Prove that the numerator of $H_{p-1}$ in reduced form is a multiple of $p$ for $p$ an odd prime Prove that for any odd prime $p$
$$
H_{p-1}=1+\frac{1}{2} + \cdots + \frac{1}{p-1}
$$
contains a multiple of $p$ in the numerator when written in reduced form, i.e. $\frac{a}{b}$ where $\mathrm{gcd}(a,b)=1$.
 A: The operation of reciprocation in $\mathbb{Z}/p \mathbb{Z}$ is a one-to-one map, thus reciprocals of all positive integers $1,2,\ldots,p-1$ would be permutations thereof. The sum of permutated numbers is the same as the sum of ordered numbers, i.e.
$$
   \sum_{k=1}^{p-1} \frac{1}{k} \equiv \sum_{i=1}^{p-1} i \equiv p \cdot \frac{p-1}{2} \equiv 0 \mod p
$$
A: Write it as
$$H_{p-1} = \frac{\frac{(p-1)!}{1} + \dots + \frac{(p-1)!}{p-1}}{(p-1)!}.$$
Then the denominator is not divisible by $p$, so it is enough to prove that the numerator is. Now $\mathbb{Z}_p$ is a field so we have actually
$$\frac{(p-1)!}{1} + \dots + \frac{(p-1)!}{p-1} = (p-1)!(1 + \dots + (p-1)) = 1 + \dots + (p-1) = \frac{p(p-1)}{2}$$
which is $0$.
Here we have repeatedly used the fact that $\mathbb{Z}_p \setminus \{0\}$ is a group under multiplication. Taking inverses or multiplying by a group element are bijective operations.
A: Because $p$ is odd, the indices in the sum can be grouped into $(p-1)/2$ pairs $\{ i , p-i \}$ and in each pair the sum is divisible by $p$.
Stronger statements mod $p^2$ and $p^3$ are known as Wolstenholme's congruences.
http://en.wikipedia.org/wiki/Wolstenholme%27s_theorem
