# Subset of order > 1/2 of order G

If G is a finite goup and let $A \subseteq G$. if $|A|> |G|/2$ then prove that for every $g \in G$ there exists $$h,k \in A$$ such that $hk=g$ (Headbang)
If G is a finite goup and let $A \subseteq G$. if $|A|> |G|/2$ then prove that for every $g \in G$ there exists $$h,k \in A$$ such that $hk=g$ (Headbang)
For any $x$ there exists a unique $y$ so that $xy=g$, so we will define $\widehat x=y$. Assume that for any $a\in A$ we have $\widehat a\not \in A$ i.e. for any $a\in A \implies \widehat a \in G-A$. Since $|A| > |G-A|$ by pigeonhole principle there exists two elements, $a,b\in A, a\not = b$ that satisfy $\widehat a = \widehat b$ but this means $a=b$, a contradiction.