Let $\displaystyle S$ be a proper subgroup of $\displaystyle G$. If $\displaystyle G-S$ is the complement of $\displaystyle S$, prove that $\displaystyle <G-S>=G$.
Thanks in advance.
clearly $\displaystyle <G-S> \subseteq G.$ now let $\displaystyle g \in G.$ if $\displaystyle g \in G - S,$ then $\displaystyle g \in <G-S>.$ if $\displaystyle g \in S,$ then choose $\displaystyle x \in G-S.$ then $\displaystyle x^{-1} \in G-S$ and $\displaystyle gx=y \in G-S.$ so: $\displaystyle g=yx^{-1} \in <G-S>.$ Q.E.D.