I want to prove if $M$ is maximal subgroup of $G$ and $H$ is a subgroup of $G$, then $H\leq M$ or $\langle H\cup M\rangle=G.$
So there are two cases to consider. When $H\leq M$ and $H\not\leq M$. We need to show if $H\not\leq M$, then $\langle H\cup M\rangle=G$. To show this, I tried to use a contradiction that if there exists an element $g\in G$ such that $g\not\in \langle H\cup M\rangle$, then a contradiction. However I couldn't reach a contradiction. Any help is appreciated.