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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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