Ah from my previous question, If $|a|$, $|b|$ is finite then $|ab|$ is finite? I made stupid assumption that $G$ is finite.
Assume arbitrary $G$. of finite order $a,b$ i mean,
Let $a,b\in G$, Assume the order of element $|a|$, $|b|$ is finite, then what i want to know is $|ab|$ finite?
No: consider the modular group $G=\mathrm{PSL}_2(\mathbb{Z})$ and the elements defined by the matrices $$ S=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$$ and $$ T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$$
One can check that $S^2=(ST)^3=1$, but $T=S(ST)$ has infinite order.
Knowing that $|a|=|b|=2$ tells us nothing about $|ab|,$ because every permutation is the product of two involutions; this is Exercise 10.1.17 on p. 259 of W. R. Scott's Group Theory.
For a concrete example, define two permutations $a,b$ of $\mathbb Z$ as follows: $$a(n)=n-(-1)^n$$ $$b(n)=n+(-1)^n$$ Then $a(a(n))=b(b(n))=n,$ i.e., $|a|=|b|=2;$ but $|ab|=\infty,$ since $a(b(n))=n+2$ for even $n.$
