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Let $A=\mathbb{Z}_{360}\oplus\mathbb{Z}_{150}\oplus\mathbb{Z}_{75}\oplus\mathbb{Z}_{3}$. I need to calculate the number of subgroups of $A$ which is isomorphic to $\mathbb{Z}_{5}\oplus\mathbb{Z}_{5}$

I know that every such subgroup is generated by $a,b$ which satisfies $ord\left(a\right)=ord\left(b\right)=5,a\notin\left\langle b\right\rangle$, and I know how to count them. But the same subgroup can be generated by many pairs - my problem is to count the number of pairs which generated the same subgroup.

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Sketch of the solution:

First count the number of the elements of order $5$. Let say this is $n$.

Then $\dfrac{n}{\phi(5)}=\dfrac{n}{4}=k$ is the number of the subgroup of order $5$.

Note that every $Z_p\times Z_p$ has $p+1$ subgroup of order $p$.

Thus, $\dfrac{k}{5+1}$ is the number of the subgroup which is ismorphic to $Z_5\times Z_5$.

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