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?

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

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


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

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 $GL_2(\mathbb{R})$.


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