Let $G$ be a group and $a$ an element of $G$ of order $n$.
Prove that: If $a^k = e$, then $n$ divides $k$
abstract-algebragroup-theory
asked 2011-11-10
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@querty89, Please consider accepting answers to your previous questions. That is considered an important feedback in this site. – 2011-11-10
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I want to understand about cyclic group more futher – 2011-11-10
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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