$\mathcal{B}$ is maximal Abelian subalgebra iff it concides with its commutant

Question

I'm reading the book Hilbert Space Operators in Quantum Physics and I come to Chapter 6: Operator sets and algebras. Before reading this chapter, I need to first read the Appendix B, where I met my problem, which is Proposition B.1.2.

Definition. A maximal Abelian subalgebra is $\color{red}{\text{an Abelian subalgebra(after editting)}}$ such that it is not a proper subalgebra of an Abelian subalgebra.

Problem. Suppose that $\mathcal{A}$ is an algebra, i.e. a complex associative linear algebra, and $\mathcal{B}$ is a subalgebra in $\mathcal{A}$. Then $\mathcal{B}$ is maximal Abelian iff $\mathcal{B}=\mathcal{B}'$. Here $\mathcal{B}'$ is the commutant: $$\mathcal{B}'=\{a\in\mathcal{A}: ab=ba,b\in\mathcal{B}\}.$$

The sufficiency is OK for me. If $\mathcal{B}=\mathcal{B}'$ and $\mathcal{B}\subset\mathcal{T}$ for some Abelian subalgebra $\mathcal{T}$, then $$\mathcal{T}\subset\mathcal{T}'\subset\mathcal{B}'=\mathcal{B},$$ so $\mathcal{B}=\mathcal{T}$. By definition, $\mathcal{B}$ is maximal Abelian.

I'm stuck on the neccessity so I'm asking for some help here. If $\mathcal{B}$ is maximal Abelian, $\style{text-decoration:line-through}{\text{I wonder whether $\mathcal{B}$ is Abelian. If this is ture (though I can't prove it),}}$ then we know that $\mathcal{B}''$ is Abelian and $\mathcal{B}\subset\mathcal{B}''$ implies that $\mathcal{B}=\mathcal{B}''$. Also, if $\mathcal{B}$ is Abelian, then $\mathcal{B}\subset\mathcal{B}'$. But whatever, we cannot conclude that $\mathcal{B}=\mathcal{B}'$.

I'm new to this algebraic topic. If this question is too stupid, forgive me. Any help would be appreciated.

EDIT: According to the comments, I misunderstood the meaning of the book about the definition of maximal Abelian subalgebra (fixed now). So now my problem is how to prove the neccessity part.

Answer 1

Suppose that $B$ is maximal Abelian. Take $x\in B^\prime$. Then $xa=ax$ for all $a\in B$. Thus the algebra $C$ generated by $x$ and $B$ is Abelian. By maximality, $C= B$. As $x\in B^\prime$ was arbitrary, $B^\prime \subseteq B$. The converse inclusion follows from the fact that $B$ is Abelian.