Commutator of two element in GL(2,5)

[Silversonic]

Homework Statement

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.

The Attempt at a Solution

My professor's answers say

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

But I get the answer

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

I won't go into 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.

 

[Dick]

Homework Statement

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.

The Attempt at a Solution

My professor's answers say

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

But I get the answer

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

I won't go into 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.

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.

 

[Silversonic]

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.

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.

 

[Dick]

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.

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