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Math Help - containment of intersection problem

  1. #1
    nhk
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    containment of intersection problem

    Suppose K is a normal subgroup of a finite group G and S is a p-Sylow subgroup of G. Prove that K intersect S is a p-Sylow subgroup of K. So I know that K is a unique p-sylow group by definition, is that enough to prove that the intersection of K with S is a p-sylow subgroup of K?
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    Quote Originally Posted by nhk View Post
    Suppose K is a normal subgroup of a finite group G and S is a p-Sylow subgroup of G. Prove that K intersect S is a p-Sylow subgroup of K. So I know that K is a unique p-sylow group by definition, is that enough to prove that the intersection of K with S is a p-sylow subgroup of K?
    It is true that a unique Sylow-p subgroup of G is a normal subgroup of G, but it is not necessarily true that a normal subgroup of G is a unique Sylow-p subgroup of G.

    Since S is a Sylow-p subgroup of G and K \cap S is a subgroup of S, we see that K \cap S is a p-subgroup of K. It remains to show that K \cap S is a maximal p-subgroup of K, which implies that K \cap S is a Sylow p-subgroup of K. This can be shown by using that [K:K \cap S] is coprime to p.

    We know that [G:S] is coprime to p. Since [K:K \cap S]=[KS:S] and [G:S]=[G:KS][KS:S], we see that [K:K \cap S] is indeed coprime to p.

    Thus K \cap S is a Sylow p-subgroup of K.
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    Since S is a Sylow-p subgroup of G and K intersect S is a subgroup of S, we see that K intersect S is a p-subgroup of K.
    Is that because K intersect S has to be either the identity e or S?

    It remains to show that K intersect S is a maximal p-subgroup of K, which implies that K intersect S is a Sylow p-subgroup of K. This can be shown by using that [K:K intersect S] is coprime to p.
    I am not seeing how we know that [K:K intersect S] is coprime to p.

    We know that [G:S] is coprime to p.
    what if |G|=p^2, then [G:S] would not be coprime to p right?
    Last edited by aabsdr; May 16th 2010 at 12:35 PM.
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    Quote Originally Posted by aabsdr View Post
    Since S is a Sylow-p subgroup of G and K intersect S is a subgroup of S, we see that K intersect S is a p-subgroup of K.
    Is that because K intersect S has to be either the identity e or S?

    It remains to show that K intersect S is a maximal p-subgroup of K, which implies that K intersect S is a Sylow p-subgroup of K. This can be shown by using that [K:K intersect S] is coprime to p.
    I am not seeing how we know that [K:K intersect S] is coprime to p.

    We know that [G:S] is coprime to p.
    what if |G|=p^2, then [G:S] would not be coprime to p right?
    K intersect S is a subgroup of both K and S. Since S is a Sylow p-subgroup of G, K intersect S is a p-subgroup of G, which in turn it is a p-subgroup of K (Check the Lagrange's theorem).

    In your example without choosing K, |S|=p^2 and [G:S]=1, which is still coprime to p.
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  5. #5
    nhk
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    We know that [G:S] is coprime to p.

    Since [K:K intersect S]=[KS:S] and [G:S]=[G:KS][KS:S], we see that [K:K intersect S] is indeed coprime to p.
    I know that KS is a group/subgroup since K is normal, but I am not seeing where you are getting the order facts that [K:K intersect S]=[KS:S]. I am also not understanding where you are getting that:[G:S]=[G:KS][KS:S] and [G:S]= [G:KS][KS:S].
    [K:K intersect S] = [KS:S] means that [K:K intersect S]= [G:S]/[G:KS], but I don't see how this implies that [K:K intersect S] is coprime (relatively prime) it P.
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    Quote Originally Posted by nhk View Post
    We know that [G:S] is coprime to p.

    Since [K:K intersect S]=[KS:S] and [G:S]=[G:KS][KS:S], we see that [K:K intersect S] is indeed coprime to p.
    I know that KS is a group/subgroup since K is normal, but I am not seeing where you are getting the order facts that [K:K intersect S]=[KS:S]. I am also not understanding where you are getting that:[G:S]=[G:KS][KS:S] and [G:S]= [G:KS][KS:S].
    [K:K intersect S] = [KS:S] means that [K:K intersect S]= [G:S]/[G:KS], but I don't see how this implies that [K:K intersect S] is coprime (relatively prime) it P.
    See Theorem 4.7 in Hungerford (p39) for the proof of [K:K intersect S]=[KS:S].

    The second part is a simple number theory result. If a positive integer n is coprime to p, then the divisors of n are coprime to p.
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