Is $SL_2(\mathbb{C})$ a solvable group?

Question

Recently we showed in class that the group $SL_2(\mathbb{R}) $ is not solvable since the commutator subgroup of $SL_2(\mathbb{R}) $ is equal to $SL_2(\mathbb{R}) $, implying that the group $SL_2(\mathbb{R}) $ is not solvable. However, it involved some tedious calculations and correct guessing of the generators of $SL_2(\mathbb{R}) $.

Anyway, I was wondering whether we can reach the same conclusion for $SL_2(\mathbb{C}) $, meaning that $SL_2(\mathbb{C}) $ is also the commutator subgroup of $SL_2(\mathbb{C}) $ and thus not solvable.

Answer 1

Actually, the proof works for all fields except for $\Bbb F_2$ or $\Bbb F_3$.

Theorem: Let $K$ be a field with at least $4$ elements. Then the group $SL_2(K)$ is perfect, i.e., its commutator subgroup equals $SL_2(K)$.

Proof: See here, Proposition $2.4$. Note that $SL_2(\Bbb F_4)\cong A_5$.

Answer 2

Following @DerekHolt's suggestion, let $N$ be the commutator subgroup of $SL_2(\Bbb{C})$. Then $SL_2(\Bbb{R})\subseteq N$ because $SL_2(\Bbb{R})$ is its own commutator subgroup and a subgroup of $SL_2(\Bbb{C})$. $$ \begin{pmatrix} \exp(i\theta)&0\\ 0&\exp(-i\theta) \end{pmatrix}= \begin{pmatrix} \frac{1+I}{2}&\frac{-1+I}{2}\\ \frac{1+I}{2}&\frac{1-I}{2} \end{pmatrix} \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix} \begin{pmatrix} \frac{1-I}{2}&\frac{1-I}{2}\\ \frac{-1-I}{2}&\frac{1+I}{2} \end{pmatrix}\tag{1} $$ This proves $\begin{pmatrix}\exp(i\theta)&0\\0&\exp(-i\theta)\end{pmatrix}\in N$ because $\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\in SO(2)\subseteq SL_2(\Bbb{R})$ and the conjugation in $(1)$ is by the matrix $\begin{pmatrix}\frac{1+I}{2}&\frac{-1+I}{2}\\\frac{1+I}{2}&\frac{1-I}{2}\end{pmatrix}\in SU(2)\subseteq SL_2(\Bbb{C})$.

Next, Let $\lambda>0$ (in $\Bbb{R}$). Then $$ \begin{pmatrix} \lambda&0\\ 0&\frac{1}{\lambda} \end{pmatrix}\in SL_2(\Bbb{R})\subseteq N\tag{2} $$ Combining $(1), (2)$, $$ \begin{pmatrix} \lambda \exp(i\theta)&0\\ 0&\lambda^{-1}\exp(-i\theta) \end{pmatrix}\in N\tag{3} $$

now let $A\in {SL}_2(\Bbb{C})$. If $A$ is diagonalizable, then its eigenvalues are $\lambda \exp(i\theta), \lambda^{-1} \exp(-i\theta)$ for some $\lambda>0,\theta\in\Bbb{R}$. Then $A$ is similar via some $S A S^{-1}$ to the matrix in $(3)$ for $S\in GL_2(\Bbb{C})$. By multiplying $S$ by some complex scalar, we can get $\det(S)=1$ so $S\in SL_2(\Bbb{C})$. Hence $A$ is a conjugate of the matrix in $(3)$ so $A\in N$.

Finally, if $A$ is not diagonalizable, its Jordan form is $$ \begin{pmatrix} z&1\\ 0&z \end{pmatrix}\tag{4} $$ where $z\in\Bbb{C}$ and $z^2=\det A=1$. So $z\in\{\pm 1\}$ and the matrix in $(4)$ is in $SL_2(\Bbb{R})$ and therefore $N$. By conjugation we again get that $A\in N$. This completes the proof that $N=SL_2(\Bbb{C})$.