Today, in class, my algebra professor stated this particular general result.
Theorem. Let $R$ be a ring of order $pq$ where $p,q$ are two primes with $p\gt q$ and $q\not\mid (p-1)$. Then, $R$ is a commutative ring.
My question is:
- Do we require the condition $q\not\mid (p-1)$ ?
Here's a proof attempt where we don't consider the above condition:
Proof.
Since $(R,+,.)$ is a ring, we know that $(R,+)$ is an abelian group.
Now, since $p,q$ are primes and $(R,+)$ is a group, we know by Cauchy's theorem that there exists two elements $a,b\in R$ such that $|a|=p$ and $|b|=q$. Now, since $p\gt q$, we know that they are distinct primes and hence $\gcd(p,q)=1$. So, we have,
$$|a+b|=|a||b|=pq=|R|\implies (R,+)=\langle a+b\rangle$$
Denote $a+b$ by $g$. Then, since $(R,+)$ is cyclic, we have,
$$\forall x,y\in R~\exists m,n\in\Bbb Z\mid x=mg~,~y=ng\\~\\ xy=(mg)(ng)=mn(gg)=nm(gg)=(ng)(mg)=yx$$
using the distributive laws.
Hence, $(R,+,.)$ is a commutative ring.
So, does my proof work and the additional condition is not required? Or is there any error in my thinking?
Your conclusion and your proof is true. This Wikipedia page said there are four non-isomorphic rings of order $pq$ exists , and all of them are commutative and underlying group of them is $\mathbb Z_{pq}$. (for the proof see this article) .
In this page the author classified some rings of order less than $100$ and said the classification of finite rings would have to be solved for the following types of prime factorizations:
$p^3,p^4,p^5,p^6,p^2q,p^3q,p^4q,p^5q,p^2q^2,p^3q^2,p^2qr $
https://oeis.org/A209401 is about number of noncommutative rings with n elements .