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Exercise (The Order Of A In UP Divides P-1)

Let $p$ be a prime. Let $a\in U(p)$. Then $|a|$ divides $p-1$.

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There are several ways to prove this result.

  1. Since we know $a^p\pmod p = a$, we konw $a^{p-1}\pmod p = 1$. We know that $a^k=e$ in any group only when $k$ is a multiple of the order of $|a|$, so this means $k=p-1$ is a multiple of $|a|$.
  2. We now that $|U(p)|=p-1$. Hence every subgroup of $U(n)$, in particular $\left<a\right>$, must have an order that divides $p-1$. Since we know $|a|=|\left<a\right>|$, we know that $|a|$ divides $p-1$.