Generating a finite group with a subset of more than $\frac{1}{p}$ elements Sorry if this is a duplicate, but I couldn't find any related question. I'm working on the following exercise 

Let $G$ be a finite group with more than $1$ element, en $S\subset G$ a subset such that $\#S>\frac{1}{p}\#G$, with $p$ the smallest prime divisor of the order of G. Prove that $\langle S\rangle=G$.

I managed to prove this in the kind of trivial case where $2|\#G$: observe that $S$ and $S^{-1}$ cannot be disjoint, so for an arbitrary $x\in G$, $\exists a\in xS^{-1}\cap S$; that is: for certain $s_1\in S^{-1}$, $s_2\in S$, $xs_1^{-1}=a=s_2$, so that $x=s_1s_2$, and since $x\in G$ was arbitrary, $\langle S\rangle=G$.
However, how do i prove this for the general case that $\#S>\frac{1}{p}\#G$, with $p$ the smallest prime divisor of the order of G?

Put $H=\langle S \rangle$ then $H$ is a subgroup and $S \subseteq H$. If $H \subsetneq G$, then the index $|G:H| \gt 1$. But $|G:H| \leq |G|/\#S \lt p$. Since $p$ is the smallest prime dividing $|G|$ and $|G:H|$ divides $|G|$, this is a contradiction.
