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Suppose that $d \in \mathbb{N}$ satisfies the property that given any two Abelian groups $A_1, A_2$ of order $d$, $A_1 \cong A_2$. Prove that given any prime $p$, $p^2 \nmid d$. What can one say for the converse statement?

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up vote 3 down vote accepted

Suppose $d = p^2 n$, where $p \geq 2$ is prime.

Then there are non-isomorphic abelian groups $C_d$ and $C_{p} \times C_{pn}$.

These groups are clearly non-isomorphic, by the way, since the former is cyclic and the latter is not ($p$ and $pn$ share a factor of $p$).

What have you tried for the converse?

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