# 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?

• This follows from Langrange's Theorem - $\#\langle S\rangle\mid \#G$, so what are the possibilities for $\#\langle S\rangle$? – Robert Chamberlain Feb 28 '18 at 15:10
• I read somewhere else already about a hint on Lagrange's theorem, which is also covered in the course I'm following. However, we're assuming $S$ is a subset and not a subgroup, so can this theorem be applied directly? – Václav Mordvinov Feb 28 '18 at 15:13
• You can't apply it to $S$ but you can apply it to the subgroup generated by $S$. – Robert Chamberlain Feb 28 '18 at 15:14
• $S$ might be a subset but $\langle S \rangle$ sure is a subgroup – krirkrirk Feb 28 '18 at 15:15
• Thanks, I see why it works now! :) – Václav Mordvinov Feb 28 '18 at 15:20

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.