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Let $G$ be a residually finite group. Show that if $G$ has finitely many conjugacy classes of elements of finite order then $G$ has a torsion free finite index subgroup.

Not sure how to get started on this one!

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Let $x_1, \cdots, x_r$ be the representives for these conjugacy classes of finite orders. Since $G$ is residually finite, there exists $N_i \lhd G$ of finite index in $G$ such that $x_i \not \in N_i$ for each $i$. Let $N=\cap N_i$, then $N$ is what you need.

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I suppose that should be $\;x_{\color{red}i}\notin N_i\;$ ...? – DonAntonio Feb 11 '14 at 11:14
Yes, fixed. Thanks. – Wei Zhou Feb 11 '14 at 11:30
Good. Nice and helpful hint for the OP to reach the solution. +1 – DonAntonio Feb 11 '14 at 11:35

Hint: Start by constructing (finitely many) homomorphisms to finite groups which do not send representatives of conjugacy classes of finite order elements to 1. Now think about kernels of these homomorphisms.

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