Question

Let H be a Sylow p-subgroup of G. Prove that H is the only Sylow p-subgroup of G
contained in N(H).

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Note that if |G|=pkm,p/|m and H is p-sylow subgroup then |H|=pk. Also H is normal in N(H) therefore H is a normal p-sylow subgroup of N(H) and any p-sylow subgroup of N(H) is of the form gHg−1 where g∈N(H) by 2nd sylow theorem. But H is normal in N(H)⟹ all conjunction of H by elements of N(H) is same as H. That is H is the only normal subgroup of N(H).

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