Lie Algebra of su(2)

This is a simple exercise before jumping into the general theory of Lie algebra classification. The commutator relations in Lie algebra $\mathfrak{su}(2)$ are \begin{equation}\Big[{\cal{J}}_a\,,{\cal{J}}_b\Big]=i\,\epsilon_{abc}\,{\cal{J}}_c\end{equation} for $a,b,c=1,2,3$. We construct an irreducible representation in the following steps:

1.  Pick any operator, say ${\cal{J}}_3$, and denote its eigenvector by $|m\rangle$: \begin{equation}{\cal{J}}_3\,|m\rangle = m\,|m\rangle\,.\tag{1}\end{equation}
2. For the remaining ${\cal{J}}_1$ and ${\cal{J}}_2$, construct a set of new operators \begin{equation}{\cal{J}}_{\pm}\equiv {\cal{J}}_1 \pm i {\cal{J}}_2\,.\tag{2}\end{equation} The commutator relations become \begin{eqnarray}\Big[{\cal{J}}_3\,,{\cal{J}}_{\pm}\Big]=\pm {\cal{J}}_{\pm}\,,\quad \Big[{\cal{J}}_+\,,{\cal{J}}_-\Big]=2\,{\cal{J}}_3\,.\tag{3}\end{eqnarray}
3. From the relation ${\cal{J}}_3\,{\cal{J}}_{\pm}\,|m\rangle = \Big[{\cal{J}}_3\,,{\cal{J}}_{\pm}\Big]\,|m\rangle + {\cal{J}}_{\pm}\,{\cal{J}}_{3}|m\rangle=(m\pm 1){\cal{J}}_{\pm}\,|m\rangle$, we know that \begin{equation} {\cal{J}}_{+}\,|m\rangle = a_m\, |m+1\rangle\,,\quad  {\cal{J}}_{-}\,|m\rangle = b_m\, |m-1\rangle \end{equation} for constants $a_m$ and $b_m$ to be determined. ${\cal{J}}_{+}$ and ${\cal{J}}_{-}$ are the raising and lowering operator on $|m\rangle$.
4. For a finite representation, there must be a state $|j\rangle$ such that \begin{equation}{\cal{J}}_{+} |j\rangle =0\,,\tag{4}\end{equation} which suggests that $a_j=0$.
5. Solve $a_m$ and $b_m$: 
1. Note that $\langle m+1|{\cal{J}}_{+}\,|m\rangle=\langle m|{\cal{J}}_{-}\,|m+1\rangle^*$, and thus $a_m =b_{m+1}^*$. 
2. Then from $\langle m|\Big[{\cal{J}}_+\,,{\cal{J}}_-\Big]\,|m\rangle=2 \langle m|{\cal{J}}_3\,|m\rangle$, we obtain a recursion relation $\left|a_{m-1}\right|^2-\left|a_{m}\right|^2=2m$.
3. The above recursion form suggests the ansatz $\left|a_{m}\right|^2=x m^2 + y m + z$. By submitting this ansatz into the recursion relation and using the boundary condition $a_j=0$, we can solve $x=y=-1$ and $z=j(j+1)$. As a result, we solve \begin{equation}{\cal{J}}_{\pm}|m\rangle =\sqrt{j(j+1)-m(m\pm 1)}\,|m\pm 1\rangle\,.\tag{5}\end{equation}
6. For completeness, define ${\cal{J}}^2\equiv {\cal{J}}^2_1+{\cal{J}}^2_2+{\cal{J}}^2_3={\cal{J}}^2_3={\cal{J}}_{+}{\cal{J}}_{-}-{\cal{J}}_3+{\cal{J}}^2_3$, then we have \begin{equation}{\cal{J}}^2|j\rangle = {\cal{J}}_{+}{\cal{J}}_{-}|j\rangle + (j^2-j)|j\rangle=j(j+1)\,|j\rangle\,.\tag{6}\end{equation}

Remarks:

1. There is no invariant subspace since any vector in the subspace will spread to the entire representation space when applied by ${\cal{J}}_{\pm}$ operators. Therefore, the representation is irreducible.
2. Since the representation is finite, there must be another state $q$ such that ${\cal{J}}_{-}\,|q\rangle=0$. By Eq. (5), we have $\sqrt{j(j+1)-q(q-1)}=0$, from which we can solve $q=-j$. Therefore, $m$ takes values from $-j$ to $j$ with increment 1 and thus $j$  must take values of integers or half integers.
3. The above construction can be generalized to the so-called highest weight method for general (semisimple) Lie algebra:
• We start by picking a set of compatible operators, similar to ${\cal{J}}_3$ in Eq. (1) in $\mathfrak{su}(2)$, which forms the so-called Cartan subalgebra
• One can construct the raising and lowering operators, similar to ${\cal{J}}_{\pm}$ in Eq. (2) in $\mathfrak{su}(2)$, with the help of adjoint representation. Similar commutator relations to Eq. (3) in $\mathfrak{su}(2)$ hold as well.
• The irreducible representation is also constructed from a state of "highest weight", similar to $|j\rangle$ in Eq. (4) in $\mathfrak{su}(2)$.
• The proof that $j$ in $\mathfrak{su}(2)$ must be integer or half-integer can be generalized to prove that there are only a few different (semisimple) Lie algebra. 
• There is the so-called Casimir operator, similar to ${\cal{J}}^2$ in Eq. (6) in $\mathfrak{su}(2)$.