Maximal subgroup of a finite group contains all other subgroups.

Question

A maximal subgroup $M$ of a finite group $G$ is a proper subgroup of $G$ such that no proper subgroup $H$ exists such that $H$ contains $M$. I want to prove that all other proper subgroups of $G$ except $M$ are contained in $M$. I want to prove it by contradiction. Take an arbitrary proper subgroup $H$ of $G$ not contained in $M$. We have a contradiction if we can show that $H\cup M$ is a subgroup of $G$. Now, we can use this lemma:

For a finite group $G$, a subset $H$ of $G$ is a subgroup of $G$ if and only if $H\neq \emptyset $ and $x, \; y \in H$ implies $xy\in H$.

In $H\cup M$, there are 9 types of elements. Let's consider them separately:

1. $x, \; y\in H\backslash M$. This implies $x,\; y\in H$ and from the lemma $xy\in H$. Now, note that $x,\; y \in H\cup M$ and $xy\in H\cup M$. So, from the lemma, $H\cup M$ is a subgroup of $G$.
2. $x\in H\backslash M, \; y \in H\cap M$. This implies $x, \; y \in H$ and the proof follows from the previous case.
3. $x\in H\backslash M, \; y \in M\backslash H$. This is where I'm stuck. Any suggestion would be helpful.

Answer 1

I want to prove that all other proper subgroups of $G$ except $M$ are contained in $M$.

No, it is false. Counter-example, Klein 4 group. $G=\{e, a,b,c \}$, where $\{e,a \}$, $\{e,b\}$, $\{e,c\}$ are maximal subgroup, and none of them contains each other.