[proofplan]
We use the defining quotient $L(\lambda)=M(\lambda)/J(\lambda)$ and the uniqueness of the maximal proper submodule of the Verma module. First, any nonzero proper submodule of $L(\lambda)$ pulls back to a proper submodule of $M(\lambda)$ strictly containing $J(\lambda)$, contradicting maximality. For uniqueness, a simple highest weight module $V$ of highest weight $\lambda$ receives a surjective map from $M(\lambda)$ by the universal property of the Verma module; the kernel is a maximal proper submodule, hence equals $J(\lambda)$, so $V$ is the same quotient as $L(\lambda)$.
[/proofplan]
[step:Show that the quotient by $J(\lambda)$ has no nonzero proper submodules]
Let
\begin{align*}
\pi:M(\lambda)\to L(\lambda)
\end{align*}
denote the quotient homomorphism, so $\ker \pi=J(\lambda)$. Suppose $N\subset L(\lambda)$ is a $\mathfrak g$-submodule. Define its inverse image
\begin{align*}
\widetilde N:=\pi^{-1}(N)\subset M(\lambda).
\end{align*}
Because $\pi$ is a $\mathfrak g$-[module homomorphism](/page/Module%20Homomorphism), $\widetilde N$ is a $\mathfrak g$-submodule of $M(\lambda)$, and $J(\lambda)\subseteq \widetilde N$.
If $N$ is proper in $L(\lambda)$, then $\widetilde N$ is proper in $M(\lambda)$; otherwise $\widetilde N=M(\lambda)$ would imply $N=\pi(\widetilde N)=L(\lambda)$. By [citetheorem:9378], $J(\lambda)$ is the unique maximal proper submodule of $M(\lambda)$. Hence every proper submodule of $M(\lambda)$ containing $J(\lambda)$ is equal to $J(\lambda)$. Therefore $\widetilde N=J(\lambda)$, and so
\begin{align*}
N=\pi(\widetilde N)=\pi(J(\lambda))=0.
\end{align*}
Thus the only proper submodule of $L(\lambda)$ is $0$, and $L(\lambda)$ is simple.
[guided]
Let
\begin{align*}
\pi:M(\lambda)\to L(\lambda)
\end{align*}
be the canonical quotient map. By the definition
\begin{align*}
L(\lambda):=M(\lambda)/J(\lambda),
\end{align*}
we have $\ker\pi=J(\lambda)$.
To prove that $L(\lambda)$ is simple, we must show that it has no nonzero proper $\mathfrak g$-submodule. Let $N\subset L(\lambda)$ be a $\mathfrak g$-submodule, and define
\begin{align*}
\widetilde N:=\pi^{-1}(N)\subset M(\lambda).
\end{align*}
This inverse image is a $\mathfrak g$-submodule of $M(\lambda)$ because $\pi$ is a $\mathfrak g$-module homomorphism: if $m\in \widetilde N$ and $x\in\mathfrak g$, then $\pi(xm)=x\pi(m)\in N$, since $N$ is stable under the action of $\mathfrak g$.
Also $J(\lambda)\subseteq \widetilde N$, because $J(\lambda)=\ker\pi$ and $0\in N$. Now suppose that $N$ is proper in $L(\lambda)$. Then $\widetilde N$ must be proper in $M(\lambda)$: if $\widetilde N=M(\lambda)$, then applying $\pi$ gives
\begin{align*}
N=\pi(\widetilde N)=\pi(M(\lambda))=L(\lambda),
\end{align*}
contradicting the assumption that $N$ is proper.
By [citetheorem:9378], the submodule $J(\lambda)$ is the unique maximal proper submodule of $M(\lambda)$. Since $\widetilde N$ is a proper submodule and contains $J(\lambda)$, maximality forces
\begin{align*}
\widetilde N=J(\lambda).
\end{align*}
Applying $\pi$ gives
\begin{align*}
N=\pi(\widetilde N)=\pi(J(\lambda))=0.
\end{align*}
Thus every proper submodule of $L(\lambda)$ is zero. Since $L(\lambda)$ is a quotient of the nonzero Verma module by a proper submodule, it is nonzero. Therefore $L(\lambda)$ is simple.
[/guided]
[/step]
[step:Obtain a surjective map from the Verma module to any simple highest weight module]
Let $V$ be a simple highest weight $\mathfrak g$-module of highest weight $\lambda$. By the meaning of highest weight module, choose a highest weight vector $v\in V$ such that $v\ne 0$,
\begin{align*}
xv=0 \quad \text{for every } x\in\mathfrak n^+,
\end{align*}
and
\begin{align*}
hv=\lambda(h)v \quad \text{for every } h\in\mathfrak h,
\end{align*}
and such that $V$ is generated by $v$ as a $\mathfrak g$-module.
By [citetheorem:9375], applied to the $\mathfrak g$-module $V$ and the vector $v$, there exists a unique $\mathfrak g$-module homomorphism
\begin{align*}
\varphi:M(\lambda)\to V
\end{align*}
such that $\varphi(v_\lambda)=v$, where $v_\lambda$ denotes the highest weight vector of $M(\lambda)$. Since $M(\lambda)$ is generated by $v_\lambda$, the image $\operatorname{im}\varphi$ is the $\mathfrak g$-submodule of $V$ generated by $v$. Hence
\begin{align*}
\operatorname{im}\varphi=V.
\end{align*}
Thus $\varphi$ is surjective.
[/step]
[step:Identify the kernel with the maximal proper submodule]
Define
\begin{align*}
K:=\ker\varphi\subset M(\lambda).
\end{align*}
Since $\varphi$ is a $\mathfrak g$-module homomorphism, $K$ is a $\mathfrak g$-submodule. Since $\varphi(v_\lambda)=v\ne 0$, the kernel $K$ is proper.
The quotient $M(\lambda)/K$ is isomorphic to $V$ by the [first isomorphism theorem for modules](/theorems/862), because $\varphi$ is surjective. Since $V$ is simple, $K$ is a maximal proper submodule of $M(\lambda)$. By the uniqueness of the maximal proper submodule in [citetheorem:9378],
\begin{align*}
K=J(\lambda).
\end{align*}
[/step]
[step:Conclude that every simple highest weight module is the quotient $L(\lambda)$]
Since $\ker\varphi=J(\lambda)$, the [first isomorphism theorem](/theorems/791) gives a $\mathfrak g$-module isomorphism
\begin{align*}
M(\lambda)/J(\lambda)\cong V.
\end{align*}
By the definition of $L(\lambda)$, this is
\begin{align*}
L(\lambda)\cong V.
\end{align*}
Therefore every simple highest weight $\mathfrak g$-module of highest weight $\lambda$ is isomorphic to $L(\lambda)$, completing the proof.
[/step]
