Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in the following which I believe is known:

Let $S$ be a subset of a finite group $G$ containing more than half of $G$'s elements. Then $S+S = G$.

I have been looking but can not find a reference. Does anyone know of one? Or know a nice proof?


share|cite|improve this question
up vote 7 down vote accepted

$\newcommand{\Size}[1]{\lvert #1 \rvert}$Let $g \in G$, and write $G$ multiplicatively.

By the inclusion-exclusion principle $$ \Size{A \cup B} = \Size{A} + \Size{B} - \Size{A \cap B}, $$ the sets $A = g S^{-1}$ and $B = S$ must have a non-trivial intersection...

[EDIT: the ending] so there are $s_{1}, s_{2} \in S$ such that $g s_{2}^{-1} = s_{1}$ or $g = s_{1} s_{2} \in S \cdot S$.

share|cite|improve this answer
So $g{s}^{-1}$ is in $S$... If $S$ were a subgroup I'd agree that we're done, but I don't see why we are in the case where $S$ is just a subset... Can you explain? – user160371 Jan 23 at 15:11
$gs^{-1} \in S$ is equivalent to $g \in S + S$, which is what you are trying to prove. I am guessing that by $S+S$ you mean $\{ st : s,t \in S \}$. – Derek Holt Jan 23 at 15:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.