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 $m_2 = 0$.
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.