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Let $A$ be an Abelian group of prime power order. It can be expressed as a (unique) direct product of cyclic groups of prime power order:

$A = \mathbb{Z}_{p^{n_1}} \times \cdots \times \mathbb{Z}_{p^{n_t}}$

where $p$ is a prime, and $n_1, \ldots, n_t $ are positive integers.

Let $B$ be a subgroup of $A$.

Question Is it always the case that:

$B \cong \mathbb{Z}_{p^{m_1}} \times \cdots \times \mathbb{Z}_{p^{m_t}} \ \ \ \ \ \ \ \ \ \ $ [Equation 1]

where $m_i \le n_i$ for $i=1, \ldots, t$ ?

Can anyone provide a counterexample, a proof, or a reference to a proof? It looks 'obviously' true, but I've spent days trying to construct a proof (based on the orders of the elements), but to no avail.

Note Equation 1 contains an isomorphism sign, not an equals sign. So, for example, $A = \mathbb{Z}_2 \times \mathbb{Z}_2$ and $B=\{ (0,0), (1,1) \}$ is not a counterexample, because $B \cong \mathbb{Z}_2 \times \{0\}$, and so we could take $n_1 = n_2 = m_1 = 2$ and $m_2 = 0$.

azimut
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Michael Pounds
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  • As noted by Michael Pounds, this question is exactly the first part of my question http://math.stackexchange.com/questions/528404, which additionally asks for quotient groups of finite abelian $p$-groups. – azimut Apr 09 '14 at 14:18

1 Answers1

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I belive this is true: we can assume $n_1 \ge n_2 \ge \cdots \ge n_t$ ad $m_1 \ge m_2 \ge \cdots \ge m_t$. Assume that $m_j \le n_j$ for $j=1, \cdots, r$ but $m_{r+1}>n_{r+1}$. Let $G^k$ donote that $\{ g^{p^k}|g \in G\}$, just for the simplicity of symbols. Let $l=m_{r+1}-1$. Since $B \le A$, then $B^l \le A^l$. But $A^l=(Z_{p^{n_1}})^l \times \cdots \times (Z_{p^{n_r}})^l$ and $B^l=(Z_{p^{m_1}})^l \times \cdots \times (Z_{p^{m_{r+1}}})^l \times \cdots$. Now $B^l$ has more elements of order $p$ than that of $A^l$. So we get a contradiction.

Michael Pounds
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Wei Zhou
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  • Thank you. How to prove that $A^l$ and $B^l$ are groups? – Michael Pounds Apr 08 '14 at 15:13
  • This is abelian group. So it is groups. – Wei Zhou Apr 08 '14 at 16:09
  • I understand that $A$ and $B$ are groups. But it is not obvious to me that $A^l = { a^{p^k} | a \in A } $ is a group. – Michael Pounds Apr 08 '14 at 16:20
  • I have found a reference which justifies that $A^l$ is a group. (I am still an undergraduate, so sorry for being slow...) P. Hall, A contribution to the theory of groups of prime-power order, Proc. London Math. Soc. (2) 36 (1933), 29–95. – Michael Pounds Apr 09 '14 at 02:48
  • Just to clarify - are there some missing symbols in your expressions for $A^l$ and $B^l$? I have highlighted what I think might be missing symbols : $A^l=(Z_{p^{n_1}})^l \times \cdots \fbox{$\times$} (Z_{p^{n_r}})^l$ and $B^l=(Z_{p^{m_1}})^l \times \fbox{$\cdots \times$} (Z_{p^{m_{r+1}}})^l \fbox{$\times$} \cdots$. Am I correct? I just want to make sure I understand. Thanks! – Michael Pounds Apr 09 '14 at 03:02
  • Yes, you are right. – Wei Zhou Apr 09 '14 at 03:58
  • Each term $\mathbb{Z}_{p^{n_i}} (i=1, \ldots , r)$ in the expression for $A^l$ is a non-trivial cyclic group, containing precisely 1 subgroup of order $p$, and therefore having $p-1$ elements of order $p$. There are therefore $(p-1) \times r$ elements of order $p$ in $A^l$. – Michael Pounds Apr 09 '14 at 12:46
  • Now let's compare this to $B^l$. Each term $\mathbb{Z}{p^{n_i}} (i=1, \ldots , r+1)$ in the expression for $B^l$ is a non-trivial cyclic group, containing precisely 1 subgroup of order $p$, and therefore having $p-1$ elements of order $p$. There are therefore _at least $(p-1) \times (r+1)$ elements of order $p$ in $B^l$. – Michael Pounds Apr 09 '14 at 12:49
  • Am I correct? Thanks for your help! – Michael Pounds Apr 09 '14 at 12:55
  • You are almost there. In $A^l$, there are exactly $p^r-1$ elements of order $p$. For example, $Z_p \times Z_p$ has $p^2-1$ elements of order $p$. Similarly, you can count in $B^l$. – Wei Zhou Apr 09 '14 at 13:13
  • Ah, of course! Thanks. So $B^l$ has at least $p^{r+1} - 1$ elements of order $p$. I say at least because $\mathbb{Z}{p^{m{r+2}}} = { 0 } $ if and only if $m_{r+2} \le l$. Correct? – Michael Pounds Apr 09 '14 at 13:25
  • @WeiZhou: Maybe you are also interested in this question: http://math.stackexchange.com/questions/528404, which in addition also asks for quotient groups. It didn't get a correct answer so far. – azimut Apr 09 '14 at 13:54
  • @Michael. You are right, and thanks for editing my answer to make it more clear. – Wei Zhou Apr 09 '14 at 23:10
  • @WeiZhou Thanks for your help. I've just posted another, related question: http://math.stackexchange.com/questions/748531/subgroups-of-direct-products – Michael Pounds Apr 10 '14 at 17:29