# Subgroups of abelian $p$-groups

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 $m_2 = 0$.

• As noted by Michael Pounds, this question is exactly the first part of my question math.stackexchange.com/questions/528404, which additionally asks for quotient groups of finite abelian $p$-groups. – azimut Apr 9 '14 at 14:18

## 1 Answer

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.

• Thank you. How to prove that $A^l$ and $B^l$ are groups? – Michael Pounds Apr 8 '14 at 15:13
• This is abelian group. So it is groups. – Wei Zhou Apr 8 '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 8 '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 9 '14 at 2:48
• 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 9 '14 at 3:02