Sylow subgroups of a finite group which are contained in some other subgroup are also Sylow there

Let G be a finite group and p a prime. Prove that if P \in \mathsf{Syl}_p(G) and P \leq H \leq G, then P \in \mathsf{Syl}_p(H). Give an example to show that, in general, a Sylow p-subgroup of if a subgroup of G need not be a Sylow p-subgroup of G.

If P \leq H \leq G is a Sylow p-subgroup of G, then p does not divide [G : P]. Now [G : P] = [G : H] [H : P], so that p does not divide [H : P]; hence P is a Sylow p-subgroup of H.

A trivial counterexample to the converse statement is the subgroup \langle r \rangle \leq D_8; \langle r \rangle is a Sylow 2-subgroup of itself, but clearly not a Sylow 2-subgroup of D_8 since |D_8| = 2^3.

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  • Steven  On December 6, 2011 at 3:48 pm

    I have another example for you, I’m not 100% this works, though: consider G = Z_{3} \times Z_{3} \times Z_{2}, and H = Z_{3} \times Z_{2} \leq G. Then Z_{3} can be identified inside H as the Sylow 3-subgroup of H, however the order of Z_3 disqualifies it from being the 3-subgroup of G.

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