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I'm trying to solve the following problem from a past exam.

Find the number of the subgroups of $P:=\Bbb Z_4 \times \Bbb Z_6 \times \Bbb Z_9$ of index 3.

Here $\Bbb Z_m$ denotes $\Bbb Z/m\Bbb Z$.

I tried to solve it by using this characterization of subgroups of a direct product from an answer to my previous question. $P \cong \Bbb Z_6 \times \Bbb Z_{36}$, by CRT. Since the subgroup defined by $(G_1, G_2, H_1, H_2, \phi)$ in the characterization has order $|G_1||H_2|$, we have $(|G_1|,|H_2|) = (2, 36), (6,12)$. In the first case, $H_1 = H_2 = \Bbb Z_{36}$ and $G_1 = G_2 \cong C_2$, so we have just one subgroup of $P$ corresponding to it. In the second case, we have $H_2 \cong C_{12}$ and $G_1 = \Bbb Z_6$. We have the following possibilities: $H_1/H_2 \cong C_1, C_2, C_3, C_6$ and $G_2\cong C_6, C_3, C_2, C_1$, respectively. (We can't have $H_1/H_2\cong C_4$ because 4 doesn't divide the order of $\Bbb Z_6$.) The number of the isomorphism corresponding to each case is 1, 1, 2 and 2, resp. In conclusion we have 7 subgroups of $P$.

I have the following questions now:

  1. Is my solution above correct?
  2. Is there any better way to solve it? The characterization I used is a powerful one, so I feel it's a kind of cheat to use it to solve this (seemingly) elementary problem. I am looking for an elementary solution to it.

I would be grateful for your help.

ADDENDUM: In the second case, $H_1/H_2$ cannot be isomorphic to $C_2, C_6$. So the correct answer should be 4.

Community
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Pteromys
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1 Answers1

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If $|P:H|=3$, then $x^3 \in H$ for all $x \in G$, so $K < H$, where $K = \{x^3 : x \in P \}$. Now $K \cong {\mathbb Z}_2 \times {\mathbb Z}_{12}$ and $H/K \cong {\mathbb Z}_3 \times {\mathbb Z}_3$, so the problem is equivalent to finding the number of subgroups of index 3 (or equivalently, of order $3$) in ${\mathbb Z}_3 \times {\mathbb Z}_3$. It is easy to see that there are $4$ such subgroups. I have not tried to figure out where you went wrong with your estimate of $7$.

Derek Holt
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  • I checked my solution and I now believe there are indeed 4 subgroups. I would be grateful if you clarify what the symbols $H, K, G$ signify. – Pteromys Sep 11 '13 at 23:43