My question arised from an explanation to the proof of another problem. Here is the original problem, posted here:
For a finite elementary abelian group, the number of subgroups of A of order p equals the number of subgroups of A of index p.
And the proof was also attached to the previous link, where I quote here:
Both $A^p$ and $A/A^p$ have exponent $p$, so, as they have the same order, they are both isomorphic to $C^n_p$ 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.
What I don't understand is the argument
two of these homomorphisms have the same kernel iff they differ by a scalar factor
I can see if these two homomorphisms differ by a scalar factor then they have the same kernel, but I cannot show the opposite direction. Can anyone help?