# Prove that number of subgroups of order p equals the number of subgroups of index p.

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 $p$.

• What's part b)? – Angina Seng Nov 11 '17 at 16:28
• I've edited my question to include the actual question. – gHem Nov 11 '17 at 16:32

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.
• @gHem Every index $p$ subgroup of $A$ contains $A^p$. – Angina Seng Nov 11 '17 at 16:57
• @LordSharktheUnkown +1 How do we know that there are $p^n-1$ of these homomorphisms? – Algebraist Nov 14 '19 at 11:43
• I think it's because $n$ should be $k$ :-) @Pascal'sWager – Angina Seng Nov 14 '19 at 18:40
• @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. – Angina Seng Nov 15 '19 at 2:19