I am stuck with the following problem, from C. Pinter's "A Book of Abstract Algebra", p. 153 ex. C 6:

Let $G$ be an abelian group, and $H\_p$ the subset of $G$ such that the order of every $x \in H\_p$ is a power of $p$. Prove that $H\_p$ is a subgroup of $G$, and that $G/H\_p$ has no elements whose order is a non-zero power of $p$.

So, to prove that $H\_p$ is a subgroup of $G$, I need to show (among others) that it is closed under the group operation. Let $a$, $b \in H\_p$, by hypothesis $\mathrm{ord}(a) = p^m, \mathrm{ord}(b) = p^n$ for some integers $m, n$. Now consider $(ab)^{p^{mn}}$: since $G$ is abelian, this is equal to $a^{p^{mn}}b^{p^{mn}}$, which is in turn equal to $a^{{(p^m)}^n}b^{{(p^n)}^m}$, which is equal to the identity $e$, since $p^m = \mathrm{ord}(a)$ and $p^n = \mathrm{ord}(b)$.

In other words, $(ab)^{p^{mn}} = e$, hence $\mathrm{ord}(ab)$ divides $p^{mn}$.

Now here is where I am puzzled: nothing is assumed about $p$. For if $p$ were assumed to be prime (as the choice of the letter "p" seems to indicate), it would indeed seem to follow that $\mathrm{ord}(ab)$ must itself be a power of $p$.

Conversely, there seem to be counterexamples when $p$ is not prime, e.g. if we take $G = (\mathbb{Z}\_{12},+)$ and set $p = 4$, then $H\_4$ contains $3$ ($\mathrm{ord}(3) = 4$), but $3 + 3 = 6$ has order $2$, which is not a power of $4$. In other words, $H\_4$ does not appear to be closed under the group operation ($+$, in this case).

Any comments? Thanks very much.