I think I have spotted an error in the proof of Theorem 4.5.18 (line 139) on Dummit & Foote's Abstract Algebra. Or maybe I am missing something?
On the proof of $ r \equiv 1 \pmod {p} $, it is assumed that $Q$ is a conjugate of $P$ (on the action by $G$) and therefore WLOG we can consider that $Q = P_1$ and from there, we prove the above assertion.
However, in the next paragraph, which is about the proof that any P-subgroup $Q$ is a conjugate of the p-subgroup $P$, it leads to the result that $p \mid r$ and therefore $r \not\equiv 1 \pmod {p}$. That's a contradiction and therefore $Q$ must be a conjugate.
Isn't that sort of a circular argument? Because it was proved that $ r \equiv 1 \pmod {p} $ with the assumption that $Q$ (the $p$-subgroup of $G$ that acts by conjugation on $S$) is part of $S$ and they use this fact to prove the exact opposite. Am I missing something here?
Best Answer
The argument is not circular. As DF specifies multiple times, $r$ does not depend on the choice of $Q$. Thus, right after equation $(4.1)$, starting with "We are now in a position...", they choose a smart $Q$ (indeed, $Q=P_1$), to show something about $r$, that $r=1$ mod $p$. Again, this result is independent of the choice of $Q$.
In the middle of page $141$, at "We now prove parts (2) and (3).", they return to the general setting before the particular choice of $Q=P_1$. Now, they want to say that $Q$ must be contained in some conjugate of $P$, on proceed by contradiction as you described. In this part, it is important that you have not specified a further choice of $Q$, as you expressed in your concerns above, but this is not a problem. They only used the particular choice of $Q$ to show $r=1$ mod $p$, and afterwards returned to the general setting.