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Are $10 \mathbb Z$ and $20 \mathbb Z$ isomorphic to each other or not as rings?

Since both of $10 \mathbb Z$ and $20 \mathbb Z$ are cyclic groups. So if some isomorphism $f$ between them do exist then either $f(10)=20$ or $f(10)=-20$. First let us assume that $f(10)=20.$ Now since $f$ is a ring homomorphism so let us calculate $f(100)$ in two different ways $:$

$(1)$ $f(100)=10f(10)=200$.

$(2)$ $f(100)=f({10}^2)=\{f(10) \}^2=400$.

Which implies that $200=400$, which is absurd. Similarly if we take $f(10)=-20$ then by the same way we arrive at a contradiction (in this case we have $-200=400$, which is again absurd). Hence there exists no epimorphism from $10 \mathbb Z$ onto $20 \mathbb Z$ and cosequently they are not isomorphic also.

Does the above reasoning hold good? Please check it.

Thank you in advance.

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    Isomorphism: in which category? As rings without identity? If $R^2$ denotes the subgroup generated by the products in $R$, then one of these has $|R:R^2|=10$ and the other doesn't. – Angina Seng Nov 20 '17 at 06:10
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    You proof is great! The only place I would go into more detail is $f(100)=f(10+\dots+10)=f(10)+\dots+f(10)=10f(10)$. – Mike Earnest Nov 20 '17 at 06:36

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Consider a "ring without identity $R$". Let $R^2$ denote the additive subgroup of $R$ generated by the $ab$ with $a$, $b\in R$. A ring isomorphism $\phi:R\to R'$ maps $R^2$ to $R'^2$ and induces an isomorphism $R/R^2\to R'/R'^2$. For $R=n\Bbb Z$ with $n\in\Bbb N$, $R^2=n^2\Bbb Z$ and so $|R/R^2|=n$. So $n\Bbb Z\cong n'\Bbb Z$ iff $n=n'$.

Angina Seng
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