$\DeclareMathOperator{\GL}{GL}$
Let $ℂ$ denote complex numbers and let $G=\GL(2, ℂ)$ be the group of all invertible $2 × 2$ matrices with complex coefficients. Show the following
- Any finite abelian subgroup $A$ of $\GL(2, ℂ)$ is simultaneously diagonalizable, i.e., there exists $P ∈ G$ such that $P A P^{-1}$ is a subgroup consisting of all diagonal matrices.
- There exists an infinite abelian subgroup of $\GL(2, ℂ)$ that is not diagonalizable.
- There exists a finite subgroup $H$ of $\GL(2, ℂ)$ that is not diagonalizable.

Proof.
- If $A$ consists of all diagonal matrices, then there's nothing to prove. So let $S ∈ A$ and assume $S$ is not diagonal. Since $S$ is contained in a finite group, the minimal polynomial for $S$ divides $x^n-1$ for some $n ∈ ℤ$. This implies, since $S$ is not diagonal, that $S$ has precisely two distinct eigenvalues, say $λ$ and $η$. Let $V_λ$ and $V_η$ be the eigenspaces for $λ$ and $η$, respectively. Both $V_λ$ and $V_η$ are 1-dimensional spaces generated by, say, $\vec{v}_λ$ and $\vec{v}_η$, respectively. Let $T ∈ A$. Since $S$ and $T$ commute, we have
$$
TS\vec{v}_λ=λ T\vec{v}_λ=ST\vec{v}_λ
$$
So $T \vec{v}_λ ∈ V_λ$. Since $V_λ$ is 1-dimensional, it follows that $\vec{v}_λ$ is an eigenvalue for $T$, as well. Similarly, $\vec{v}_η$ is also an eigenvector for $T$. Since $T$ was arbitrary, we have that $\vec{v}_λ$ and $\vec{v}_η$ are non-zero linearly independent eigenvectors for all matrices in $A$. Hence $A$ is simultaneously diagonalizable.
- Take the cyclic group generated by an invertible Jordan block of size 2,
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
The minimal polynomial for the above matrix is $(x-1)^2$, hence the group generated by this matrix is infinite, and since it's cyclic, it's definitely abelian.
- Finally, let $H$ be the subgroup generated by
$$
A=\begin{pmatrix}
ζ & 0 \\
0 & ζ^2
\end{pmatrix} \text{ and } B=\begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}
$$
where $ζ$ is a 3-rd root of unity. It's easy to verify that $B A B^{-1}=A^2$. So the group generated by $A$ and $B$ is non-abelian. Moreover, the order of the group generated by $A$ and $B$ is of order $6$, hence, is isomorphic to $S_3$. This group is non-diagonalizable since if it were diagonalizable, then it would be isomorphic to an abelian group by Q1.

# A finite group which is not isomorphic to a subgroup of $\GL(2,ℂ)$

$C_2×C_2×C_2$ is the smallest such group. To see this, note that abelian subgroups of $\GL(2,ℂ)$ are diagonalizable. Thus if $G$ is an abelian subgroup of $\GL(n,ℂ)$ then $G$ is generated by at most $n$ elements.