# How to show that these two groups are isomorphic?

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 (which is for $p=2$). In fact, I'm tempted to call it a duplicate. – Arturo Magidin Mar 7 '12 at 5:18

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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Another way to see things, but the map from $A$ to $A^p$ doesn't break everything down uselessly... still a good idea though. – Patrick Da Silva Mar 7 '12 at 3:50
Uselessly breaking things down into «indecomposable» parts is quite not a useless approach... – Mariano Suárez-Alvarez Mar 7 '12 at 3:53
I am not saying that =) I am saying that here the breaking down is not necessary. Perhaps the word useless was rude. =P – Patrick Da Silva Mar 7 '12 at 4:07
@Patrick: As far as I can tell, you do need to break up the group into cyclic factors. The isomorphism $A/A_p\to A^p$ is natural, but the isomorphism $A/A^p\to A_p$ is not. – Arturo Magidin Mar 7 '12 at 5:41
@PatrickDaSilva: See this question (for which I voted this as a duplicate) and the answers for some comments on it. – Arturo Magidin Mar 7 '12 at 6:31

$\bf Hint:$ 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 am afraid that I do not get your hint, could you please show more details? Thank you. – John Smith Mar 7 '12 at 4:05
This yields an isomorphism $A/A_p \cong A^p$. The OP might be confused as to how to obtain the desired isomorphism from this. – Dylan Moreland Mar 7 '12 at 4:15
Here is exactly where I am stuck.... – John Smith Mar 7 '12 at 4:50
I don't see a way around using the classification of finite abelian groups as in Mariano's answer, but I'm willing to be surprised. – Dylan Moreland Mar 7 '12 at 4:56