I bumped on the following question.
Let $N$ be a normal subgroup of index $m$ in $G$. Prove that $a^m \in N$ for all $a \in G$.
Proof $\;\;$ Let $a\in G$. Since $[G:N]=m$, then $|G/N|=m$. From Lagrange's Theorem, it follows that $(aN)^m=a^mN=eN=N$. Hence $a^m \in N$.
I'm trying to provide an example that this statement is true only if $N$ is normal. Usually such examples can be easily found in $S_n$, but this time my attempts were unsuccessful.
consider the subgroup $\{e,(1,2)\}\subseteq S_3$, it has index $3$. And $(2,3)^3=(2,3)\not \in \{e,(1,2)\}$