# Abstract Nonsense

## Groups of Order pq (Pt. I)

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

$\text{ }$

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.

$\text{ }$

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:

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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. $\blacksquare$

$\text{ }$

$\text{ }$

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 $G\cong\mathbb{Z}_p\times\mathbb{Z}_p$

$\text{ }$

$\text{ }$

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

$\text{ }$

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:

$\text{ }$

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. $\blacksquare$

$\text{ }$

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

$\text{ }$

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 $G\cong\mathbb{Z}_p\times\mathbb{Z}_q$. $\blacksquare$

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

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

$\text{ }$

$mp+1=q$

$\text{ }$

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. $\blacksquare$

$\text{ }$

$\text{ }$

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 $G\cong\mathbb{Z}_{pq}$.

$\text{ }$

$\text{ }$

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 -

## 7 Comments »

1. […] Point of Post: This is a continuation of this post. […]

Pingback by Groups of Order pq (Pt. II) « Abstract Nonsense | April 19, 2011 | Reply

2. […] sixteen are abelian (i.e. the only remotely non-trivial one of these is and since this has been previously established). Probably the most fantastic part of this theorem is how incredibly easy it is to prove using […]

Pingback by Representation Theory: The Number of Conjugacy Classes of a Finite Group of Odd Order is Equivalent to The Order of The Group Modulo Sixten « Abstract Nonsense | April 23, 2011 | Reply

3. […] Since we know that or . Thus, it assumes to prove this is true for those two groups. So, first define […]

Pingback by University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) ( January 2003)) « Abstract Nonsense | May 1, 2011 | Reply

4. […] works towards proving not only that every group of order where for any (greatly generalizing the statement that a group of for primes with is cyclic) but also that numbers of this form are the only […]

Pingback by A Classification of Integers n for Which the Only Groups of Order n are Cyclic (Pt. I) « Abstract Nonsense | October 28, 2011 | Reply

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