Question

Let G₁ ⊂ G₂ be groups whose orders are divisible by p, and let H₁ be a Sylow p-subgroup of G₁. Prove that there is a Sylow p-subgroup H₂ of G₂ such that H₁ = H₂ ⋂ G₁.

Answer 1

We want to prove that there exists a Sylow p-subgroup H₂ of G₂ such that H₁ = H₂ ∩ G₁.

Step-by-step explanation:

Let G₁ and G₂ be groups whose orders are divisible by p and let H₁ be a Sylow p-subgroup of G₁. We want to prove that there exists a Sylow p-subgroup H₂ of G₂ such that H₁ = H₂ ∩ G₁.

We want to prove that there exists a Sylow p-subgroup H₂ of G₂ such that H₁ = H₂ ∩ G₁.

Consider the group G₂/G₁, where G₂ is the group of order divisible by p and G₁ is a subgroup of G₂. By Lagrange's theorem, the order of G₂/G₁ is also divisible by p.

Let K be a Sylow p-subgroup of G₂/G₁. Then H₂ = (K ∩ G₁) * G₁ is a Sylow p-subgroup of G₂ and satisfies H₁ = H₂ ∩ G₁.