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Exercise (If $a^k=e$, Then The Order Of $a$ Divides $k$)
Suppose that $a$ is a group element with order $n$. If $a^k=e$, prove that $k$ is a multiple of $n$.
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Suppose $a^k=e$. This means $a^k=a^0$, which from the previous problem is true if and only if $k-0$ is a multiple of $n$. This shows that $k=k-0$ is a multiple of $n$.