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...

Any suggestion? Thanks in advance.


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.

  • 1
    $\begingroup$ I don't think that $\;J\;$ of yours is a Jordan block...A $\;2\times2\;$ Jordan block is of the form $\;\begin{pmatrix}a&1\\0&1\end{pmatrix}\;,\;\;a\in\Bbb C\;$ ... $\endgroup$ – DonAntonio Oct 18 '18 at 16:35
  • $\begingroup$ Yep, sorry, I confused myself. $\endgroup$ – user593746 Oct 18 '18 at 16:37
  • $\begingroup$ 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. $\endgroup$ – 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


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