0
$\begingroup$

The set of all diagonal matrices with nonzero determinant is a normal subgroup of GL2(R).

I know you need to prove the conjugate in order for it to be a normal subgroup, but I am not sure where to go from there. I am thinking that this can be disproved. Do I just pick any diagonal matrix and another 2 x 2 matrix and plug them into the conjugate AHA^-1 to show the end result is not a diagonal matrix?

$\endgroup$
  • $\begingroup$ If you want to disprove something, it suffices to give a counterexample. $\endgroup$ – Bulberage Mar 26 '14 at 0:38
1
$\begingroup$

As you suspect, this is false (think about diagonalisable matrices that are themselves nor diagonal).

Let

$$A = \begin{bmatrix} 1& 0\\ 0&2\end{bmatrix}, \; P = \begin{bmatrix}-1& 1\\ 1& 0\end{bmatrix}.$$

Then $PAP^{-1} = B$ where

$$B = \begin{bmatrix}2& 1\\ 0& 1\end{bmatrix}$$

so we've conjugated $A$ to something outside this subgroup and it's hence not normal in $GL_2(\mathbb{R})$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.