Finitely many conjugacy classes of finite subgroups of given order Let $G$ be a periodic locally soluble  group with finite Sylow p-subgroups for all primes p.
It is know that in these conditions $G$ is residually finite. Moreover it can be proved that $G$ has only finitely many conjugacy classes of finite subgroups of given order.
Let $L$ be a subgroup of $G$ such that $L=HN$, where $H$ is a finite subgroup of $G$ and $N$ is a normal subgroup of $G$.
Why the index $|N_G(L):N_G(H)N|$ is finite?
I know this is true since this is stated at Lemma 1.6 of the paper 
"Locally inner endomorphisms of SF-Groups" by Belyaev 

You know that there are only finitely many conjugacy classes of subgroups of $HN$ that are isomorphic to $H$.
Supose $g_1,g_2 \in N_G(L)$ and $H^{g_1}$ and $H^{g_2}$ are in the same conjugacy class in $HN$. Since $g_1 \in N_G(L)$, we have $HN=H^{g_1}N$, so there exists $n \in N$ with $H^{g_1n}=H^{g_2}$, so $g_1ng_2^{-1} = g_1g_2^{-1}(g_2ng_2^{-1}) \in N_G(H)$ and hence, since $N$ is normal in $G$, $g_1 \in N_G(H)Ng_2$.
So the number of cosets of $N_G(H)N$ in $N_G(L)$ is at most equal to the number of conjugacy classes of subgroups of $HN$ that are isomorphic to $H$, which is finite.
