Question

If K is a Sylow p-subgroup of G, prove that$\mathbf{N}\left(N\left(K\right)\right)\mathbf{=}\mathbf{N}\left(K\right)$

Short answer

It is proved that,$\mathrm{N}\left(\mathrm{N}\left(\mathrm{K}\right)\right)=\mathrm{N}\left(\mathrm{K}\right)$.

Step-by-step solution

Given information

It is given that K is a Sylow p-subgroup of G.

Prove that N(NK)=N(K)

Since, every subgroup is contained in its normalizer,$\mathrm{N}\left(\mathrm{K}\right)\subseteq \mathrm{N}\left(\mathrm{N}\left(\mathrm{K}\right)\right)$.

Let$\mathrm{x}\in \mathrm{N}\left(\mathrm{N}\left(\mathrm{K}\right)\right)$

Then,${\mathrm{x}}^{-1}\mathrm{N}\left(\mathrm{K}\right)\mathrm{x}=\mathrm{N}\left(\mathrm{K}\right)$

This implies,${\mathrm{x}}^{-1}\mathrm{Kx}\subseteq \mathrm{N}\left(\mathrm{K}\right)$

But K is normal in N(K), so it is the only Sylow p-subgroup of N(K).

Thus,${\mathrm{x}}^{-1}\mathrm{Kx}=\mathrm{K}$

This implies,$\mathrm{x}\in \mathrm{N}\left(\mathrm{K}\right)$

Thus,$\mathrm{N}\left(\mathrm{N}\left(\mathrm{K}\right)\right)\subseteq \mathrm{N}\left(\mathrm{K}\right)$

Hence, it can be concluded that$\mathrm{N}\left(\mathrm{N}\left(\mathrm{K}\right)\right)=\mathrm{N}\left(\mathrm{K}\right)$