Show an example of a finite abelian $p$-group that has $p^2 + p + 1$ subgroups of order $p$

Question

Show an example of a finite abelian $p$-group that has $p^2 + p + 1$ subgroups of order $p$

My Path: I tried with the abelian group of $\mathbb{Z}_{63}=\mathbb{Z}_{9} \times \mathbb{Z}_{7} \times \mathbb{Z}_{1}$ but I'm not sure how to say if it's an abelian group

Answer 1

Take the elementary abelian group $\;V:=\left(\Bbb Z/3\Bbb Z\right)^3 =\Bbb Z/3\Bbb Z\times\Bbb Z/3\Bbb Z\times\Bbb Z/3\Bbb Z\;$ .

This is also a vector space of dimension $\;3\;$ over the field $\;\Bbb F_p\cong\Bbb Z/3\Bbb Z\;$, and its subspaces are the same as its subgroups (when we flip from the group to the linear space structure or back).

Let us see how many subspaces of dimension $\;1\;$ this linear space has: for any non-zero vector $\;v\in V\;$, and there are $\;p^3-1\;$ of these, you get that Span$\,v\;$ is a linear subspace of dimension 1.

To count all the different subspaces of dimension $\;1\;$ (subgroups 0f order $\;p\;$), we must divide by all the non-zero multiple scalars of $\;v\;$, when $\;0\neq v\in V\;$ ...well, there you go.

The idea, of course, is for you to complete details in the above.