# $G$ finite abelian. $A \times B$ embedded in $G$. Is $G=C \times D$ such that $A$ embedded in $C$, $B$ embedded in $D$?

$G$ 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 $B$ –  anonymous Nov 17 '10 at 11:17

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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Thanks! How is this lemma proved? (a reference could be good) –  user3533 Nov 17 '10 at 11:32
To prove such a result, assume that the $s_i$ are decreasing, and then for each $t$ count the number of elements of order $p^t$ in $G$ and the purported subgroup. –  Robin Chapman Nov 17 '10 at 11:37
Tried to follow your direction. If I hadn't erred, it shows that what needs to be proven is that if F(K,t)=(number of elements of K with order $p^{t}$), then for each t: F(G,t)/(F(G,0)+..+F(G,t-1)) >= F(H,t)/(F(H,0)+..+F(H,t-1)). Is this what you meant? If so, what should I do from here? –  user3533 Nov 18 '10 at 0:08
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? –  user3533 Nov 18 '10 at 1:10

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 $G \cong C \times D$.

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