# Dummit & Foote's Abstract Algebra: Error on proof of Sylow's theorem (Theorem 4.5.18)

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?

• Use $x\equiv y\pmod{p}$ for $x\equiv y\pmod{p}$. Commented Apr 27, 2023 at 11:27
• Use $p\mid r$ for $p\mid r$. Commented Apr 27, 2023 at 11:28
• Similar question here: math.stackexchange.com/questions/839012/… Commented Apr 27, 2023 at 11:47

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.
• I see the point. So basically, because $r$ is invariant when picking $Q$, they first make an "exploratory" trial to calculate it and afterward they are using the results for (2) and (3), right? Wow, that's smart! Thanks for your answer! Commented Apr 27, 2023 at 12:14