $ and $|H|=n$ are isomorphic if and only if they have same number of elements of order $m$,$1\le m\le n$

Proof : Let $G$ and $H$ are isomorphic but there exists an element an element $g \in G$ whose order is $k$ but there is no element in $H$ whose order is $k$. So $g^1 \neq g^2\neq,\cdots,\neq g^{k-1} \neq e$. Now as there is an isomorphism between $G$ and $H$ lets say $\phi$. Assume that $\phi(g) = h\in H$. Now if $h^k$ should be equal to identity. Otherwise $\phi$ is not an isomorphism.

I don't this proof is correct or wrong. Please spot errors so that I can fill those gaps.

Answer 1

Use the structure theorem for a Abelian groups, write Both groups as direct product of cyclic groups, if their decomposition is not same then number of elements of given order would be different.