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Let $G$ be a group and $a$ an element of $G$ of order $n$.

Prove that: If $a^k = e$, then $n$ divides $k$

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@querty89, Please consider accepting answers to your previous questions. That is considered an important feedback in this site. – Srivatsan Nov 10 '11 at 3:38
I want to understand about cyclic group more futher – qwerty89 Nov 10 '11 at 3:50
Suppose $a^k = e$, then we know $k = nq + r$, $0 \leq r < n$, plugging this in we see $a^k = a^{nq + r} = a^{nq}a^{r}$ but $a^{nq} = e$ since $|a| = n$, thus $a^{nq + r} = a^{r}$, or $a^k = a^r$ but $a^k = e$ so $a^r = e$ but $r$ is strictly less than the order of $a$, so it must be that $r = 0$, and hence $k = nq$, so $n|k$. – Deven Ware Nov 10 '11 at 4:19
Please start accepting answers to your earlier questions. In Math.SE, this is considered important feedback for answerers. You can accept an answer by clicking the green tick/check mark under it. – Did Nov 13 '11 at 10:45

Here's a hint: Try writing $k=qn+r$ with $0\leq r<n$ by the division algorithm.

If $r\neq 0$, can you find a contradiction?

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Why use a contradiction? Use the definition of order as "smallest positive integer $m$ such that $a^m=e$. – Arturo Magidin Nov 10 '11 at 6:09
@Arturo You're right, I should be more to the point, but that's how I always justified it to myself. When I get to the step $a^r=e$, I conclude $r=0$, otherwise $r$ would be a positive integer smaller than $n$ such that $a^r=e$, a contradiction to the fact that $a$ has order $n$. – yunone Nov 10 '11 at 6:28

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