# Sufficient Conditions for $(R,+,.)$ to be a commutative ring

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?

• It seems fine to me. Commented Mar 8, 2016 at 15:55
• What is true is that if $G$ is a group of order $pq$ where $p,q$ are two primes with $p\gt q$ and $q\not\mid (p-1)$, then $G$ is a commutative (actually cyclic) group. Is it possible that your P stated this for groups and not for rings? Commented Mar 8, 2016 at 16:42
• You can see math.stackexchange.com/questions/67129/…
– user217174
Commented Mar 8, 2016 at 18:54

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 .

• So, are you agreeing with my proof or are you saying there's an error in it? Commented Mar 8, 2016 at 18:23
• @learner I have trouble with |a+b|=|a||b| . From which theorem you conclude this?
– user217174
Commented Mar 8, 2016 at 18:32
• @learner If $q \nmid p-1$ then the underlying group is isomorphic to $\mathbb Z_{pq}$ and this proves your professor statement.
– user217174
Commented Mar 8, 2016 at 18:57
• I don't know if there's a name for the theorem or not but the statement is: If $a,b\in G$ and $(G,\circ)$ is an abelian group with $\gcd(|a|,|b|)=1$, then $|a\circ b|=|a||b|$ Commented Mar 8, 2016 at 19:25
• Oh ! Yes , its true . I forgot that the group is abelian!
– user217174
Commented Mar 8, 2016 at 19:31