Groups of Order pq (Pt. II)
Point of Post: This is a continuation of this post.
Case 3 and
Lemma 1: Let be a prime, then .\ (where is the group of units of the field )
Proof: Define by . This is evidently a homomorphism since it’s injective since is a generator of and it’s surjective because given any one has that is an automorphism of .
Remark: Note there was no need for to be prime here, the above proof shows that for any one has that .
Thus, it remains to prove that . To do this (since ) it suffices to prove that is cyclic. To do this we note that if there does not exist an element of of order then there exists some such that for every . But, this implies that the polynomial has more solutions than its degree which is impossible. The conclusion follows.
Remark: The above proof that is cyclic is a little wishywashy. A more rigorous, and more general proof can be found in this post of mine.
Lemma 2: Let be a homomorphism. Suppose there are maps and such that (where is the inner autormorphism induced by in ). Then, .
Proof: This is mostly just grunt work. Namely, define by . As indicated, is an isomorphism. Indeed, to see it’s a homomorphism we note that
But, evidently is a bijection since a quick check shows that serves as an inverse.
Lemma 3: If is cyclic and are such that is conjugate to then .
Proof: By assumption for some and there exists some such that . Thus, for some generator of the cyclic . Since is a generator we know that and we may evidently then assume that . We can then define and . But, a quick check shows that and so the lemma follows from lemma 2.
We are now prepared to prove our main theorem. Namely:
Theorem: Let and be primes such that . Then, there exists a unique, up to isomorphism, nonabelian group of order .
Proof: It follows by our previous argument that there is precisely one Sylow subgroup of and Sylow subgroups. If represents the latter and one of the former then we have that is trivial, , and . Thus, for some homomorphism . Thus, we desire to seek how many of these homomorphisms define nonisomorphic nonabelian structures . But, since and this evidently is equivalent to finding the number of nonisomorphic nonabelian for some homomorphism . But, by our first lemma we know that and so evidently every nontrivial must have image with order . But, since is cyclic there is precisely one subgroup of order . Thus, for every nontrivial one has that and so by lemma 3 we know that . The conclusion follows.
Thus, every group with where are primes with is either isomorphic to or where is any nontrivial (and since there are evidently of them) homomorphism.
References:
1. Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print
2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Math. Soc., 1996. Print
April 19, 2011  Posted by Alex Youcis  Algebra, Group Theory, Representation Theory  Algebra, Classifying, Group Theory, Groups of order pq, Representation Theory
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My name is Alex Youcis. I am currently a senior a first year graduate student at the University of California, Berkeley.
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