## Review of Group Theory: Alternate Proof of the Sylow Theorems (Pt. II)

**Point of post: **This is a continuation of this post.

We continue now with a proof what was previously called the third Sylow theorem. Namely, that the number of Sylow -subgroups of a group is congruent to modulo . Namely:

**Theorem: ***Let be a finite group with where . Then, if denotes the set of all Sylow -subgroups of then .*

**Proof: **Let (we know that by our last theorem). Then, consider the -action on given by

for and . Evidently , and we claim that this is the only orbit under this action with one element. To see this suppose that , , and . Consider then that for each and we have that by assumption. Thus the map

where is the variant of the inner automorphism given by is well-defined. Thus, consider the group

Note then that the map

is a homomorphism. Indeed,

Note though that and and since we clearly must have that

since and . Otherwise or and since they are of the same cardinality it would follow that they are equal, contradicting our assumption. Note though that as a corollary of the First Isomorphism Theorem we have that

and thus it follows that for some . But, recalling that we may then conclude that where . But, by Lagrange’s Theorem we may conclude that

but this clearly implies that , which is a contradiction. It follows then that for every with we have that . But, by the Orbit-Stabilizer Theorem

And so in particular we have that for every we have that with . But, from previous comment we know that for with we have that and so in conjunction with the previous sentence we have that with . It follows then from the Orbit Decomposition Theorem that

from where the conclusion follows.

1. Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Mathematical Society, 1996. Print.

2. Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 200

3.Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

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