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Let $A$ be a finite abelian group and $p$ be a prime, $p$ divides the order of $A$. Define: $A^p=\{a^p | a\in{A}\}$ and $A_p=\{x\in{A}|x^p=1\}$, where $1$ is the identity in $A$.

Show that $A/A^p\cong A_p$.

This is a homework problem. I am thinking of applying the First Isomorphism Theorem. I tried to define a surjective homomorphism $A\to A_p$ such that the kernel is $A^p$. But I am having trouble finding such a map.

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    This is *very* closely related to [this question](http://math.stackexchange.com/q/107203/742) (which is for $p=2$). In fact, I'm tempted to call it a duplicate.2012-03-07

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Since $A$ is a direct sum of cyclic groups and the operations $(-)^p$ and $(-)_p$ on groups distribute over direct sums, it is enough to prove this for cyclic groups.

This is easy :)

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    @PatrickDaSilva: See [this question](http://math.stackexchange.com/q/107203/742) (for which I voted this as a duplicate) and the answers for some comments on it.2012-03-07
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$\bf Hint:$ Consider the map $f:A\to A^p$ given by $f(a)=a^p$ which is a homomorphism since $A$ is Abelian.

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    I don't see a way around using the [classification](http://en.wikipedia.org/wiki/Finitely-generated_abelian_group#Classification) of finite abelian groups as in Mariano's answer, but I'm willing to be surprised.2012-03-07