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Count the subgroups of order $25$ in $G=C_{75} \times C_{10}$

Is there a general way to tackle this kind of question? That is, find the number of subgroups of a direct product of cyclic groups?.

In this case, it's simple to count the number of elements of order $25$: we have $100$ elements of order $25$ in $G$. So the number of cyclic subgroups of $G$ is $100/ \varphi(25)=5$. But there could be other non cyclic subgroups, which I'm not sure how to count. Morover, it seems that a subgroup of a direct product of cyclic groups doesn't need to be a direct product of subgroups, which would be easier to count.

blundered_bishop
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  • Note that ${\rm Sym}(4)=S_4$ is not abelian and hence not cyclic. For subgroups of direct products see for example this post. – Dietrich Burde Nov 18 '21 at 10:27
  • @DietrichBurde I know that, but I imagined that $Sym(4)$ doesn't contribute to any subgroups of order $25$, so it is enough to count those of $C_{75} \times C_{10}$. I'm going through that post, thanks, but it seems more complicated than what it needs to be, at least in this case. – blundered_bishop Nov 18 '21 at 10:38
  • You could edit your post, deleting $S_4$ then, and searching again for duplicates for the new version, e.g., starting with this post. – Dietrich Burde Nov 18 '21 at 10:40
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    Note that $C_{25}\times C_{10}\cong C_{5^2}\times C_5\times C_6$. – Dietrich Burde Nov 18 '21 at 11:22

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