Abstract Nonsense

Crushing one theorem at a time

Groups of Order pq (Pt. I)


Point of Post: In this post classify groups of order pq where p and q are primes.

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Motivation

We make a pretty cool undertaking in this post. Namely, we’ll classify all groups of order pq where p and q 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.

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The Classification

Case 1: p=q

Our first case is a familiar one, namely we’ve already proven that every group of order p^2 where p is prime is abelian. But, we reprove this (for didactic purposes) using representation theory:

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Theorem: Let G be a finite group with |G|=p^2 where p is prime. Then, G is abelian.

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

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\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|=p^2\quad\mathbf{(1)}

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But, the dimension theorem tells us that each d_\alpha\mid |G|=p^2 and so d_\alpha=1,p,p^2. But, since there is always the degree of \alpha_{\text{triv}} which is d_{\alpha_{\text{triv}}}=1 we may conclude from \mathbf{(1)} that d_\alpha< p and so d_\alpha=1. The conclusion follows. \blacksquare

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Thus, as stated we know that every group of order p^2 is abelian and, so from the structure theorem we may conclude that if |G|=p^2 with p prime then G\cong\mathbb{Z}_{p^2} or G\cong\mathbb{Z}_p\times\mathbb{Z}_p

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Case 2: q>p and q\not\equiv 1\text{ mod }p

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

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Lemma (Pure Group Theory): Let G be a finite group and P a Sylow p-subgroup, then P\unlhd G if and only if n_p=\#\left(\text{Syl}_p\left(G\right)\right)=1.

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

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Theorem: Let G be a group with |G|=pq where p<q are primes and q\not\equiv 1\text{ mod }p. Then, G is abelian.

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Proof (Pure Group Theory): Let n_p=\#\left(\text{Syl}_p\left(G\right)\right) (i.e. then number of Sylow p-subgroups) and n_q=\#\left(\text{Syl}_q\left(G\right)\right). We know that n_p\mid pq and so n_p=1,p,q,pq. But, from the same theorem we know that n_p\equiv 1\text{ mod }p. Evidently then n_p\ne p,pq and also by assumption we have that n_p\ne q (otherwise q\equiv 1\text{ mod }p) and thus n_p=1. Similarly, we have that n_q\mid pq and n_q\equiv 1\text{ mod }q. By the first of these statements we have that n_q=1,p,q,pq but by the second one clearly n_q\ne p,q,pq. Thus, n_q=1. It thus follows from our previous lemma that if P is the Sylow p-subgroup of G and Q the Sylow q-subgroup of G then P,Q\unlhd G. We thus have by a previous theorem that G\cong P\times Q but since |P|=p and |Q|=q (and p,q are primes) that P\cong\mathbb{Z}_p and Q\cong\mathbb{Z}_q so that G\cong\mathbb{Z}_p\times\mathbb{Z}_q. \blacksquare

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Proof (Representation Theory): Just as before it suffices to prove that d_\alpha=1 for every \alpha\in\widehat{G}. Now, we know that

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\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|=pq\quad\mathbf{(3)}

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And since each d_\alpha\mid |G|=pq we have that d_\alpha=1,p,pq. But, since d_\alpha<\sqrt{pq} we may conclude that d_\alpha\ne q,pq (not q since q>p so that q^2>qp). Now, if we let m be the number of \alpha\in\widehat{G} such that d_\alpha=p and n the number of \alpha\in\widehat{G} with d_\alpha=1 we may rephrase \mathbf{(3)} as

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mp^2+n=pq\quad\mathbf{(4)}

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since n\mid pq we have that n=1,p,q,pq. Clearly n\ne 1 otherwise 1\equiv 0\text{ mod }p. If n=q then \mathbf{(4)} implies mp^2\equiv 0\text{ mod }q and so q\mid m but this implies that q\geqslant m and so mp^2>pq which is impossible since n\geqslant 1. Lastly, to suppose that n=p then \mathbf{(4)} implies

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mp+1=q

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or q\equiv1\text{ mod }p contradictory to assumption. Thus, n=pq and so every irrep of G is of degree one. The conclusion follows. \blacksquare

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Thus, either proof proves that if G is a finite group with |G|=pq then by the structure theorem G\cong\mathbb{Z}_{pq} or G\cong\mathbb{Z}_p\times\mathbb{Z}_q but since (p,q)=1 we may conclude that \mathbb{Z}_p\times\mathbb{Z}_q\cong\mathbb{Z}_{pq} so that G\cong\mathbb{Z}_{pq}.

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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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April 19, 2011 - Posted by | Algebra, Group Theory, Representation Theory | , , , , , ,

7 Comments »

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  5. 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 | Reply

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

      Comment by mike | February 28, 2012 | Reply

      • 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 U(\mathbb{Z}_q)\cong\mathbb{Z}_{q-1} 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


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