# Prove that if $[2]$ is a generator for $\mathbb{Z_{9}^{\times}}$, then $[5]$ is as well.

It's easy to prove explicitly that for $\mathbb{Z_{9}^{\times}}$, the invertible elements of $\mathbb{Z_9}$ with respect to multiplication, $[2]$ and $[5]$ are the only generators.

Is there a way to prove that $[2]$ is a generator $\implies [5]$ is a generator without computation?

My guess is to use something from a recently covered theorem with several parts:

Let $G$ be a finite cyclic group. Let $n=|G|$ and let $a$ be a generator of $G$. Then,

(1) Every subgroup of G is cyclic and is equal to $\left\langle a^d \right\rangle$

(2) $a^m$ is a generator of $G\iff$ $\gcd(m,n)=1$

(3) $\forall m\in\mathbb{Z}$, $\left\langle a^m\right\rangle=\left\langle a^{\gcd(m,n)}\right\rangle$

• Hint: What is the algebraic relationship between $9$ and $5$? – basket Oct 16 '16 at 19:36
• Well, $gcd(5,9)=1$. But I'm not sure how that may help @basket – user322548 Oct 16 '16 at 19:38

We have $[2][5]=[10]=[1]$, so that $[5]=[2]^{-1}$. But in any cyclic group $G$, if $g$ is a generator, then $g^{-1}$ is as well.