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$G$ is a finite abelian group. $A \times B$ can be embedded in $G$. Does this mean there exist $C$, $D$ such that $G=C \times D$ and $A$ can be embedded in $C$ and $B$ can be embedded in $D$?

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    @user3533: What are $A$ and $B$2010-11-17

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Yes, this reduces to the case of finite $p$-groups. Use the lemma that if $G=Z_{p^{r_1}}\times Z_{p^{r_2}}\times\cdots\times Z_{p^{r_k}}$ where $r_1\ge r_2\ge\cdots \ge r_k$ then each subgroup of $G$ is isomorphic to $Z_{p^{s_1}}\times Z_{p^{s_2}}\times\cdots\times Z_{p^{s_k}}$ where each $s_i\le r_i$.

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    I think I got it. Adding 1 to both sides, the inequality becomes: [G[p^{t}]:G[p^{t-1}]] >= [H[p^{t}]:H[p^{t-1}]] so it's enough to show that $H[p^{t}]/H[p^{t-1}]$ embeds in $G[p^{t}]/G[p^{t-1}]$ and that's due to the second isomorphism theorem: $H[p^{t}]/H[p^{t-1}] = H[p^{t}]/(H[p^{t}]\cap G[p^{t-1}]) = (H[p^{t}]G[p^{t-1}])/G[p^{t-1}] \leq G[p^{t}]/G[p^{t-1}]$. Is this right?2010-11-18
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My first reaction was that the answer is no, but I had misinterpreted the question.

A different question is: If $A$ and $B$ are subgroups of the abelian group $G$ such that $A \times B$ embeds in $G$, is it true that there are subgroups $C$ and $D$ of $G$ with $A \subseteq C$, $B \subseteq D$ and $G \cong C \times D$.