What is meant by 'runs through'? I'm independently studying abstract algebra for fun (not my forte...) and I'm reading Herstein.  He has a question in the chapter on rings:
Let $p$ be an odd prime and let$$\sum_{k=1}^{p-1}\frac{1}{k}=\frac{a}{b}$$ where $a,b\in{\mathbb{Z}}$.  Show $p|a$. (Hint: As $a$ runs through $U_p$, so does $a^{-1}$.
$U_p=\{[a]\in\mathbb{Z}_n|(a,n)=1\}$
What does he mean "run through"?    I've seen this terminology before and it has always perplexed me.  And how does it apply to the proof?  
I've done it this way for a problem solving class:
$$\sum_{k=1}^{p-1}\frac{1}{k}=\sum_{k=1}^{\frac{p-1}{2}}\frac{p}{k(p-k)}=p\sum_{k=1}^{\frac{p-1}{2}}\frac{1}{k(p-k)}=\frac{a}{b}$$. 
 A: It means that for any $x\in U_p$, there exists $a\in U_p$ such that $a^{-1}=x$.
"Runs through" essentially means "takes on arbitrary values in". It's kind of like a verbal version of set theory notation:
$$\{3n^2\mid n\in\Bbb N\}$$
The above could be verbally expressed as "$3n^2$ as $n$ runs through $\Bbb N$".
A: 
As $a$ runs through $U_p$, so does $a^{-1}$.

This means that the function $f:U_p\to\Bbb Z_p:a\mapsto a^{-1}$ is actually a bijection from $U_p$ onto $U_p$. Imagine calculating $f(a)$ one at a time for each $a\in U_p$ and listing the outputs as they appear: your list will be a listing of $U_p$, though in general in a different order from your input listing.
A: As far as how the hint applies: we have 
$$b(\sum_{k=1}^{p-1} \frac{(p-1)!}{k}) \equiv a(p-1)! \pmod p$$ 
where $(p-1)! \equiv -1 \pmod p$ by Wilson's theorem. The integer $\frac{(p-1)!}{k}$ modulo $p$ is the negative reciprocal of $k$ in $\mathbb{Z}/(p)$, and since the map $a \mapsto -a^{-1}$ is a bijection on $U_p$, the sum in parentheses is the same as the sum of $\sum_{k=1}^{p-1} k$ modulo $p$. 
