about maximal proper subgroup

Question

If in group $\;G\;$ there exists an element $\;g\in G\;$ and a proper subgroup $\;H\le G\;$ s.t. $\;g\notin H\;$, then there exists a maximal proper subgroup $\;M\le G\;\;s.t.\;\;g\notin M\;$ . is this true ? or is this true for abelian group ?

Answer 1

There exists a subgroup $M$ which is maximal with respect to the property $g\notin M$, but the subgroup may not be maximal in the “absolute sense”.

The proof uses Zorn's lemma. Consider the family $\mathscr{M}$ of all subgroups $H$ of $G$ such that $g\notin H$. The family is not empty by assumption.

Since the union of a chain (totally ordered set) of subgroups is a subgroup and if none of them contains $g$ neither their union does, Zorn's lemma provides the existence of a maximal element in $\mathscr{M}$.

However, this subgroup need not be maximal among all (proper) subgroups: just take a group that has no maximal subgroups to begin with.

There are also examples in the finite case: take the group of elementary quaternions $Q$; then $-1$ is the example you are looking for, because every nontrivial subgroup contains $-1$.