Let $G$ be a group and $M$ be a maximal abelian subgroup of $G$. Prove $C_M(g)\lneq M$ for every $g\in G-M$. Question: Let $G$ be a group and $M$ be a maximal abelian subgroup of $G$.  Prove $C_M(g)\lneq M$ for every $g\in G-M$.
Thoughts: I figured I could appeal to Zorn's Lemma here, but I was really trying not to do that.  I "see" why this is true... $M$ must be maximal among abelian subgroups, so if we take any element not in $M$, then the centralizer over $M$ must be proper in $M$.... but I suppose I never thought of "how" this is true, even though I've used it.  Again, I would like a more direct argument (even though I suppose it is) over Zorn's Lemma, but I can't quite think of anything that has led to anything useful.  Any help is greatly appreciated!

Let $g\in G$. Note that because $M$ is abelian, $\langle C_M(g),g\rangle$ is abelian. If $C_M(g)=M$, then $M\leq \langle C_M(g),g\rangle$, so the maximality of $M$ gives $M=\langle C_M(g),g\rangle$, hence $g\in M$.
By contrapositive, if $g\notin M$, then $C_M(g)\neq M$.
