If a is a $\mathbb Z / p\mathbb Z$ generator, if $k$ is not a multiple of $p-1$, $a ^ k \not\equiv 1\pmod p$

Question

If $a$ is a $\mathbb Z / p\mathbb Z$ generator, if $k$ is not a multiple of $p-1$, $a ^ k \not\equiv 1\pmod p$.

I don't understand why.

What does "$a$ is a $\mathbb Z / p\mathbb Z$ generator" mean?

Please tell me.

Answer 1

If $a$ is a generator, then $a,a^2,\ldots,a^{p-1}$ are all different. Because if there were two of them equal, in that sequence there would be fewer than $p-1$ different elements.

We know by little Fermat's Theorem that $a^{p-1}=1$.

So take $k$ s.t. $a^k\neq 1$. Now use Euclidean division $k/(p-1)$: $k=q(p-1)+r$. Then $$a^k=a^{q(p-1)+r}=(a^{p-1})^q\cdot a^r=1^q\cdot a^r=a^r$$

Since $k$ is not a multiple of $q-1$, $r\neq 0$ and $r<p-1$. So $a^k\neq 1$.

Answer 2

HINT:

If you're question is "Let $a$ be a generator of $\Bbb Z/p \Bbb Z$, if $k$ is not a multiple of $p-1$, then $a^k \not \equiv 1$ (mod $p$)".

Then this is the same as asking "Let $a$ be a generator of $\Bbb Z/p \Bbb Z$, if $a^k \equiv 1$ (mod $p$), then $k$ is a multiple of $p-1$."

Answer 3

Since $\mathbb Z/p\mathbb Z$ has order $p$, $\operatorname{ord}(a)=p$. But this means that, in particular, that $a^k\neq1$ if $k\in\{1,2,\ldots,p-2\}.$