If $G$ be a cyclic group of prime order $p$, prove that every non-identity element of $G$ is a generator of $G$.

Question

If $G$ be a cyclic group of prime order $p$, prove that every non-identity element of $G$ is a generator of $G$.

WE know that no of generator= $\phi{(p)}=p-1$.

We have, if $a$ is a generator of $G$ then $a^k$ is a generator iff $gcd(p,k)=1$. Here as $p$ is prime, then $k=1, 2,3,\cdots, p-1$ as gcd of each of $k$ with $p$ is $1$. Therefore, generator of $G$ are $a, a^2, a^3, \cdots a^{p-1}$. They are $p-1$ in number.

I am not satisfied with my above argument. I need more significant proof. Please help me.

Answer 1

Hint: Recall that by Lagrange's thereom, $\vert g \vert$ divides $\vert G \vert $ $\forall$ $g$ $\in $ $G$.So what can you about $\vert g\vert $ ?

Edit: let $g \in G$,consider the subgroup $H=<g>$ and by Lagrange's theorem $\vert H \vert$ divides $\vert G \vert $,hence $\vert g \vert$ divides $\vert G \vert$.

Answer 2

We know that for all g in G, <(g)> is a subgroup of G, so by Lagrange's Theorem, |<(g)>| = |g| divides |G|. Since |G| = p, a prime, for all g in G, |g| is 1 or p. But, if |g| = 1, then g=e, thus for all non-identity g in G, |g|=p, so |<(g)>|=|G|, and <(g)> is a subset of G, and <(g)> and G are finite, so <(g)>=G, for all non-identity g in G.