How many subgroups of order $p^2$ does the abelian group $\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$ have? Hint: Every cyclic group of order n contains a unique subgroup of order m for all divisors m of n.
Proof:there are two types of groups of order $p^2$: cyclic, and non-cyclic.
non-cyclic: such a subgroup must be of the form HxK, where H is a subgroup of order $p$ in $\mathbb{Z}_{p^3}$, and K is a subgroup of order $p$ in $\mathbb{Z}_{p^2}$.
since there is only ONE such subgroup in each factor, we conclude the sole non-cyclic group of order $p^2$ in $\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$ is: $<p^2> x <p>.$
cyclic: subgroup K of order $p^2$ of $\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$,
where K = <(a,b)>, contains either $a$ as order $p^2$ (and b is arbitrary), or $b$ is of order $p^2$ and $a$ is of order $1$ or $p$.
there are $p^3-p^2$ elements of order $p^2$ in $\mathbb{Z}_{p^3}$, and $p^2-p$ elements of order $p^2$ in $\mathbb{Z}_{p^2}$.
this means we have a total of $p^4-p^3 + p^3-p^2 = p^4-p^2$ elements of order $p^2$ in $\mathbb{Z}_{p^3} \oplus \mathbb{Z}_{p^2}$.
since each distinct subgroup accounts for $p^2-p$ of these elements (a cyclic group of order $p^2$ has $p^2-p$ elements of order $p^2$), we must have:
$\frac{p^4 - p^2}{p^2 - p} = \frac{p^2(p^2 - 1)}{p(p-1)} = p^2+p$
cyclic subgroups of order $p^2$.
this gives us $p^2+p+1$ subgroups of order $p^2$ in all.
Is this right?
