Let $N$ be a normal subgroup of a finite group $G$, and $a \in G$ is an element of order $o(a)$. Prove that the order $m$ of $aN$ in $G/N$ is a divisor of $o(a)$.
Here what I did:
$(aN)^{o(a)}=a^{o(a)}N=eN=N$ but is the least power such that $(aN)^m=N$. I then assumed that $m$ will have to divide $o(a)$ which is apparently wrong. Here's what I did to prove the assumption. $(aN)^{o(a)}=(aN)^{mq+r} 0\le r<m\implies ((aN)^m)^{-q}(aN)^{o(a)}=(aN)^r \implies N=(aN)^r$ but $r<m$ then $r=0$ hence $mq=o(a)$. Is this right? I know another proof exists but I'm trying to do this in my own way .
Best Answer
Here's a slightly more general statement that might actually isolate the fact better.
Let $\phi:G\rightarrow H$ be a homomorphism. Then the order of $\phi(g)$ divides the order of $g$.
In your case, you are just talking about the projection $\phi:G\rightarrow G/N$.
And if you didn't know (or learned and forgot) that $g^k=1$ implies that the order of $g$ divides $k$, I just wanted to remind you with this sentence.
I see that you had some of the elements of the proof of this fact in your proof. If you are already aware of this result, you don't have to repeat the steps using the division algorithm in this proof. You just refer the reader to that result, rather than reproving it.