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I am working on this problem: let R be a ring with identity 1 not equal 0. Show that if the order of R is less than 8, then R is commutative.

I observed that ring with 1,2 or 3 elements are commutative but I was stuck at 4. Can anyone give me a hint? I feel like this is not the right approach.

Thanks!

Quy Xuan Cao
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    Please provide more details, for instance, what are your thoughts about the question, what you have tried, where you are stuck. – Sahiba Arora Jun 27 '17 at 19:56
  • $8$ elements ? ... say "I am trying to classify all of the rings with less than $8$ elements" .... also edit into your question you know that for $2$ elements there is only $\mathbb{F}_2$ ... etc ... – Donald Splutterwit Jun 27 '17 at 20:08
  • I made it clearer now. Thank you for your input! – Quy Xuan Cao Jun 27 '17 at 20:28

2 Answers2

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For 6 you need an abelian group which is cyclic by default. Same for 5 and 7. Why does cyclic group imply commutative ring? Any two elements are $na$, $ma$, product is $(na)(ma)$ which is $nma^2=mna^2$ which is $(ma)(na)$. In fact, if only one abelian group exists of a particular order then any ring of that order is commutative. Here $a$ is the generator of cyclic group and $m,n$ are natural numbers.

user26857
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CoffeeCCD
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If the additive group is cyclic then call one of its generator as $a$. Then any element of the underlying set is of the form $na$ for $n \in \mathbb{Z}$. Hence for the multplication of such elements we have: $(na).(ma)= (a+a... n times).(a+a+.. m times)$. Now use the distributive property of the ring to to conclude that it is equal to $(nm)a^{2}=(ma).(na)$. Hence commutative. This works almost for all cases for $n\leq 7$. Since for $n=1,2,3,5,6,7$, the additive group is cyclic. What about n=4? See Ring of order $p^2$ is commutative..

Riju
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