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Let both $G_1, G_2$ are finite Abelian groups of equal order. Let $n_d(G)$ be the total number of elements of order $d$ in $G$.

Earlier in the following post of MSE

If two finite Abelian groups posses equal number of elements of some particular order, then justify they are isomorphic. it was discussed if $n_d(G_1)=n_d(G_2)$ holds for some particular positive divisor $d>1$ of $|G_1|$, we cannot conclude the groups are isomorphic. The suitable example is $\mathbb{Z}_{12}, \mathbb{Z}_2\times \mathbb{Z}_6$ because $n_3(\mathbb{Z}_{12})=2$, $n_3(\mathbb{Z}_2\times \mathbb{Z}_6)=2$ but the groups are not isomorphic.

However, I am willing to know, if we restrict ourselves $d$ to be the highest possible order, will then the isomorphism still work ? For example, if both $G_1, G_2$ have highest possible order as 16 and $n_{16}(G_1)=n_{16}(G_2)$, can we establish $G_1\simeq G_2$ ?

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No, e.g. $\Bbb Z_8\oplus\Bbb Z_4$ and $\Bbb Z_8\oplus\Bbb Z_2\oplus\Bbb Z_2$ have an equal number of elements of order $8$.

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  • $\begingroup$ Beautiful! And again I have to search for alternatives :-D But anyway, thanks $\endgroup$
    – KON3
    Commented Jul 30, 2022 at 19:05

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