Let $H$ be a normal subgroup of $G$ and $[G: H] = m$, then for any $g \in G$, $g^m \in H$.

Question

Let $H$ be a normal subgroup of $G$ and $[G: H] = m$, then for any $g \in G$, $g^m \in H$.

My attempt: Since $H$ be a normal subgroup of $G$, the quotient group $G/H$ is meaningful. Now we have $\frac{|G|}{|H|} = m$, so $m$ divides $|G|$.

But I cannot proceed further.

Please help me.

Answer 1

Let $xH$ be an element of $G/H$. Since $[G:H] = m ={\rm ord}(G/H)$ we have $x^m H = (xH)^m = 1H$ so $x^m \in H$.