# Understanding Sylow's First Theorem Using Double Cosets

Remark on Double Cosets

Sylow's First Theorem

The above hyperlinks are on the proof I'm referring to.

The above is the proof of: All Sylow $p$-groups are conjugate.

It is assumed: Let $G$ be a finite group such that $|G| = p^{n}q$, where $p$ is prime and $gcd(p,q) = 1$.

I don't understand the proof from the following line: $|AxB| = p^{2n-m}$, where $2n-m \geq n + 1$. Since $p^{n+1} \nmid |AxB|$ $\forall x \in G$.

How is this so?

Since, how I understand the above line is: that since $2n-m \geq n + 1$ then $p^{n+1} | |AxB|$ $\forall x \in G$ and since $|G| = p^{n}q$ where $p^{n}$ is the highest power of $p$ that divides $G$, and from this we get our contradiction!

Thanks in advance for the kind help!

• If I've read it correctly, then I think it is a typo. I think it should be, $p^{n+1}$ divides $\mid AxB\mid$, rather than not divides, since that was indeed the point of the sentences that came before. Jul 24 '17 at 20:48

The source sentence has poor placement of the negation. Where it says

Since $p^{n+1} \nmid |AxB|$ for every $x \in G$ $\ldots$

Since it is $\underline{\textbf{not true}}$ that $p^{n+1} \mid |AxB|$ for every $x \in G$ $\ldots$
Since it is not true that $\forall x\in G\left(\;p^{n+1} \mid |AxB|\; \right)$ $\ldots$