Two subgroups $A,B \le G$ permute if $AB = BA$. All groups considered are finite. I have a question on some points of the following proof of Lemma 2.
Lemma 1: Let $H \le G$, and let $\mathcal X$ be a collection of subgroups of $G$, each of which permutes with $H$. Then $\langle \mathcal X \rangle$ permutes with $H$.
Lemma 2: Let $Q$ and $H$ be respectively a $q$-subgroup and a $\pi$-subgroup of $G$, where $q$ is prime and $\pi$ is a set of primes. Suppose that $H$ permutes with every conjugate of $Q$ in $G$. Then there exists a $(\pi \cup \{ q \})$-subgroup $M$ containing $Q$ and normalized by every conjugate of $H$.
Proof: Observe that all conjugates of $H$ in $G$ permute with all conjugates of $Q$. Let $\mathcal X$ be the family of all $(\pi \cup \{q\})$-subgroups of $G$ that are generated by conjugates of $Q$, and note that $Q \in \mathcal X$. Let $M \ge Q$ be a maximal member of $\mathcal X$, and let $U$ be an arbitrary conjugate of $H$ in $G$. Since the conjugates of $Q$ that generate $M$ all permute with $U$, if follows by the above Lemma 1 that $M$ permutes with $U$. Then $MU$ is a subgroup with order dividing $|M||U| = |M||H|$, and since $M$ is a $(\pi \cup \{ q\})$-group, so too is $MU$. If $t \in MU$, therefore, the subgroup $\langle M, M^t \rangle$ is a $(\pi\cup \{q\})$-group generated by conjugates of $Q$, and so $\langle M, M^t \rangle$ lies in $\mathcal X$. This subgroup, therefore, cannot properly contain $M$, and hence $M^t = M$. It follows that $M \unlhd MU$, and hence $U$ normalizes $M$. $\square$
The points I do not understand:
1) "Observe that all conjugates of $H$ in $G$ permute with all conjugates of $Q$.
I do not see that?
2) "If $t \in MU$, therefore, the subgroup $\langle M, M^t \rangle$ is a $(\pi \cup \{q\})$-group generated by conjugates of $Q$, and so $\langle M, M^t \rangle$ lies in $\mathcal X$".
Why is $\langle M, M^t \rangle$ a $(\pi \cup \{q\})$-group generated by conjugates of $Q$, I do not see the connection?
1) $H^{x}Q^{y} = (HQ^{yx^{-1}})^{x} = (Q^{yx^{-1}}H)^{x} = Q^{y}H^{x}$ for any $x,y \in G.$
2) $\langle M,M^{t} \rangle$ is a subgroup of $MU$, so is certainly a $(\pi \cup \{q \})$-group. Since $M$ is generated by conjugates of $Q,$ as is $M^{t},$ for it is conjugate to $M$. Hence $\langle M,M^{t} \rangle$, being generated by $M$ and $M^{t}$, is still generated by conjugates of $Q.$