Cyclic groups under mod p multiplication? I have a quick question:
How does a cyclic group form under $\bmod {p}$ multiplication? 
(Reading from: http://dogschool.tripod.com/cyclic.html)
I really think I should know this, but, I really don't understand it, at the moment.

This is stated a little awkwardly: I think what you are asking is "Why do the integers modulo $p$, excluding the zero element (i.e. the integers divisible by $p$), form a cyclic group under multiplication?"
The key to showing that this is a group is Bezout's lemma, which states that if $m$ and $n$ are integers with greatest common divisor $d$, then there exist integers $r$ and $s$ such that $rm + sn = d$. In particular, if $m = p$ and $n$ is not divisible by $p$, then $d = 1$ and this says precisely that $n$ has a multiplicative inverse modulo $p$.
As for proving that this group is cyclic, the point is that all the nonzero elements of $\mathbb{Z}/p\mathbb{Z}$ are roots of unity (meaning some power is $1$). Try proving that any finite subgroup of the nonzero complex numbers $\mathbb{C}^{\times}$ under multiplication is cyclic: the argument is identical. If you need more hints, let me know.
