# 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.