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)?