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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.
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 $p$.
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.
