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How many subgroups does $H = \mathbb{Z}_6 \times\mathbb{Z}_6 \times\mathbb{Z}_6 \times\mathbb{Z}_6 $ have?

$\mathbb{Z}_6$ has 4 subgroups (including itself), so the answer is at least $4^4$. But, not all subgroups of $H$ are products of subgroups of $\mathbb{Z}_6$, for example the group generated by $(1,1,1,1)$ is not. Any ideas how to count this type of groups in a simple way?

Andrey S
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    Are you counting $\mathbb{Z}_6\times \mathbb{Z}_1\times\mathbb{Z}_1\times\mathbb{Z}_1$ (where $\mathbb{Z}_1$ is a convenient notation for the trivial grounp) as distinct from $\mathbb{Z}_1\times\mathbb{Z}_1\times\mathbb{Z}_1\times\mathbb{Z}_6$, for instance? (i.e., do you want the number of isomorphism classes of subgroups or the number of distinctly embedded subgroups?) – Steven Stadnicki Sep 30 '13 at 04:19
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    @steven Stadnicki: Yes he count different, that why the number $4^4$ has come into picture – D. N. Sep 30 '13 at 04:24
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    Hint: $Z_6 = Z_2\times Z_3$ implies $H = Z_2^4\times Z_3^4$. The subgroups of $Z_2^4$ and $Z_3^4$ you should be able to determine. As $Z_2^4$ and $Z_3^4$ have coprime order, what can you say for $U\le H$ about $(U\cap Z_2^4)\times (U\cap Z_3^4)$? – j.p. Sep 30 '13 at 06:38
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    If $H$ is a subgroup of the group $H_1\times H_2\times \dots \times H_n$, then $p_i(H)$ form a subgroup of $H_i$, where $p_i$ is the projection map from $H$ to $H_i$. – D. N. Sep 30 '13 at 07:26
  • Thanks, using your hint and math.stackexchange.com/q/142589 I've got 212x67 – Andrey S Sep 30 '13 at 15:16

1 Answers1

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Let $G=G_1\times G_2$ be a direct product of finite groups, $|G_1|=n_1$, $|G_2|=n_2$ and $(n_1,n_2)=1$. Then for every subgroup $H$ of $G$ there exists $H_i$ subgroup of $G_i$ such that $H=H_1\times H_2$.

Since $H\le G$ we have $|H|=k_1k_2$ with $k_i\mid n_i$. Let $H_i=p_i(H)$, where $p_i:G\to G_i$ is the canonical projection. Obviously $H\subseteq H_1\times H_2$. But $|H_i|\mid n_i$ and $|H_i|\mid |H|=k_1k_2$. It follows that $|H_i|\mid k_i$, so $|H_1\times H_2|\le k_1k_2$. On the other side, $|H_1\times H_2|\ge |H|=k_1k_2$, therefore they are equal.

In your case $G_1=\mathbb Z_2^4$ and $G_2=\mathbb Z_3^4$. In general, the number of subgroups of $\mathbb Z_p^n$ ($p$ prime) equals the number of $\mathbb Z_p$-vector subspaces of $\mathbb Z_p^n$ and this is well known.

  • Where do you need that $G_1$ and $G_2$ are abelian in your proof? – j.p. Oct 01 '13 at 09:47
  • @j.p. The conclusion actually needs even less restrictive conditions in the general case. All subgroups being direct products like this is equivalent to the two groups $G_1$ and $G_2$ not having any common subquotients (which is then equivalent to their orders being coprime when they are abelian). – Tobias Kildetoft Oct 01 '13 at 10:44
  • @TobiasKildetoft: I just wanted to get rid off of the unnecessary word "abelian". – j.p. Oct 01 '13 at 14:52