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Q : Let $A$ be a finite abelian group and let $p$ be a prime. Prove that the number of subgroups of $A$ of order $p$ equals the number of subgroups of $A$ of index $p$.\\

A : Define a homomorphism $\varphi:A\to A$ by $a\mapsto a^p$. Let $A^p:=Im(\varphi)$ and $A_p:=ker(\varphi)$. Then for any $a\in A_p$, $a^p =1$ and as $A$ is a finite abelian group, $A_p\simeq(\Bbb Z/p)^n$ for some $n$. Also for $a A^p\in A/A^p$, $(a A^p)^p = A^p$ so $A/A^p\simeq(\Bbb Z/p)^m$ for some $m$. By the first isomorphism theorem, $A/A_p\simeq A^p$ so that $|A/A^p| = |A_p|$. Hence, $n =m$ i.e. $A/A^p\simeq A_p$ as an elementary abelian group.\\

First prove for the case $(\Bbb Z/p)^n$. Note any nontrivial element in $(\Bbb Z/p)^n$ has order $p$ and by Lagrange theorem, they form a partition. Hence, the number of order $p$ subgroup of $(\Bbb Z/p)^n$ is $(p^n-1)/(p-1)$. The index $p$ subgroup of $(\Bbb Z/p)^n$ corresponds to the kernel of the epimorphism $(\Bbb Z/p)^n\to \Bbb Z/p$ (note that such group is normal as $(\Bbb Z/p)^n$ is a $p$-group). Since there are total $p^n$ homomorphisms and only the trivial map is not an epimorphism and two epimorphism has the same kernel if and only if they are differ by a scalar factor, there are total $(p^n-1)/(p-1)$ epimorphism.\\

Now prove for the general case. Note that if $B<A$ has index $p$ then $A/B\simeq\Bbb Z/p$ so that $(A/B)^p =0$ so $A^p<B$. So if $B<A$ is an index $p$ subgroup then there is a corresponding subgroup of index $p$ in $A/A^p$ by the correspondence theorem. Now if $P<A$ is a subgroup of order $p$ then $P<A_p\simeq A/A^p$ so the number of subgroup of $A$ of order $p$ equals the number of subgroups of $A/A^p$ whose index is $p$ as $A/A^p\simeq A_p$ is an elementary abelian group. Hence the number of subgroups of order $p$ equals the number of subgroups of index $p$ in $A$.