Center of a non-abelian subgroup of $GL(2, \mathbb{C})$

I'm trying to do the following exercise:

Let be $$G$$ a non-abelian subgroup of $$GL(2, \mathbb{C})$$. Prove that the center of $$G$$ is contained in the center of $$GL(2, \mathbb{C})$$

My (very partial) attempt:

Suppose that there is a matrix $$A$$ in the center of $$G$$ but not in that of $$GL(2, \mathbb{C})$$. But the center is a normal subgroup, so it contains all the conjugates of $$A$$. In particular the center of $$GL(2, \mathbb{C})$$ does not contain the matrix $$J$$, the canonical Jordan form of $$A$$. Instead there is a subgroup isomorphic to $$G$$ that contains $$J$$.

My idea is to prove that it is abelian against the hypothesis. I know that $$J$$ could have only one of these two form: diagonal (but not a multiple of identity) and a Jordan block. But I don't know how to continue...

First, suppose that $$J$$ is diagonal but not scalar. That is, $$J=\begin{pmatrix}a&0\\0&b\end{pmatrix}$$ with $$a\neq b$$. Let $$X=\begin{pmatrix}p&q\\r&s\end{pmatrix}$$ be an element of $$G$$. Then, $$XJ=JX$$ yields $$\begin{pmatrix}ap&bq\\ar&bs\end{pmatrix}=\begin{pmatrix}ap&aq\\br&bs\end{pmatrix}.$$ That is, $$q=0$$ and $$r=0$$. Thus, every matrix in $$G$$ is diagonal, and so $$G$$ is abelian.

Finally, we suppose that $$J$$ is a nontrivial Jordan block. Then, $$J=\begin{pmatrix}k&1\\0&k\end{pmatrix}=kI+E$$ with $$k\neq 0$$, where $$E=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$. Again, let $$X=\begin{pmatrix}p&q\\r&s\end{pmatrix}$$ be an element of $$G$$. Then, $$XJ=JX$$ yields $$\begin{pmatrix}kp&p+kq\\kr&r+ks\end{pmatrix}=\begin{pmatrix}kp+r&kq+s\\kr&ks\end{pmatrix}.$$ This shows that $$r=0$$ and $$p=s$$, so $$X=pI+qE$$. Hence, $$G$$ contains only polynomials in $$E$$, and they all commute.

• I don't think that $\;J\;$ of yours is a Jordan block...A $\;2\times2\;$ Jordan block is of the form $\;\begin{pmatrix}a&1\\0&1\end{pmatrix}\;,\;\;a\in\Bbb C\;$ ... – DonAntonio Oct 18 '18 at 16:35
• Yep, sorry, I confused myself. – user593746 Oct 18 '18 at 16:37
• I supposed so, yet in this case all the computations in the second part change and that's why it is important to edit your answer. – DonAntonio Oct 18 '18 at 16:38

Observe that $$\;\Bbb C\;$$ is algebraically closed, thus the characteristic polynomial of any element $$\;A\in K:=GL(2,\Bbb C)\;$$ splits over $$\;\Bbb C\;$$ , which means the basic forms of equivalence classes are of the form

$$T_1:=\left\{\;\begin{pmatrix}a&0\\0&b\end{pmatrix}\;/\;a,b\in\Bbb C\;\right\}\;,\;\;T_2:=\left\{\;\begin{pmatrix}a&1\\0&a\end{pmatrix}\;/\;a\in\Bbb C\;\right\}$$

Suppose now that a matrix $$\;A\;$$ of the form $$\;T_2\;$$ (i.e., non diagonalizable and thus with only one eigenvalue) is in $$\;Z(G)\;,\;\;\text{with}\;\;G\le K\;$$ non-abelian, and let $$\;B=\begin{pmatrix}x&y\\z&w\end{pmatrix}\in G\;$$ , then:

$$AB=\begin{pmatrix}ax+z&ay+w\\az&aw\end{pmatrix}=\begin{pmatrix}ax&x+ay\\az&z+aw\end{pmatrix}\implies z=0,\,x=w$$

so in fact $$\;B=\begin{pmatrix}x&y\\0&x\end{pmatrix}\;$$ ...but a group with this kind of elements is abelian, as you can check at once.

You can check that the same outcome is true if an element of $$\;T_1\;$$ , with $$\;a\neq b\;$$ , is in the center of $$\;G\;$$ . Thus, only an element of $$\;T_1\;$$ with $$\;a=b\;$$ can be in the center of the subgroup...but these are precisely the elements in the center of $$\;K\;$$ ...!

Now fill in details