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I have this problem:

Let $G$ be an abelian group of order $72$. Show that $G$ has exactly one subgroup of order $8$.

I've seen how to find all abelian groups (up to isomorphism) of order $n$, and I know that if $m$ divides the order of a finite abelian group $G$, then $G$ has a subgroup of order $m$.

I also know that $72 = 2^33^2$. However, I feel like I'm missing something to put it all together. Can someone give me a hint or tell me the theorem I need to apply here?

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  • $\begingroup$ Do you know the Sylow-Theorems? $\endgroup$ – user114885 Feb 7 '14 at 15:19
  • $\begingroup$ @user114885 Not yet. $\endgroup$ – Newb Feb 7 '14 at 15:23
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Suppose there are two subgroup $H, K$ of order 8. Since $G$ is abelian, $HK=KH$. and $HK$ is a subgroup of $G$. But now $|HK|=8^2/|H \cap K|=8\cdot m$, where $m=2, 4, 8$. But this contradicts to Lagrange theorem.

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