Prove that $GL_n(\mathbb{C}) $ is not solvable

Question

How should I prove that $GL_n(\mathbb{C}) $, the group of invertible matrices over the complex numbers, is not solvable?

I have no idea how to prove this by supposing that $GL_n(\mathbb{C}) $ is solvable and deducing a contradiction. Does anyone have ideas?

Any hints or advices will help a lot!

Answer 1

Let $K=\Bbb{C}$. The derived group of $GL_n(K)$ is $GL_n(K)^{(1)}=SL_n(K)$, which is perfect for $n\ge 2$. Hence the derived series cannot end with the trivial group, because $$ GL_n(K)^{(k)}=SL_n(K)\neq 1 $$ for all $k\ge 1$.

Answer 2

Soluble groups are closed under taking subgroups. So work in $GL_2(\mathbb{Z})$, which is a subgroup of $GL_n(\mathbb{Z})$, and try and find your favourite non-soluble group as a subgroup. What this subgroup is depends on your background...

My favourite non-soluble subgroup of $GL_2(\mathbb{C})$ is the subgroup generated by the matrices $$ \begin{align*} \left(\begin{array}{cc}1&2\\0&1\end{array}\right),\:\: \left(\begin{array}{cc}1&0\\2&1\end{array}\right) \end{align*} $$ which is free of rank two.

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An alternative approach is to use finite groups (but this is easiest for $n\geq5$):

The $n\times n$ permutation matrices are those $n\times n$ matrices that have exactly one entry of $1$ in each row and each column and $0$s elsewhere. For each $n$, these form a group isomorphic to $S_n$. Hence, $S_n$ embeds into $GL_n(\mathbb{C})$. For $n\geq5$, $S_n$ is not soluble and so we've found the required non-soluble subgroup.

Answer 3

I filled the detail of Dietrich Burde's answer as below:

Note that every commutator of $G=GL_n(\mathbb{C})$ has determinant 1, so $[G,G]\le SL_n(\mathbb{C}).$ Also, by appropriately scaling the elements in $G$(such scaling is justified since $\mathbb{C}$ is algebraically closed.), one can easily check that $[G,G]=[SL_n(\mathbb{C}),SL_n(\mathbb{C})]$. Now it remains to prove that $[SL_n(\mathbb{C}),SL_n(\mathbb{C})]=SL_n(\mathbb{C})$. Since $SL_n(\mathbb{C})$ is generated by the matrices of the form $E_{ij}(\lambda)=I_n+\lambda e_{ij}$, $i\neq j,\:\lambda\in \mathbb{C}$, where $e_{ij}$ is the matrix with the $(i,j)$ entry is 1 and the other entries are zero, it suffices to show that $E_{ij}(\lambda)$ is a commutator of two elements in $SL_n(\mathbb{C})$.

If $n=2$, then

$$ \begin{pmatrix} 1 & \lambda \\ 0 & 1 \\ \end{pmatrix}= \begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix} \begin{pmatrix} 1 & -\lambda/2 \\ 0 & 1 \\ \end{pmatrix} \begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix} \begin{pmatrix} 1 & \lambda/2 \\ 0 & 1 \\ \end{pmatrix} $$

and

$$ \begin{pmatrix} 1 & 0 \\ \lambda & 1 \\ \end{pmatrix}= \begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ -\lambda/2 & 1 \\ \end{pmatrix} \begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix} \begin{pmatrix} 1 & 0 \\ \lambda/2 & 1 \\ \end{pmatrix} $$

and if $n\ge 3$, for each $i\neq j$ we can choose $l$ distinct from both $i$ and $j$, and observe that

$E_{ij}(\lambda)=E_{il}(\lambda)E_{lj}(1)E_{il}(-\lambda)E_{lj}(-1)$

and we are done. For the calculation of the RHS above, the relation $e_{ij} e_{kl}= \delta_{jk}e_{il}$, where $\delta_{jk}$ is the Kronecker delta, will help.