Show that no ring of order 6 is an integral domain. Problem Statement:
Show that no ring of order $6$ is an integral domain.
Some Definitions:
Integral Domain: a commutative ring with identity and no zero divisors.
$\mathbb{Z}_n$: group of the elements ${0, 1, 2,...,n−1}$ under addition modulo n.
My Attempt:
Let $(R, +, \cdot)$ be a ring of order $6$.
Clearly, $(R, +)$ is $\mathbb{Z}_6$ (upto isomorphism), since $R$ must be an abelian group to be a ring.
I recognize that if I can show $1\in\mathbb{Z}_6$ is the identity of $R$, then I am done [$\because (1+1)(1+1+1)=6=0$]. However, if this statement is true, I am unable to figure out a way to show this.
How do I proceed?
I am aware that this can be proved in better ways [like by using the fact that every finite integral domain is a field and that every finite field has a prime power order], however, I am interested to know if this can be proved the way I am going.
 A: You are aware that this collection has the additive structure of $\mathbb{Z}_6$, but you are wondering if a multiplicative identity exists. But you don't need that. You can do it just using the additive generator of $\mathbb{Z}_6$. I think in the comments some users are writing $1$ for the additive generator of $\mathbb{Z}_6$, and that's fine, but it also comes with the connotation of being the multiplicative identity which is what you are wondering about. So use something else like $a$ instead.
As an additive group, you have $\{0,a,2a,3a,4a,5a\}$. Now what is $a\cdot a$? It is in there somewhere: $a\cdot a=ka$.
So what is $(2a)(3a)$? That's $(a+a)(a+a+a)=\cdots=6a^2=6(ka)=(6k)a=0$.
A: The possible additive orders of elements in the underlying abelian group are $1, 2,3, 6$.
Cauchy's theorem says there must be elements of additive order $2$ and $3$. The sum of two such elements must have additive order $6$.  Call such an element $c$.
Then $2c$ and $3c$ are nonzero and $(2c)(3c)=(6c)c=0$, showing the ring is not a domain.
