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 .

  • $\begingroup$ then I assumed that $m$ will divide o(a) which is apparently wrong Yes, you are not allowed to assume the conclusion of what you are trying to prove. Here's what I did to prove the assumption You don't need to prove assumptions! At the very least, I think you are having a little trouble expressing yourself... it comes out a bit mixed up. $\endgroup$ – rschwieb Jan 16 '13 at 21:23
  • $\begingroup$ I guess my wording is a bit off, I meant that because I showed $(aN)^m=N$ then $m$ will divide $o(a)$ but the teacher marked it as wrong hence I'm trying to prove it. $\endgroup$ – user10444 Jan 16 '13 at 21:26

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.

  • $\begingroup$ So this means saying $(aN)^o(a)=N \implies $ $m$ divides $o(a)$ is actually correct? $\endgroup$ – user10444 Jan 16 '13 at 21:27
  • $\begingroup$ If $m=o(aN)$, yes. In symbols, $o(aN)|o(a)$. That was the conclusion of the question you posted :) $\endgroup$ – rschwieb Jan 16 '13 at 21:37
  • $\begingroup$ Thank you for your time, I'm going to argue for my five grades back. $\endgroup$ – user10444 Jan 16 '13 at 21:39
  • $\begingroup$ Disregard a comment I had here a moment ago, I see the rearrangement you did is fine :) Details could be a little clearer, though $\endgroup$ – rschwieb Jan 16 '13 at 21:51
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    $\begingroup$ You are right, this was the last problem and I was rushing through it. Thank you for your help $\endgroup$ – user10444 Jan 16 '13 at 21:54

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