3
$\begingroup$

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

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

  • 5
    @querty89, Please consider accepting answers to your previous questions. That is considered an important feedback in this site.2011-11-10
  • 0
    I want to understand about cyclic group more futher2011-11-10
  • 3
    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 $n|k$.2011-11-10

1 Answers 1

7

Here's a hint: Try writing $k=qn+r$ with $0\leq r

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

  • 0
    Why use a contradiction? Use the definition of order as "smallest positive integer $m$ such that $a^m=e$.2011-11-10
  • 1
    @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 $n$.2011-11-10