# 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

• Doesn't this follow from the fact that $HN$ has only finitely many conjugacy classes of subgroups isomorphic to $H$ - and hence only finitely many such that are conjugate to $H$ in $N_G(L)$? – Derek Holt Sep 29 '17 at 16:47
• Yes, it should follow from these facts, but how? I mean, yes, $H$ has finitely many conjugacy classes in $N_G(L)$, but I cannot connect this fact to the index $|N_G(L):N_G(H)N|$. Could you be more explicit? – W4cc0 Sep 29 '17 at 18:27

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.