If $p$ is prime, then $\mathbb{Z}_{p}^{\times}, \cdot $ is cyclic

Question

If $p$ is prime, then $\mathbb{Z}_{p}^{\times}, \cdot $ is cyclic. How can this be proven using the following statement?: "Let $K$ be a field and let $f(x)$ be a non-constant polynomial of degree $n$ with coefficients in $K$. Then $f(x)$ has at most $n$ roots in $K$."

I don't really see how this could be linked to a group being cyclic. Can someone enlighten me?

Do you know the structure theorem for finite Abelian groups? — user49640, Jun 14 '17 at 19:14
I have not learned about that theorem yet, so I do not think it necessarily has to be used here. :) — simp, Jun 14 '17 at 19:21

Asinomás · Answer 1 · 2017-06-14T19:38:07.780

0

If an abelian group has an element $a$ of order $x$ and $b$ of order $y$ then there is an element of order $lcm(x,y)$. To prove this notice that $ac$ has exactly this order, where $c=b^{gcd(x,y)}$.

Let $m$ be the $lcm$ of all the orders of elements in $G$, we have proved that there is an element of order $m$ in $G$.

We now move to the setting where $G$ is the multiplicative group of $\mathbb Z_p$

Clearly $m\leq n$. Now, the polynomial $x^m-1$ has exactly $n$ roots and so $n\leq m$, and so $m=n$.

edited Jun 14 '17 at 19:38

answered Jun 14 '17 at 19:32

Asinomás

105,651

Thanks, that's a clear explanation! The only thing that I don't readily get is why $x^{m}-1$ must have $n$ roots. – simp Jun 14 '17 at 20:08
$x^m$ is clearly equal to $1$ for every element $x$ of $G$, because $m$ is the lcm of all of the orders. – Asinomás Jun 14 '17 at 20:09
Of course, thanks again! – simp Jun 14 '17 at 20:12

If $p$ is prime, then $\mathbb{Z}_{p}^{\times}, \cdot $ is cyclic

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