If a finite group has a normal Sylow p-subgroup, then all subgroups also have a normal Sylow p-subgroup

Let P be a normal Sylow p-subgroup of a finite group G and let H \leq G. Prove that P \cap H is the unique Sylow p-subgroup of H.


Note that P \cap H is normal in H, and P \cap H \leq P is a p-subgroup.

Suppose now that P \cap H is not a Sylow p-subgroup of H; then there exists an element x of p-power order in H which is not in P. However, as the unique Sylow p-subgroup of G, this is a contradiction because all elements of p-power order in G are contained in P. Thus P \cap H is a Sylow p-subgroup of H.

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Comments

  • Gobi Ree  On December 21, 2011 at 9:42 pm

    Another method to show that P \cap H is a Sylow p-subgroup of H: |H:P\cap H|=|HP:P| is a divisor of |G:P|=|G:HP||HP:P|, so p does not divide |H:P \cap H|.

    • nbloomf  On January 19, 2012 at 10:15 am

      Very nice!

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