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I have an infinite group $G$ with $H \leq G$ and $[G:H] < \infty$. But i can not see why there are finite conjugates $gHg^{-1}$. I know that $\{gHg^{-1} \mid g \in G\} \cong \{gN_G(H) \mid g \in G \}$ and $H \leq N_G(H)$. Note that $N_G(H)$ is the normalizer of $H$ in $G$. How can I solve this ?

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Recall that if $G\ge N\ge H$ then $[G:H]=[G:N][N:H]$. If the LHS is finite, what does that tell you about $[G:N]$? –  anon Oct 20 '12 at 17:13
$[N:H]$ is finite because $[G:H]$ is finite and so $[G:N]$ must be finite. –  André Oct 20 '12 at 17:22
@André: Are you searching for a group satisfiyng $H\le G$ with $[G:H]<\infty$? –  Babak S. Oct 20 '12 at 17:55
I think what André is going after is the nice theorem: if a subgroup has finite index in a group then there is a normal subgroup of finite index in G which, btw, is contained in the first subgroup. –  DonAntonio Oct 20 '12 at 18:49
@André: nobody seems to be sure exactly what you are asking! Could you clarify? –  Derek Holt Oct 20 '12 at 19:53

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