On order of elements of a infinite group Let $a$ and $b$ be elements of finite order of an infinite  group $G$. Then do we can say that order $ab$ is finite?
I think this is true for Nilpotent Groups. 
 A: No, we can't. Take for example the modular group. This group can be generated by elements $S^2 = 1$ and $(ST)^3 = 1$. Yet $S(ST) = S^2 T = T$ is a translation which generates (a group isomorphic to) $\mathbb Z$.
A: Another one is $GL(2,\mathbb Q)$. Take $A=\begin{pmatrix}
  0 & -1 \\
 1 & 0
\end{pmatrix}, B=\begin{pmatrix}
  0 & 1 \\
 -1 & -1
\end{pmatrix}$. Try to show that $A^4=B^3=E$ but $AB$ has infinite order. This shows that if the group $G$ is not abelian then $tG$, including the torsion elements of $G$, may not be a subgroup.
A: Consider the group of euklidean movements of the plane.
Reflections at lines have finite order $2$, their product can be an arbitrary translation or rotation.
A: As has been pointed out in the other answers, an infinite group can have two torsion elements which multiply to give a non-torsion element. However, this is not the case if the group is nilpotent.
The following is Theorem 5.2.7. from Derek J. S. Robinson's fine text "A Course in the Theory of Groups". It clearly answers your question.
Theorem: Let $G$ be a nilpotent group. Then the elements of finite order in $G$ form a fully-invariant subgroup $T$ such that $G/T$ is torsion-free and $T=Dr_pT_p$ where $T_p$ is the unique maximum $p$-subgroup of $G$.
I will omit the proof - look up Robinson's book, or any other advanced text which has something to say about Nilpotent groups.
To see that the examples are not nilpotent groups you have to know that $D_{\infty}$ and any non-cyclic free group are not nilpotent, and that subgroups of nilpotent groups are nilpotent. Hagen von Eitzen's group clearly contains $D_{\infty}$ while the other two contain free groups of arbitrary rank.
