If $N \lhd G, (G : N) = 100, a \in G, a^{23} = e$, show that $a \in N$.

Question

I'm having some trouble with a question that my instructor suggested to think about: Let $N \lhd G$ be a normal subgroup of $G$, the index $(G : N) = 100, a \in G, a^{23} = e$. Show that $a \in N$.

My first approach was to use a theorem stating that if $N \lhd G$ where $(G : N) = m$, then $a^m \in N$. From this result, it obviously follows that $a^{100} \in N$.

However, I am a little stuck continuing and would appreciate any hint to guide my way (I am not looking for a complete solution as I'd love to learn this on my own too :D).

Thanks and regards.

Answer 1

If $a \not\in N$, then $aN \neq N$. What can you say about the order of the element $aN$ in $G/N$?