3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

You were having a good idea. If there is $g \not \in \langle H \cup M\rangle$, then $\langle H \cup M\rangle\neq G$ is a subgroup and it strictly contains $M$. This is against the maximality of $M$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Well, if $H$ is not contained in $M$, then $\langle H\cup M\rangle$ is a subgroup of $G$ which contains $M$ as a proper subgroup. Since $M$ is maximal, it follows that $\langle H\cup M\rangle =G$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

I think a less stressful approach would be to start by observing that $H'=\langle H\cup M\rangle$ is a subgroup of $G$. Since $M$ is a maximal subgroup of $G$, that gives us either that $H'=M$ or $H'=G$. Then proceed towards your conclusion.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy