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Can someone make an example to show that the product of two elements of finite order In a non-abelian group is not necessarily of finite order.

My goal of asking this question, is introducing a nice counterexample from matrix space and I know there are so many example about this question. thanks for referencing of these examples.

Amin235
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    What is an "Annabelle group"? Do you mean abelian? – hmakholm left over Monica Mar 21 '16 at 12:17
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    I suspect this is yet another installment of "Damn you, Autocorrect!" – Asaf Karagila Mar 21 '16 at 12:19
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    http://math.stackexchange.com/questions/41303/examples-and-further-results-about-the-order-of-the-product-of-two-elements-in-a, http://math.stackexchange.com/questions/59791/do-there-exist-groups-whose-elements-of-finite-order-do-not-form-a-subgroup, http://math.stackexchange.com/questions/1497991/give-an-example-of-a-nonabelian-group-in-which-a-product-of-elements-of-finite-o, ...and so many links can be found duplicate to this. – p Groups Mar 21 '16 at 12:21

4 Answers4

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Take the group of all rigid motions of the plane, and let $p$ and $q$ be rotations by $180^\circ$ about two different points.

Then $pq$ is a nontrivial translation, which cannot have finite order.

hmakholm left over Monica
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In the group of invertible linear maps from reals to reals under composition, let $f(x)=-x$ and $g(x)=1-x,$ each of order $2.$ Then $g(f(x))=1+x$ which has infinite order (it is translation by $1$).

Since this group can be made into a matrix group by taking $f(x)=mx+b$ to be the matrix with first row $[m,b]$ and second row $[0,1]$ it gives a matrix example of your requirement. (Then $m \neq 0$ is the nonzero determinant condition for the matrix group, and one needs $m$ nonzero for considering $mx+b$ as an invertible linear map.)

coffeemath
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You can consider $\mathbb{Z}_2*\mathbb{Z}_2$ as $\langle a,b| a^2=b^2=1\rangle$. Clearly $a,b$ have finite order, but $ab$ hasn't finite order.

InsideOut
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It is a counterexample from an article called (A General Q-Matrix) by (JOHN IVIE):

http://www.fq.math.ca/Scanned/10-3/ivie-a.pdf

Define a matrix $Q_r$ by: $$ Q_r=\left({\begin{array}{ccccccc} 1&1&0&0&\cdots&0&0\\ 1&0&1&0&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&1&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&\ddots&1&0\\ 1&\vdots&\ddots&\ddots&\ddots&\ddots&1\\ 1&0&\cdots&\cdots&\cdots&\cdots&0\end{array}}\right)_{(r)\times (r)}\, . $$ Now let us define the generalized Fibonacci sequence $(f_{n,r})$ for $(r\geq 2)$ by: $$f_{n,r}=f_{n-1,r}+f_{n-2,r}+\cdots+f_{n-r,r}.$$ with $(f_{0,r}=f_{1,r}=\cdots=f_{r-2,r}=0 \, , \, f_{r-1,r}=1)$ . Note that $r = 2$ gives the Fibonacci number. we can see that $(Q_r^n)$ has a closed form based on $(f_{n,r})$. We now use the matrix $Q_r$ to show that the product of two elements of finite order In a non-abellan group is not necessarily of finite order. Let: $$ R_{r}=\left({\begin{array}{ccccccc} -1&0&0&\cdots&0&0\\ 0&1&0&\ddots&\ddots&\vdots\\ \vdots&\ddots&1&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\ddots&\ddots&\ddots&1&0\\ 0&0&0&\cdots&0&1\end{array}}\right)$$ $$ S_{r}=\left({\begin{array}{ccccccc} -1&-1&0&\cdots&\cdots&0\\ 1&0&1&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&\ddots&\vdots\\ \vdots&\vdots&\ddots&\ddots&\ddots&0\\ \vdots&\vdots&\ddots&\ddots&\ddots&1\\ 1&0&0&\cdots&\cdots&0\end{array}}\right) $$ be elements of the group of Invertible square matrices, then: $$R_r^2=S_r^{r+1}=I_r\, .$$ so $(R_r)$ and $(S_r)$ are of finite order, but ${(R_rS_r)}^n=Q_r^n \neq I_r$ for all n, so that $(R_rS_r)$ Is not of finite order.

Amin235
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