If $\sigma\in S_p$ with $|\sigma| = p$, why is $\sigma$ a $p-$cycle, and why is $|\sigma^r| = p$?

Question

If $\sigma\in S_p$ with $|\sigma| = p$, why is $\sigma$ a $p-$cycle, and why is $|\sigma^r| = p$ for each $r$, $1\leq r < p$? ($p$ is a prime)

I guess I am just having a hard time understanding the way the symmetric group works. I know that since $\sigma$ has order $p$, it can be written as a product of disjoint cycles, the least common multiple of whose lengths must be $p$, but why does this imply that $\sigma$ is a $p-$cycle? (And how do I get to the second conclusion?)

Answer 1

Let $n=|\sigma|$, then $n|p$ so $n=1$ or $p$. If $\sigma\ne e$, then $|\sigma|=p$.

Now since $1\le r<p$, $r$ and $p$ are coprime and so $|\sigma^r|=p$. Finally let $i\in\{1,\ldots, p\}$. The orbit of $i$ consists of $i,\sigma(i),\sigma^2(i),\ldots,\sigma^{p-1}(i)$, therefore $\sigma$ is a $p$-cycle.