Subgroups lattices of groups of order $p^3$

Question

Where could I find the subgroups lattices of the groups of order $p^3$? Of course the main problem are the two non abelian groups with $p>2$ (the extraspecials), that I can't determinate easily by myself. I'd need a referenceable source (books, academic literature,...). Thank you.

Answer 1

I do not know a reference book; but, I think it is easy to find if we come from top to bottom in these groups.

There are $p+1$ subgroups of order $p^2$ in both; intersection of all of them (or any two) is the center (as well as commutator) of original group.

So, it remains to determine what are the subgroups inside each maximal subgroup other than the center. For this, one can use structure of groups of order $p^2$.

If $G$ is of exponent $p$, then each of the maximal subgroups contains $p$ many subgroups of order $p$ other than the center of $G$. You can then complete the lattice.

If $G$ is of exponent $p^2$, then among $p+1$ maximal subgroups, $p$ many are cyclic and one is non-cyclic. It is the non-cyclic which gives new members in lattice - $p$ many subgroups of order $p$ other than the center of $G$. (The cyclic maximal subgroup has unique subgroup of order $p$, so it must be center of $G$)

Answer 2

Let $p>2$ be a prime. There is one non-abelian group $P$ of order $p^3$ and exponent $p$. There are just $\frac{p^3-1}{p-1}$ subgroups of order $p$ and $p+1$ (maximal) subgroups of order $p^2$. Now let $P$ be non-abelian of exponent $p^2$. Here, there are $p+1$ subgroups of order $p$ and their union is a non-cyclic subgroup of order $p^2$. All other ($p$) subgroups of order $p^2$ are cyclic.