I know that if $H$ is a normal subgroup of $G$ with index $m$ ,then $g^m \in H, \forall g\in G$. But is the converse true in this case when m is a prime $p$?
Let $G$ be a group and $H$ is a subgroup of $G$ with index $p$. If $g^p \in H , \forall g\in G$ then $H \triangleleft G$
I found the answer to my question in this paper On Subgroups of Prime Index by T. Y. Lam
Yes. Consider the action of $G$ by multiplication on the (left or right, whichever you prefer) cosets of $H$ in $G$. This action is transitive, and so the image of the action has order divisible by $p$.
I claim that this image has order exactly $p$, which implies that the action is regular and $H$ is normal in $G$. If not, then the stabilizers in the action are non-trivial. Since the action is transitive, we can choose $g$ in such a stabilizer that does not fix the coset $H$. So the cycle of $g$ on the coset $H$ has order $m$ with $1 < m < p$, and since $p$ is prime, $m$ does not divide $p$, so $g^p$ does not fix $H$. That is, $g^p \not\in H$, contrary to assumption.
I expect there is a more elegant way of writing that proof!
Let $h\in H$ and suppose that $h\not\in\text{Core}_G(H)$. If $g\in G$ then $(h^g)^p=(h^p)^g\in H$ and hence $h^p\in\text{Core}_G(H)$. By Cayley's theorem $|G/\text{Core}_G(H)|$ divides $p!$, and since $|G:H|=p$ it follows that $|H:\text{Core}_G(H)|$ divides $(p-1)!$. As $\text{Core}_G(H)\unlhd H$ and $o(h\text{Core}_G(H))=p$, by Lagrange's theorem we deduce that $p$ divides the order of $H/\text{Core}_G(H)$, which is not possible. Therefore $h\in\text{Core}_G(H)$ and $H\unlhd G$.
