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Let $U \le G$ be a subgroup of the finite group $G$ of odd order such that $|N_G(U) : U| = 2$ and different conjugates of $U$ intersect trivially, i.e. $U^g \cap U = 1$ for $g \notin N_G(U)$. Suppose $N \unlhd G$ and $N \cap U \ne 1$, then there exists an involution $t$ such that with $T = \langle t \rangle$

(a) $G = TUN$

(b) if $t \notin N$, then $UN$ is a Frobenius group with $U$ as complement, and $TU$ has a normal complement in $G$;

(c) if $t \in N$, then $t$ centralizes $U/(U\cap N)$, and $U \cap N$ has the same properties in $N$ as $U$ in $G$, i.e. it has index two in its normalizer $|N_N(U\cap N) : U\cap N| = 2$ and $$(U\cap N)^n \cap (U\cap N) = 1$$ if $n \in N \setminus N_G(N\cap U)$.

Hint: for (a) use the Frattini argument on $U \cap N$, and for (c) use it on $T$.

As $U$ has odd order and index two in its normalizer, we find an involution $t \notin U$ normalizing $U$ such that $N_G(U) = TU$. For (a) if $P \in \mbox{Syl}_p(U \cap N)$, as $U \cap N \unlhd U$ we can apply the Frattini argument, which gives $$ U = N_U(P)(N \cap U) \le N_U(P)N $$ but I do not see why this gives $G = TUN$?

For (b), the first claim follows if no element from $N \setminus U$ can normalize $U$. If $U^n = U$ then $n \in TU$, and we have to exclude $n = tu$, but I do not see how to do that? Also for the second claim, $N$ could not be the this complement for $TU$, as $1 \ne N \cap U \le N \cap TU$ has nontrivial intersection. So what should be this complement?

Just part (c) is clear to me. I can supply a proof if wanted of (c) [but I had not used the Frattini argument in my proof of (c)].

But how to prove part (a) and (b)? I do not see how to fill the gaps!?

StefanH
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  • A Sylow $p$-subgroup $P$ of $U \cap N$ must be a Sylow $p$-subgroup of $N$, since otherwise it would be strictly contained in its normalizer in $N$, contradicting the trivial intersection of $U$ with its conjugates. So $G = N_G(P)N$, and $N_G(P) \le N_G(U)=TU$, giving (a). – Derek Holt Jan 09 '16 at 21:24
  • @DerekHolt Thank you, it brought me on the right track. Now I also have a proof that $U^x = U$ with $x \notin U$ implies $x = t$, which gives (b) too. I will write up an answer, but may I ask just because I am curious. Do you have an idea how to solve (c) with Frattini? I said I have a solution by "direct computation", but I do not see how Frattini would be of any help here, and I do not see how applying Frattini to a group of order $2$ (which is not supposed to be normal) is possible or of any help here? – StefanH Jan 09 '16 at 22:05
  • It is not true that $U^x=U$ with $x \not\in U$ implies $x=t$. This is true if and only if $x=tu$ for some $u \in U$. But, in your attempt to prove (b), $n=tu \Rightarrow t \in N$. – Derek Holt Jan 09 '16 at 22:48
  • @DerekHolt But for the choosen $t$ normalising $U$ this always holds, as $U^x = U$ implies $x \in N_G(U) = TU$, hence $x = tu$ for some $u \in U$? And still would be interested how to apply Frattini for (c)... – StefanH Jan 09 '16 at 22:58
  • @DerekHolt: I thought I proved $x = tu$ with $u \in U, x \notin U$ implies $x = t$, but I noticed that my reasoning was faulty: as $x^2 \in U$ and $x = x^{-1}x^2 = tu$ I guessed as every element of $TU$ could be uniquely written that we must have $x^{-1} = t$, hence $x = t$, but that is not valid. But I do not see how to "repair" this proof, could you help me getting a sound proof? – StefanH Jan 10 '16 at 13:53
  • Okay, I see that I somehow confused what you were saying. You stated that this is false, so of course I cannot prove it. – StefanH Jan 10 '16 at 16:27
  • Sorry for asking so many questions, but for (b) why $n = tu \Rightarrow t \in N$? I still cannot see it... – StefanH Jan 10 '16 at 18:06
  • $t = nu^{-1}$ and $|U|$ is odd, so some odd power of $t$ lies in $N$. – Derek Holt Jan 10 '16 at 18:13
  • @DerekHolt Okay, I see! I wanted to write everything up: but at (a) I am still stuck: Let $P \in \mbox{Syl}_p(U \cap N)$ be nontrivial. Then $P$ is also a Sylow subgroup of $N$. For otherwise there exists $Q \in \mbox{Syl}_p(N)$ properly containing $P$, and hence $P < N_Q(P)$ in the $p$-group $Q$. Let $g \in N_Q(P)$, but not in $P$. Then $P \le (U \cap N) \cap (U\cap N)^g \le U \cap U^g$, hence $U = U^g$, which gives $g \in TU$. Also $g$ is not from $U \cap N$, hence $g \notin U$ and we have $g = tu$.This would also give $t \in N$, but I do not see any contradiction? – StefanH Jan 10 '16 at 19:46
  • Well I think you should see a contradiction from $g=tu$. – Derek Holt Jan 10 '16 at 20:12
  • @DerekHolt For what should I look for? I see that $g^2 \in U \cap N$ and for $g^2 = x \in U$ by order considerations we have $|g| = |x|$. Also $g^2 \in N_Q(P)$, and so $g^2 \in Q \cap U = P$, also $u^2 = (tgt)g \in N$, guess this leads me nowhere, so where I have to search for the contradiction? – StefanH Jan 10 '16 at 20:42
  • OK, this is definitely my last comment on this post, so please do not ask me any more questions! $g$ has odd order and $tu$ has even order. – Derek Holt Jan 10 '16 at 20:44

1 Answers1

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I just collect and write up what was written by Derek in his comments.

(a) Let $p$ be a prime dividing $U \cap N$, then every Sylow $p$-subgroup $P$ of $U \cap N$ is also a Sylow $p$-subgroup of $N$. For suppose not, then there exists $Q \in \mbox{Syl}_p(N)$ with $P < Q$, and by the nilpotency of $Q$ we have $P < N_G(P) \cap Q$. Choose some $g \in N_G(P) \cap Q$ not in $P$. As $P = P^g \le U^g$ and $P \ne 1$ this implies $U^g = U$, hence $g \in N_G(U) = TU$ where we must have $g = tu$ as $g \notin U$. But $tu$ has even order, as $tu \notin U$ and so $$ 2|U| = \frac{|\langle tu \rangle||U|}{|\langle tu \rangle \cap U|} $$ and the denominator and $|U|$ have odd order, or more simply note that $U$ is normal in $TU$ with index $2$, hence $(tu)^2 \in U$ (which is not valid for non-normal subgroups) or by direct computation $tutu = ttu'u = u'u \in U$ and by this fact. But as $p$ is an odd prime $g$ has odd order, and this contradiction shows that every Sylow $p$-subgroup of $U \cap N$ is also a Sylow $p$-subgroup of $N$.

Also as already mentioned above we have $N_G(P) \le N_G(U) = TU$, hence by Frattini $G = N_G(P)N \le TUN$.

(b) For $UN$ to be a Frobenius group with complement $U$, we must have for every $g \in UN$ not in $U$ that $U^g \cap U = 1$. We can suppose $g \in N$, and we have to exclude that $U^g = U$, then by assumption $U^g \cap U = 1$ would follow. If $U^g = U$, then $g = tu$, or $t = gu^{-1}$. Now let $k = |U|$, then $t^k = (gu^{-1})^k = (u^{-1})^kg'$ where $g' \in N$ by successively using $nu = un'$ for $n \in N$ and normality of $N$. And as $u^{-k} = 1$ and $t^k = t$ as $k$ is odd, this would imply $t \in N$, which is excluded. Hence $UN$ is a Frobenius group. Denote its kernel by $K$.

i) $TU \cap K = 1$

As $N \unlhd UN$ we have $K \le N$ or $N < K$ (this is a property of the kernel). Hence as $t \notin N$ we have $t \notin K$. Now suppose $tu = k$ with $k \in K$ and $u \in U$, then $t = ku^{-1}$ and with a similar trick as under (a) this implies $t^n \in K$ for some odd number $n$, hence $t \in K$, which is excluded. This gives i). Or simpler: If $k = tu \in K$ with $k\ne 1$, then $U^{tu} = U$ as $t \in N_G(U)$, but $U^k\cap U = 1$ as $k$ is from the kernel. Hence $TU\cap K = 1$.

ii) $K$ is normal in $G$.

We have $UN \unlhd G$ with (a) and $(UN)^t = U^t N^t = UN$, and as $K$ is characteristic in $UN$ it is normal in $G$.

iii) $G = TUK$.

Simply as $UN = UK$ we have $G = TUN = TUK$.

(c) [Without Frattini, and no idea how Frattini might be of use here...] If $t \in N$ then $[t,u] \le N$, and $[t,u] \le U$ because $U^t = U$, hence $[t, U] \le U \cap N$ and $t$ centralizes $U / (U\cap N)$. Suppose $(U \cap N) = (U\cap N)^g = U^g \cap N$, as $1 \ne N \cap U = N\cap U^g \le U \cap U^g \cap N$, this implies $U^g = U$, hence $N_G(U\cap N) = N_G(U)$. Now $2 \ge |N_G(U) \cap N : U \cap N|$ and we have $t \in N_N(U)$ but $t \notin U \cap N$, hence $2 = |N_G(U) \cap N : U \cap N| = |N_N(U\cap N) : U \cap N|$. Lastly if $n \in N \setminus N_G(N\cap U)$, then $n \notin N_G(U)$, hence $(U\cap N) \cap (U\cap N)^n = U\cap U^n \cap N = 1$.

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