# If both finite groups $G_1$ and $G_2$ (of equal order) possess equal number of elements of highest possible order, will $G_1\simeq G_2$ hold?

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 $$G_1\simeq G_2$$ ?

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 $$8$$.