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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

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  • $\begingroup$ 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 $N_G(L)$? $\endgroup$
    – Derek Holt
    Sep 29, 2017 at 16:47
  • $\begingroup$ 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? $\endgroup$
    – W4cc0
    Sep 29, 2017 at 18:27

1 Answer 1

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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.

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  • $\begingroup$ You'll not believe but I've just figure out what you meant with your comment by self and was writing to you to not worry for more explanation. But, since you come first, thanks :) $\endgroup$
    – W4cc0
    Sep 29, 2017 at 18:54

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