# Permutable subgroups of finite groups

I have a question about the following theorem that I found in some research.

How did the author conclude that $PQ=QP$ at the end of the proof?

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thank you very much. –  user28083 Jul 19 '12 at 20:31
It would be clearer if you wrote $g=xa$ rather than $g=ax$. Then $Q^{xa} \subseteq B$ implies $Q^x \subseteq B$, since $a \in B$. Then $PQ^x = QP^x$, and conjugating by $x^{-1}$ gives $PQ=QP$. –  Derek Holt Jul 19 '12 at 21:17