# Element membership for normal subgroups

Let $H$ be a normal subgroup of $G$. Let $m = (G:H)$. Show that $a^m \in H\ \forall a \in G$.

I don't know how to approach this proof. I know that since $H$ is normal we have $$aH=Ha$$ $$aha^{-1}\in H\ \forall h \in H$$

And that we need to show $a^m \in H$ which is the same as showing $a^mH = H$.

Any pointers in the right direction would be appreciated.

## 2 Answers

HINT: $aH\in G/H$; what’s the order of $G/H$?

I suppose something like the following would work:

\begin{align} aH\in G/H \\ (aH)^m = a^mH \end{align} but since $m = (G:H) = \lvert G/H \rvert$ we also have \begin{align} &(aH)^m = H \\ &\implies a^mH = H \\ &\implies a^m \in H \end{align}