Additive group of a finite ring of square free order is cyclic

Question

$R$ is a finite ring of square free order $n>1$.

How to show by the basis theorem for finite Abelian groups that additive group of $R$ is cyclic group of order $n$?

Answer 1

Any finite abelian group of square free order is cyclic. Note that $\mathbb Z/a\mathbb Z\oplus \mathbb Z/b\mathbb Z\cong \mathbb Z/ab\mathbb Z$ if $\gcd(a,b)=1$.

Answer 2

Hints:

== It is given that $\,n=p_1\cdot p_2\cdot\ldots\cdot p_k\;,\;\;p_i\;$ pairwise different primes

== From the above it follows that as an abelian group

$$R\cong C_{p_1}\times\ldots\times C_{p_n}\;,\;\;C_r=\text{ cyclic group of order r}$$

== Finally, remember (or prove) that

$$\gcd(a,b)=1\implies C_a\times C_b\cong C_{ab}$$