Review of Group Theory: Sylow’s Theorems (Pt. II)
Point of post: This is a continuation of this post.
We now prove the second Sylow theorem. But first we remark on some notation. If is a group we can think of acting on (the power set) by conjugation. In other words, . This is easily verified to be a group action with the orbit of equal to and isotropy subgroups . Thus, by earlier theorem we know that for every . If we say that is conjugate to . So, with one small lemma we are ready to prove the second Sylow theorem:
Lemma: Let be a group and such that . Then, .
Proof: By a previous theorem it suffices to show that . To do this we merely note that by definition for each one has and thus (this is easy to verify, we proved in in the post about normality) and so
from where the conclusion follows.
With this in mind:
Theorem(Second Sylow Theorem): Let be a finite group. Then:
Let be a finite group, and a prime. Then all Sylow -subgroups of are conjugate.
Proof: Let be a Sylow -subgroup of . Let then denote the orbit of under the -action of conjugation on . More directly, let
Note firstly that each element of being the image of under an inner automorphism is itself a Sylow -subgroup. Note next then that by previous comment
and since it follows that
consider acting on by for and . Note by the Orbit Decomposition Theorem we have that
note thought that from an earlier theorem we have that
and thus for each . But, if for each then one would have that
which is a contradiction. Thus, it follows that for some and thus . Note though that this implies that and thus by our lemma we know that . But, by an earlier theorem we know that
but it’s fairly evident that if
then would be a subgroup of whose order is higher power of than , but this contradicts that is a Sylow -subgroup. Thus, and thus , and so . The conclusion follows.
: Note that evidently since “being conjugate to” is an equivalence relation it suffices to show that every Sylow -subgroup of is conjugate to a fixed Sylow -subgroup . Note though that using the above methodology taking $laetx P$ as our initial Sylow -subgroup we have that for any Sylow -subgroup there exists some conjugate of such that . But, since we may conclude that . Thus, every Sylow -subgroup of is conjugate to and thus must be conjugate to one another.
Lastly, we prove the third Sylow theorem.
Theorem(Third Sylow Theorem): Let be a finite group. Then, if denotes the number of Sylow -subgroups of then and .
Proof: Fix a Sylow -subgroup of . By the second part of the second Sylow theorem we know that the set of all Sylow -subgroups is equal to
and thus .
To prove the other part of the theorem we recall that if acts on by conjugation we get that
but we know that and if then . Thus, recalling that for each we may conclude that
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