The number of index $p$ subgroup and order $p$ subgroup for elementary abelian group In my previous post, fact 2 says that

For elementary abelian group $A=(\Bbb Z/p)^n$, the number of subgroup of $A$ of order $p$ equals the number of subgroup of $A$ of index $p$.

Here's the proof (last page). But I can't understand the proof of this.

First assume $A = (\Bbb Z/p)^n$. Then fixing any non-identity element $\alpha$ of $(\Bbb Z/p)^n$, $P =\langle\alpha\rangle$ will have order $p$, and this group can be described by $p-1$ distinct elements of $(\Bbb Z/p)^n$. Now consider the composition of maps $(\Bbb Z/p)^n\to(\Bbb Z/p)^n/P\to(\Bbb Z/p)^n$ where the first map is the quotient map and the second map is an injection. So the image of the composition will be a subgroup of $(\Bbb Z/p)^n$ with index $p$. Since there are $p-1$ ways to fix representatives of $(\Bbb Z/p)^n/P$ and inject them into $(\Bbb Z/p)^n$, there are the same number of subgroups of order and index $p$.

In the proof, what is $(\Bbb Z/p)^n/P\to (\Bbb Z/p)^n$? I wonder how the map is defined. Could you help? Or is there any other way to prove this?
 A: The duality between subgroups of order $p^i$ and subgroups of order $p^{n-i}$ is non-canonical. This means that there is no way to describe a bijection between the two sets that does not depend on choices, in this case a generating set for the group.
Let $g_1,\dots,g_n$ be $n$ elements of order $p$ that generate $A$. Then the elements of order $p$ are labelled by tuples $(a_1,\dots,a_n)$, not all $a_i$ equal to $0$, where such a tuple corresponds to the element $$\sum_i a_ig_i.$$
If one requires subgroups of order $p$ rather than elements, one simply takes the tuples up to scalar multiples, or equivalently, taking those whose first non-zero entry of the tuple is equal to $1$.
For subgroups of order $p^{n-1}$, this is equivalent to kernels of maps onto a group of order $p$. Given a tuple $\mathbf a=(a_1,\dots,a_n)$, and a group element $g=\sum_i b_ig_i$, write
$$f_{\mathbf a}(g)=\sum_{i=1}^n a_ib_i.$$
The kernel of the map $f_{\mathbf a}$ is the set of group elements $g$ such that
$\sum a_ib_i=0$. As before, the kernel depends only on the tuple $\mathbf a$ up to scalar multiple.
Thus to each tuple up to scalar, we can associate both a subgroup of order $p$ and a subgroup of order $p^{n-1}$. But this bijection is entirely dependent on the choice of generating set for $A$.
