General Idea:
Assuming Sylow's 2 Theorem Holds, Prove the 1st Sylow's Theorem.
Mathematically written:
if both are true:
G is a finite Group
Given such G, and prime number p, All Sylow p-subgroups are conjugate to each other
then G has a Sylow p-subgroup
My effort in trying to solve the problem:
- I took the group $GL_n(\mathbb{F}_p)$ and showed it has a Sylow p-subgroup (denote $|A| = p^{\frac{|G|^2-|G|}{2}}$). (where $\mathbb{F}_p$ is the field of p elements)
- Then I showed That a for any finite G, exist a bijective homeomorphism $\phi:G\rightarrow GL_{|G|}(\mathbb{F}_p)$
- So if (hypothetically) G has a p-group $H\le G$ then $\phi (H) \subseteq GL_{|G|}(\mathbb{F}_p)$ is a p-subgroup as well.
- Since $|G|=n=p^m*k$ for some k and a maximal m, And Since for any $m \le \frac{|G|^2-|G|}{2}$ exist a p-subgroup $B \le A$ with $|B|=p^m$.... there might be a way of showing that B(which exists!) is conjugated to a subgroup of $\phi(G)$
- And By the bijectivity of $\phi$ we will conclude G has a Sylow p-subgroup.
I don't know how to complete the missing part, And would appreciate any help. thanks.