Proving the Maximal Property of p-Sylow Subgroups in Finite Groups

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SUMMARY

The discussion focuses on proving that the intersection of a normal subgroup K and a p-Sylow subgroup S of a finite group G, denoted as K ∩ S, is a p-Sylow subgroup of K. It is established that since K is a normal subgroup and S is a Sylow-p subgroup, K ∩ S is a p-subgroup of K. The proof hinges on demonstrating that the index [K: K ∩ S] is relatively prime to p, utilizing Lagrange's theorem and the relationships [K: K ∩ S] = [KS:S] and [G:S] = [G:KS][KS:S].

PREREQUISITES
  • Understanding of finite group theory, specifically Sylow theorems.
  • Familiarity with normal subgroups and their properties.
  • Knowledge of Lagrange's theorem and its implications on group indices.
  • Basic concepts of group intersections and subgroup properties.
NEXT STEPS
  • Study the Sylow theorems in detail, focusing on their applications in finite groups.
  • Explore the properties of normal subgroups and their significance in group theory.
  • Review Lagrange's theorem and practice problems involving group indices.
  • Investigate the concept of maximal p-subgroups and their role in group structure.
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Mathematics students, particularly those studying abstract algebra, group theory enthusiasts, and anyone interested in the properties of Sylow subgroups in finite groups.

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Homework Statement



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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The Attempt at a Solution



Since S is a Sylow-p subgroup of G and K intersect S is a subgroup of S and K, we see that K intersect S is a p-subgroup of K. 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.

We know that [G:S] is relatively prime to p.
THe hints I was given to finish this off was that [K: K intersect S] = [KS:S] and [G:S] = [G:KS][KS:S] and from that we are suppose to get that [K:K intersect S] is relatively prime to p.
I am not getting how we know that [K: K intersect S] = [KS:S] and that [G:S] = [G:KS][KS:S] and how this implies that [K: K intersect S] is relatively prime to p.
I know that KS is a subgroup of G since K is a normal subgroup of G. If anyone could help me with this problem I will be entirnaly gratefull
 
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tyrannosaurus said:
I am not getting how we know that [K: K intersect S] = [KS:S] and that [G:S] = [G:KS][KS:S] and how this implies that [K: K intersect S] is relatively prime to p.
I know that KS is a subgroup of G since K is a normal subgroup of G. If anyone could help me with this problem I will be entirnaly gratefull

A general fact is that |KS| = |K| |S| / |K intersect S|. Since [G:H] = |G|/|H| (this is Lagrange's theorem), this statement is equivalent to [K: K intersect S] = [KS:S].

The equation [G:S] = [G:KS][KS:S] is again a consequence of Lagrange. The left side is |G|/|S| and the right side is (|G|/|KS|) (|KS|/|S|). Now, since [G:S] doesn't contain a factor of p and [KS:S] is a factor of [G:S], p cannot be a factor of [KS:S] = [K: K intersect S].
 
A general fact is that |KS| = |K| |S| / |K intersect S|. I forgot about this fact! THanks so much, this makes a whole lot of more sense now.
 

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