Commutator of two element in GL(2,5)

1. Dec 20, 2013

Silversonic

1. The problem statement, all variables and given/known data

This is more about checking a solution I've been given is correct, because either my professor is consistently getting this wrong or I am.

3. The attempt at a solution

$[ \left( \begin{array}{ccc} 1 & 0 \\ 0 & 4 \end{array} \right), \left( \begin{array}{ccc} 1 & 4 \\ 0 & 1 \end{array} \right)] = \left( \begin{array}{ccc} 4 & 1 \\ 0 & 4 \end{array} \right)$

$\left( \begin{array}{ccc} 1 & 3 \\ 0 & 1 \end{array} \right)$

I won't go in to explicit detail unless someone asks because I can't be bothered to do the latex, but can anyone confirm which answer is right? My professor does a lot of these through the solutions and I seem to be getting contradictory answers in each case.

2. Dec 20, 2013

Dick

Assuming GL(2,5) means invertible 2x2 matrices mod 5 and commutator means [A,B]=AB-BA, I don't get either answer. You don't have to TeX, just explain what you mean.

3. Dec 20, 2013

Silversonic

Apologies, the the commutator of two elements x and y in a group is denoted [x,y], meaning

$[x,y] = x^{-1}y^{-1}xy$

But everything else you said was correct, this is the group of invertible 2x2 matrices mod 5.

4. Dec 20, 2013

Dick

Ah, ok. That makes more sense. If it's the group commutator not the matrix commutator, then I agree with your answer.