Question

If K is a Sylow p-subgroup of G and H is a subgroup that contains N(K), prove that $\left[G:H\right]\mathbf{\equiv }\mathbf{1}\left(\mathrm{mod}p\right)$ .

Short answer

It is proved that, $\left[G:H\right]\equiv 1\left(modp\right)$.

Step-by-step solution

Given that

Given that K is a Sylow p-subgroup of G and H is a subgroup that contains N(K).

Use Sylow Theorems

According to the second and Third Sylow Theorem, the number of conjugates of K in G is equal to the number of Sylow p-subgroups.

And that is equal to $1\left(modp\right)$.

Use Theorem 9.25

Theorem 9.25 implies that this number is equal to the index of the normalizer.

This implies, $\left[G:N\left(K\right)\right]=1\left(modp\right)$.

Since $N\left(K\right)\subseteq H$, this will be applicable to K as well.

This implies, $H:\left[H:N\left(K\right)\right]=1\left(modp\right)$.

Since the indices multiply, it is clear that $\left[G:H\right]=1\left(modp\right)$.

Therefore, $\left[G:H\right]\equiv 1\left(modp\right)$.