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Let $A$ be a finite abelian (additive) gp. and $p$ be a prime. I want to show $A/A^{p}\cong A_{p}$ where $A^{p}:=\left\{pa:a\in A\right\}$ and $A_{p}:=\left\{a\in A:pa=0\right\}$.(I want to show $A/A^{p}\cong A_{p}$ not $A/A_{p}\cong A^{p}$)

To show this, I want to find some surj. homom. $f:A\to A_{p}$ with $\ker f=A^{p}$. Let me explain my way more concretely. Where $n:=\left|A\right|$ and $n=p^{k}m$($\left(m,p\right)=1$), there exists integer $u$, $v$ s.t. $mu+pv=1$. From this, I got $p^{k+1}va=a$. I met wall in this step.

jawlang
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  • This is not that obvious. One proof that works is by using the structure theorem for finite abelian groups (which reduces the problem to the case of cyclic groups, and those can be easily solved). Probably there is something more elementary, but I guess it won't be a "formal" argument. – darij grinberg Aug 05 '15 at 00:45

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the obstacle to what it seems you wish to do is the fact that the kernel $A_p$ of the endomorphism $h_a: a \to pa$ and the image $A^p$ may not be direct summands. a simple example where $A_p \subset A^p$ is the cyclic group of order $8$. using the integers modulo $8$ to represent this, then $h_2$ has image $\{0,2,4,6\}$ and kernel $\{0,4\}$.

of course since $A^p \triangleleft A$ there is a homomorphism with kernel $A^p$ and image isomorphic to $A_p$.

David Holden
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