I'm trying to prove that given a $p$-group $G$ and any normal subgroup $A$ of $G$, the quotient group $G/A$ is also a $p$-group.
If $G$ is finite, then $A$ is finite too and the size of $G/A$ is a power of $p$ divided by a power of $p$, and hence a power of $p$ itself.
But I couldn't think of anything for the infinite case.
By definition of $p$-group, for every $g\in G$, $o(g)=p^k$ for some nonnegative integer $k$ (depending on $g$). But then $(gA)^{p^k}=g^{p^k}A=A$, and hence $o(gA)$ divides $p^k$, namely $o(gA)=p^l$ for some nonnegative integer $l\le k$ (depending on $g$, given $A$). This holds for every $g\in G$ and hence for every $gA\in G/A$, so $G/A$ is a $p$-group as well.