Finite order In a non-abelian group 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.
 A: 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.
A: 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.)
A: 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.
A: 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.
