Show that every nonzero element of $(\mathbb{Z}/p\mathbb{Z})$ has order $p$

Question

Consider $G = (\mathbb{Z}/p\mathbb{Z})$, where $p$ is prime. Show that every non-zero element has order $p$.

I have seen the proof for this in Dummit and Foote using multiplicative notation, but I'm having trouble solving this using additive notation. What I have so far:

Let $(\mathbb{Z}/p\mathbb{Z}) = \{\bar{0}, \bar{1}, \bar{2}, ..., \overline{p-1}\}$, and let $\bar{a} \in \{ \bar{1}, \bar{2}, ..., \overline{p-1}\}$. Suppose ord$(\bar{a}) = n$, that is, $na \equiv 0$ (mod $p$). We want to show $n=p$.

Since $na \equiv 0$ (mod $p$), then $p \mid na$. Given that $p$ is prime and $a < p$, we have gcd$(a,p) = 1$. These facts together imply that $p \mid n$.

I still need to show that $p=n$. I know by the definition of order that $n$ must be the least positive integer such that $n\bar{a} = \bar{0}$, i.e. $na = p$. Is this enough to say also that $n \mid p$, and thus $n=p$?

I think I am very close to solving this but I can't quite put the pieces together. Any help would be appreciated.

Answer 1

In $(\mathbb{Z}/p\mathbb{Z})$, for each element $\overline a$ you have$$\overbrace{\overline a+\overline a+\cdots+\overline a}^{p\text{ times}}=\overline{pa}=0.$$So, no element has order greater than $p$.

Answer 2

Observe that order of $g \in G$ is same as the cyclic group generated by $g$. By Lagrange's theorem, the order of the subgroup divides the order of the group. Since the order of $G$ is prime, the order of subgroup has to be $1$ or $p$. So every non-trivial element must have order $p$.