Suppose that $G_1$ has $m_1$ number of subgroups and $G_2$ has $m_2$ number of subgroups. Can we find any closed formula on the number of subgroups of $G_1\times G_2$ ? Here $G_1\times G_2$ is the external direct product of the groups $G_1, G_2$.

I tried to start with finite abelian groups $G_1, G_2$. But in vain. I mean manually step by step it is possible for me to derive the answer but I am unable to derive a closed formula. I know fundamental theory of finite abelian groups and I know how to use it to find step by step the ans. But in case, if I want to find a closed formula how shall I finish the task ?

Any help will be appreciated.

  • 2
    $\begingroup$ This short article should illustrate how hard it is to accomplish that. $\endgroup$ – user1337 Jun 18 '15 at 15:41
  • 3
    $\begingroup$ You cannot compute the result from $m_1$and $m_2$ alone: $\mathbb Z/2\mathbb Z$ and $\mathbb Z/3\mathbb Z$ have exactly two subgrops each, but $\mathbb Z/2\mathbb Z\times\mathbb Z/2\mathbb Z$, $\mathbb Z/2\mathbb Z\times\mathbb Z/3\mathbb Z$, $\mathbb Z/3\mathbb Z\times\mathbb Z/3\mathbb Z$ have different numbers of subgroups ($5$, $4$, $6$). $\endgroup$ – Hagen von Eitzen Jun 18 '15 at 15:41

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