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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!!

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    $\begingroup$ Any idea about what? You are basically given the solution. take these elements, show that they have finite order, take their product, show it does not have finite order. $\endgroup$ – Najib Idrissi Nov 8 '13 at 23:56
  • $\begingroup$ @nik - How do I show this: take their product, show it does not have finite order? Thank you! $\endgroup$ – CS1 Nov 9 '13 at 8:13
  • $\begingroup$ But their product have a finite order - 6... $\endgroup$ – CS1 Nov 9 '13 at 8:27
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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.

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  • $\begingroup$ If I'll multiply them I'll get:$\begin{bmatrix} 1 &1 \\ -1& 0 \end{bmatrix}$ and the order of this matrix is finite - is 6, so I don't understand how can I show that their product a matrix with infinite order... Thank you! $\endgroup$ – CS1 Nov 9 '13 at 8:25
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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

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    $\begingroup$ This doesn't really answer the question, it sounds just like a restatement of the original post. $\endgroup$ – Tyler Nov 14 '13 at 5:50

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