I just do not understand one step of proof of Sylow's second theorem given in my textbook Basic Abstract Algebra by P. B Bhattacharya.

The step is,

as K is a Sylow p-subgroup of G, hence $gcd(|G/N(K)|, p) = 1$.

How it comes? Here, the notation $N(K)$ represent the normalizer of $K$ and $| |$ is used for order.


2 Answers 2


The order of $K$ is the highest power of $p$ that divides the order of $G$.

This implies that the index of $K$ is coprime with $p$.

Since $N(K)$ contains $K$, the index of $N(K)$ divides the index of $K$ and so is also coprime with $p$.

  • $\begingroup$ @Ihf please if possible can you explain your answer with the help of example.for example if we consider |G| = 120 or 80 then? $\endgroup$ Sep 25, 2015 at 3:11

Suppose $G$ has order $p^n \cdot m$, where $p$ doesn't divide $m$. Then if $K$ is a Sylow $p$-subgroup of $G$, it has order $p^n$, the highest power of $p$ possible.

Since $K$ is a subgroup of its normalizer $N(K)$, the order of $N(K)$ must be a multiple of $p^n$.

I have no idea if $N(K)$ is normal in $G$, but if it is, the quotient $G/N(K)$ has order $\frac{\lvert G \rvert }{\lvert N(K)\rvert}$, where both numerator and denominator are multiples of $p^n$ (in other words, have the largest $p$-part possible).

So that quotient cannot be a multiple of $p$ (exactly the same number of $p$'s divide the top and bottom), which is exactly what the book is asserting.


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