0
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$\begin{align} &(aH)^m = H \\ &\implies a^mH = H \\ &\implies a^m \in H \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.