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I know that $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m \mathbb{Z}$ and $\mathbb{Z}/nm \mathbb{Z}$ are isomorphic as rings, if and only if, they are isomorphic as groups, which happens exactly when $\gcd(n, m) = 1$.

I am know wondering : are $\mathbb{Z}/ n\mathbb{Z} \times \mathbb{Z}/ m \mathbb{Z}$ and $\mathbb{Z} / n' \mathbb{Z} \times \mathbb{Z} / m' \mathbb{Z}$ isomorphic as groups iff they are isomorphic as rings ?

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This is true for the rings and groups also for non-coprime $n$ and $m$, using the CRT as above for coprime divisors. So $\Bbb Z/12\Bbb Z \times \Bbb Z/2\Bbb Z\cong \Bbb Z/6\Bbb Z\times \Bbb Z/4\Bbb Z$ is the same for groups and rings (applying $12=3\cdot 4$ and $6=2\cdot 3$ into coprime factors).

However, the answer is negative in general for rings and their additive groups. Isomorphic rings necessarily have isomorphic additive groups. The converse need not be true. This is perhaps not so well known for finite rings. Consider the rings $\Bbb F_p\times \Bbb F_p$ and $\Bbb F_{p^2}$ for a prime $p$. They are not isomorphic as rings, since the last one is a field, whereas the first one has zero divisors. Nevertheless, both rings have additive group isomorphic to $\Bbb Z/p\Bbb Z\times \Bbb Z/p\Bbb Z$, see here.

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    $\begingroup$ Just a note that $\mathbb{Z}/p^2 \neq \mathbb{Z}/p^2\mathbb{Z}$, in case that confuses anyone else (where the first one refers to the field of order $p^2$). $\endgroup$
    – tkw
    Commented Sep 19 at 13:38
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    $\begingroup$ The field with $p^2$ elements is not isomorphic to any ring of the form $\mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z}$. $\endgroup$
    – Geoffrey Trang
    Commented Sep 19 at 14:18
  • $\begingroup$ +1 but imo, only the first paragraph (the positive answer) is relevant and needed your explanations. The rest should be obvious even for beginners: $\Bbb R\times\Bbb R\not\cong\Bbb C$. $\endgroup$
    – Anne Bauval
    Commented Sep 19 at 14:35
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    $\begingroup$ More precisely, $\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/m\mathbb{Z}$ and $\mathbb{Z}/n'\mathbb{Z} \times \mathbb{Z}/m'\mathbb{Z}$ are isomorphic as groups iff they are isomorphic as rings iff $\mathrm{gcd}(n,m)=\mathrm{gcd}(n',m')$ and $\mathrm{lcm}(n,m)=\mathrm{lcm}(n',m')$. $\endgroup$
    – Geoffrey Trang
    Commented Sep 19 at 14:40

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