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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.

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up vote 4 down vote accepted

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.

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+1, nice result. – Marek Jan 15 '13 at 12:13
See also for a detailed proof of the result for nilpotent groups. Basic questions like this have usually been asked and answered previously! – Derek Holt Jan 15 '13 at 15:51

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) $\mathbb Z$.

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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.

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@user1729 Ah, I ignored the restriction as it came after the question amark, sorry ... – Hagen von Eitzen Jan 15 '13 at 11:41
That's okay - this is not my favourite group which has torsion elements which multiply to give non-torsion! It is much more intuitive that $D_{\infty}$! – user1729 Jan 15 '13 at 11:44
(Sorry, this is now my favourite group which has......) – user1729 Apr 26 '13 at 9:22

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.

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@user1729: Should I hold this example here or I am supposed to delete it? – Babak S. Jan 15 '13 at 11:57
Hold it here - it is a nice example (everyone knows about matrices!) and I make reference to it in my answer. – user1729 Jan 15 '13 at 11:58
Hold it here! It is a great example! +1 – amWhy Feb 15 '13 at 0:07

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