Non-Cyclic Group of Order p^2 has p+3 Subgroups

Theorem
Let $p$ be a prime number.

Let $G$ be a non-cyclic group whose order is $p^2$.

Then $G$ has exactly $p + 3$ subgroups.

Proof
By Order of Element Divides Order of Finite Group, all elements of $G$ have order in $\set {1, p, p^2}$.

But as $G$ is non-cyclic, it can have no element of order $p^2$.

By Identity is Only Group Element of Order 1, $G$ has $p^2 - 1$ elements of order $p$.

Let $m$ denote the number of subgroups of $G$ of order $p$.

From Number of Order p Elements in Group with m Order p Subgroups, $G$ has $m \paren {p - 1}$ elements of order $p$.

Then:

Together with the trivial subgroup and $G$ itself, that makes $p + 3$ subgroups.