From the text A Book of Abstract Algebra by Charles C. Pinter, chapter 15, exercise C6.
If $G$ is an abelian group, let $H_p$ be the set of all $x \in G$ whose order is a power of $p$. Prove that $H_p$ is a subgroup of $G$. Prove that $G / H_p$ has no elements whose order is a nonzero power of $p$.
I am stumped trying to prove that $H_p$ is a subgroup of $G$ without making an extra assumption that $p$ is prime. (It is possible that this was intended by the author, since the variable is named $p$.)
Presume $a, b \in H_p$. Then ${\rm ord}(a) = p^m$ and ${\rm ord}(b) = p^n$.
$$(ab)^{(p^{\max(m, n)})} = a^{(p^{\max(m, n)})}b^{(p^{\max(m, n)})} = e$$
So I showed that $ab$ to a power of $p$ is $e$, but I can't figure out how to prove that this is the order of $ab$, or that the order of $ab$ is a power of $p$ at all.
In fact, there seems to be a simple counter example. Let $G = \mathbb{Z}_8, p = 4$:
Then $H_p = \{x \in \mathbb{Z}_8 : {\rm ord}(x) = 4^n for\ some\ n \in \mathbb{Z} \} = \{0, 2, 6\}$. That is obviously not a group, since $2^2 = 4 \notin H_p$.
Is the assumption that $p$ is prime necessary for this proof?
Yes, $p$ is supposed to be a prime.
Without the assumption that $p$ is a prime, the assertion that $H_p$ is a subgroup need not follow.
For another example, with a squarefree number, take $G=C_2\times C_2\times C_3$, with the cyclic groups generated by $x$, $y$, and $z$.Look at elements whose order is a power of $6$. Among them is $(x,1,z)$ and $(1,y,z^2)$. But their product is $(x,y,1)$, whose order is $2$, not a power of $6$.