Normal subgroups in group having a t.i. subgroup which has index two in its normalizer

Let $G$ be a finite group and $U \le G$ a subgroup such that for each $g \notin N_G(U)$ we have $$U \cap U^g = 1$$ (a so called t.i. subgroup). Further suppose $|N_G(U) : U| = 2$. Also suppose $t$ is an involution normalizing $U$ and $T = \langle t \rangle$. Then $N_G(U) = TU$.

Lemma: If $N \unlhd G$ and $N \cap U \ne 1$, then

(a) $G = TUN$;

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

(c) if $t \in N$, then $t$ centralizes $U/(U\cap N)$.

Proof: Statement (a) follows from the Frattini argument applied to a Sylow subgroup of $U \cap N$; (b) is immediate; and (c) again follows from the Frattini argument applied to $T$. $\square$

I do not understand this proof. If we apply the Frattini argument to a Sylow subgroup $P$ of $U \cap N$, which is possible as $U \cap N \unlhd U$, we get $U = N_U(P)(N\cap U)$. But how does this implies $G = TUN$? In (b) what should be the normal complement? I guess $N$ is it not, as we have $1 \ne U\cap N \le TU \cap N$, i.e. the last intersection is nontrivial. And in (c), as $|T| = 2$ this group is its own Sylow $2$-subgroup, but I do not see where $T$ is normal in or some way to apply it to get (c)?

• If the prime is odd, then $P \in {\rm Syl}_p(N)$, so $G = N_G(P)N \le TUN$. If $P \cap N$ is a $2$-group, then I think it would work better to apply the Frattini argument to $TU \cap N \in {\rm Syl}_2(N)$ and again $G = N_G(TU \cap N)N \le TUN$. – Derek Holt Nov 21 '15 at 18:23
• To show that $P \in \mbox{Syl}_p(N)$ I have to show that $|N : P|$ is not divisible by $p$. We have $|N : P| = |N : N \cap U||N\cap U : P|$, so it is left to show that $|N : N \cap U|$ is not divisble by $p$. But this I do not see? I see that $|N : N \cap U| = |N : N_N(U)||N_N(U) : N\cap U|$ and that $|N_N(U) : N\cap U| \le 2$ by the assumptions, but that is all I get? Also why $TU \cap N \in \mbox{Syl}_2(N)$? As $2$ is the highest $2$-power dividing $TU$, this is equivalent to showing that $|N| = 2m$ with $(2,m) = 1$, but here I am also stuck... – StefanH Nov 21 '15 at 19:31
• All these things follow from $U$ beig a t.i. subgroup. – Derek Holt Nov 21 '15 at 19:49
• If $p$ is odd and divides $|N \cap U|$ and also $|N:N\cap U|$, then you would have a $p$-element outside of $U$ normalizing $N \cap U$, contradicting $U$ being a t.i. set. Sorry, but that's my last comment on this! – Derek Holt Nov 21 '15 at 21:11
• I will however comment briefly on (b) and (c). The normal complement in (b) is the Frobenius kernel of the Frobenius group $UN$. In (c), if $t \in N$ then clearly $[t,U] \le U \cap N$, so $t$ centralizes $U/U \cap N$. – Derek Holt Nov 21 '15 at 21:39