Question about Abelian group proof

Question

I prove that if $G$ is Abelian group so if $a,b\in G$ has a finite order so $ab$ has a finite order to.. (Maybe later I'll upload here my proof to see of she is correct....)

Now, I have to show that this is false if the group is not Abelian group with those 2 matrices: $\begin{bmatrix} 0 &-1 \\ 1& 1 \end{bmatrix} and \begin{bmatrix} 0 &1 \\ -1&-1 \end{bmatrix}$

This is the problem:

$\begin{bmatrix} 0 &-1 \\ 1& 1 \end{bmatrix}\cdot \begin{bmatrix} 0 &1 \\ -1&-1 \end{bmatrix}=\begin{bmatrix} 1 &1 \\ -1&0 \end{bmatrix}$

$ord\left(\begin{bmatrix} 1 &1 \\ -1&0 \end{bmatrix}\right)=6$

This is not an infinite order element...

An you have any idea?

Thank you!!

Answer 1

You probably saw the group $GL(2,\mathbb R)$, perhaps not under this name. It is the group of all invertible $2\times 2$ matrices with real entries. It is a group under the operation of matrix multiplication. So, this question is probably asking you to identify that the two matrices belong to that group. Then you proceed, a la nik's advise, to compute, in $GL(2,\mathbb R)$, the order of each, the product of the two, and the order of the product.

Answer 2

we are showing if group is non abelian then product two elements of group whose orders are finite but order of their product may be infinite and in abelian it does't happen. you are doing good job on groups