containment of intersection problem

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May 15th 2010, 02:24 PM
nhk

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?
May 16th 2010, 04:59 AM
TheArtofSymmetry

Quote:

Originally Posted by nhk

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.
May 16th 2010, 11:14 AM
aabsdr

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?
May 16th 2010, 02:33 PM
TheArtofSymmetry

Quote:

Originally Posted by aabsdr

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.
May 16th 2010, 04:58 PM
nhk

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.
May 16th 2010, 08:49 PM
TheArtofSymmetry

Quote:

Originally Posted by nhk

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.