This might be a stupid question since I don't have much knowledge in group theory, but I've been struggling on this for a bit of time ; I am currently reading Wehrfritz's book on linear groups, and during the proof for the corollary 4.8, Wehrfritz states that
" [...] $H$ is residually a $p$-group. Then every periodic subgroup of $H$ is a p-group"
There seems to be no additionnal hypothesis. Correct me if I'm mistaken, but since the order of the image of an element by a morphism is lower than the original order of the element, couldn't we imagine some element of order $p^mq$ (so which is of torsion) in $H$ such that the image by the morphism given by the fact that $H$ is residually a $p$-group is of order $p^n, n \leq m$ as expected ? Then we could have a periodic residually $p$-group which isn't a $p$-group ?
Assume $G$ is residually a $p$-group and also periodic. By the first assumption, $G$ embeds into a product of $p$-groups. Take an element of $G$, say $g$, and consider its image in the product. By assumption, $g$ has finite order, and each component of $g$ has also finite order and that order is a power of $p$. Now the point is that these powers of $p$ are bounded since they all divide the order of $g$. Hence, we can take a power of $p$ which is the maximum of all these. Now you can easily check that this power of $p$ must be the order of $g$. Hence, $G$ is a $p$-group.