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My question is in relation to the answer on the following post by @Biao: https://math.stackexchange.com/q/1975135

I've solved part a) myself, but I'm stuck on part b). I understand how to count the subgroups of order p and index p for the most part. However, I don't understand why we can assume A is elementary abelian. The Dummit and Foote book suggests this as a hint but I don't see how. Could someone clear this up for me?

Edit: Here's the problem.

Let $A$ be a finite abelian group and let $p$ be a prime. Let $A^{p} = \{a^{p}\mid a \in A\}$ and $A_{p} = \{x\mid x^{p} = 1\}$.
a) Prove that $A/A^{p}$ is isomorphic to $A_{p}$, and
b) prove that the number of subgroups of $A$ of order $p$ equals the number of subgroups of $A$ of index $p$.

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  • $\begingroup$ What's part b)? $\endgroup$ – Angina Seng Nov 11 '17 at 16:28
  • $\begingroup$ I've edited my question to include the actual question. $\endgroup$ – gHem Nov 11 '17 at 16:32
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Both $A_p$ and $A/A^p$ have exponent $p$, so, as they have the same order, they are both isomorphic to $C_p^n$ for some $n$.

Each order $p$ subgroup of $A$ is contained in $A_p$, and there are $(p^n-1)/(p-1)$ of these. Each index $p$ subgroup of $A$ corresponds to an index $p$ subgroup of $A/A^p$. These subgroups are kernels of non-zero homomorphisms from $A/A^p$ to $C_p$ and there are $p^n-1$ of these homomorphisms. But two of these homomorphisms have the same kernel iff they differ by a scalar factor, so there are $(p^n-1)/(p-1)$ of these kernels.

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  • $\begingroup$ Thank you. Could you clarify how each subgroup of A with index p corresponds to a subgroup of A/A^p with index p? I️ think that’s where my confusion is. $\endgroup$ – gHem Nov 11 '17 at 16:49
  • $\begingroup$ @gHem Every index $p$ subgroup of $A$ contains $A^p$. $\endgroup$ – Angina Seng Nov 11 '17 at 16:57
  • $\begingroup$ @LordSharktheUnkown +1 How do we know that there are $p^n-1$ of these homomorphisms? $\endgroup$ – Pascal's Wager Nov 14 '19 at 11:43
  • $\begingroup$ I think it's because $n$ should be $k$ :-) @Pascal'sWager $\endgroup$ – Angina Seng Nov 14 '19 at 18:40
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    $\begingroup$ @Pascal'sWager Well, there are $p^n$ homomorphisms from $(Z_p)^n\to Z_p$, but one of them is the trivial one, sending all elements to zero, so there are $p^n-1$ non-zero homomorphisms. $\endgroup$ – Angina Seng Nov 15 '19 at 2:19

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