## Groups of Order pq (Pt. I)

**Point of Post: **In this post classify groups of order where and are primes.

*Motivation*

We make a pretty cool undertaking in this post. Namely, we’ll classify all groups of order where and are primes. While far from ideal knowing this can cut a fairly large swath out of the groups (in very small cases) that need to be classified.

*The Classification*

*Case 1: *

Our first case is a familiar one, namely we’ve already proven that every group of order where is prime is abelian. But, we reprove this (for didactic purposes) using representation theory:

**Theorem: ***Let be a finite group with where is prime. Then, is abelian.*

**Proof: **We know that it suffices to prove that every irrep of is of degree one. So, we know that

But, the dimension theorem tells us that each and so . But, since there is always the degree of which is we may conclude from that and so . The conclusion follows.

Thus, as stated we know that every group of order is abelian and, so from the structure theorem we may conclude that if with prime then or

*Case 2: and *

We claim that under these circumstances any group of order is abelian. We give two proofs-one pure group theoretic and one using representation theory:

**Lemma (Pure Group Theory): ***Let be a finite group and a Sylow -subgroup, then if and only if .*

**Proof: **This follows immediately from the fact that all Sylow -subgroups are conjugate to each other.

**Theorem: ***Let be a group with where are primes and . Then, is abelian.*

**Proof (Pure Group Theory): **Let (i.e. then number of Sylow -subgroups) and . We know that and so . But, from the same theorem we know that . Evidently then and also by assumption we have that (otherwise ) and thus . Similarly, we have that and . By the first of these statements we have that but by the second one clearly . Thus, . It thus follows from our previous lemma that if is the Sylow -subgroup of and the Sylow -subgroup of then . We thus have by a previous theorem that but since and (and are primes) that and so that .

**Proof (Representation Theory): **Just as before it suffices to prove that for every . Now, we know that

And since each we have that . But, since we may conclude that (not since so that ). Now, if we let be the number of such that and the number of with we may rephrase as

since we have that . Clearly otherwise . If then implies and so but this implies that and so which is impossible since . Lastly, to suppose that then implies

or contradictory to assumption. Thus, and so every irrep of is of degree one. The conclusion follows.

Thus, either proof proves that if is a finite group with then by the structure theorem or but since we may conclude that so that .

**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

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You have only classified all abelian groups of order pq. What if p | q-1? Certainly non-abelian groups like this exist. Take p =2, then you have dihedral groups. I don’t see where you handle this case.

Comment by mike | February 28, 2012 |

and im an idiot and didn’t see the continuation of the post. good job!

Comment by mike | February 28, 2012 |

Mike,

It’s no worries–the way I have it set up is admittedly a little confusing. This may not have been an issue for you, but I was looking over this and the next post and noted that my proof that was a little wishy-washy, and so I added a remark linking to another post where I prove a more general statement, more rigorously. Regardless, I hope that you were able to find everything you needed! Feel free to ask a question if something is unclear!

Best,

Alex

Comment by Alex Youcis | February 28, 2012