Question

Question: Let K be a Sylow p-subgroup of G and N a normal subgroup of G. Prove that $K\cap N$ is a Sylow p-subgroup of N.

Short answer

It is proved that $K\cap N$ is a Sylow p-subgroup of N.

Step-by-step solution

Step 1:Referring to the result of Exercise 15(b)

$\left|HK\right|\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\frac{\left|H\right|\mathbf{ }\mathbf{ }\left|K\right|}{\left|H\cap K\right|}$

Given that K is a Sylow p-subgroup of G and N a normal subgroup of G.

Proving that K∩N is a Sylow p-subgroup of N.

Since K is a Sylow p-subgroup of G, its order must be some power of p.

The order of $K\cap N = p^k , k⩾0$

Referring to Exercise 15(b), we know, $\left|KN\right| = \frac{\left|K\right| \left|N\right|}{\left|K\cap N\right|}$.

By rearranging the equation, we have:

$\frac{\left|N\right|}{\left|K\cap N\right|} = \frac{\left|KN\right|}{\left|K\right|}$

From Lagrange’s theorem, we can write as:

$\left|N: K\cap N\right| = \left|KN:K\right|$

From Lagrange’s formula for subgroups, we can write as :

$$\begin{gathered} \left|G:K\right|=\left|G:KN\right|\left|KN:K\right| \\ \left|KN:K\right|=\left|G:KN\right|/\left|G:K\right| \end{gathered}$$

As we know, K is the Sylow p-subgroup of group G, so $\left|G:K\right|$is not divisible by p.

Therefore, $\left|NK:K\right|$ is also not divisible by p.

This implies $\left|N:K\cap N\right|$is not divisible by p.

Hence, it is proved that $K\cap N$ is a Sylow p-subgroup of N.