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Let $G$ be a finite group with subgroup $U \le G$ of odd order such that $U\ne U^g$ implies $U\cap U^g = 1$. Suppose there exists an involution $t \notin U$ such that $N_G(U) = TU$ with $T = \langle t \rangle$ (and hence $U$ has index two in its normalizer).

Let $N$ be a maximal normal subgroup disjoint from $UT$. Then $N$ contains all involutions not conjugate to $t$ and is nilpotent.

Suppose $x \notin N$ is an involution, then we have to show that $x$ is conjugate to $t$. Maybe it would be helpful to look at $M = \langle x^g : g \in G \rangle$, but I do not see how. Also for the nilpotency of $N$, I have no idea what nilpotency criterion to apply, maybe that the Sylow subgroups are normal. So why do the stated properties about $N$ hold?

EDIT: To give some context, the question arose from this paper, see the argumentation in the proof of Theorem 5. It seems to be related to Lemma 4 of the paper, where it is said:

[...] If $t$ does not centralize $U/U'$, then there is a characteristic subgroup $N$ of $G$ such that $N \cap TU = 1$ and $N$ contains all involutions not conjugate to $t$.

In the proof of Theorem 5 they state "[...] $t$ does not centralize $(M \cap U)/(M \cap U)'$ [...]", where $M$ is normal with $t \in M$. But why that is true I do not see either, see my other recent post, and why this maybe implies that $t$ does not centralize $U'/U$ [the only thing I can see is that if $t$ does centralize $U/U'$ we would have $[u,t] \in U'$ and if also $u \in M$ then $[u,t] \in U' \cap M'$ as $t \in M$, but I am not sure if that implies $[u,t] \le (U\cap M)'$, as I guess we do not have $U'\cap M' \le (U\cap M)'$]. Also I guess they take the normal subgroup whose existence is claimed by Lemma 4 and enlarge it to a maximal one, because the lemma does not say that it has to hold for all maximal normal subgroups intersecting $TU$ trivially.

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  • $\begingroup$ It's nilpotent because $NU$ is a Frobenius group. I'm not sure about the other property. $\endgroup$ – Derek Holt Jan 14 '16 at 12:03
  • $\begingroup$ @DerekHolt Okay, thank you! Yes by definition $n \in N$ implies $n \notin N_G(H)$, hence $U^n \cap U = 1$ and $UN$ is a Frobenius group. $\endgroup$ – StefanH Jan 14 '16 at 15:58
  • $\begingroup$ I am sceptical about the claim that $N$ contains all involutions not conjugate to $t$. Have you missed out any assumptions? $\endgroup$ – Derek Holt Jan 14 '16 at 17:45
  • $\begingroup$ @DerekHolt: See my edit. Sorry now I see that it probably is related to Lemma 4 from the cited paper (I put everything in the edit), I had not noticed this before. $\endgroup$ – StefanH Jan 14 '16 at 18:35
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I have written a detailed answer which entails also this question in an answer to a question I asked a while ago.

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