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I'm currently working on a proof, but didn't get the right idea, a hint would be appreciated:

Show that $\mathbb{Z}/n^2\mathbb{Z}$ is not isomorphic to $\mathbb{Z}/n \times \mathbb{Z}/n$

So, lets assume there is an isomorphism between these groups, my first idea to was to construct a contradiction with the homomorphism property. But I don't know how, there should be an element in $\mathbb{Z}/n \times \mathbb{Z}/n$ which is not in $\mathbb{Z}/n^2$ to reach this kind of contradiction. Of course there is $n \in \mathbb{Z}/n^2$ which is not in the other group but I don't know if this helps.

Thanks

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  • $\begingroup$ Think about the order of an element of greatest order in each of the two groups. $\endgroup$ Mar 6, 2018 at 8:12
  • $\begingroup$ The first group is cyclic the latter one is not, indeed the order of $(a,b)\in \mathbb{Z}_n\times \mathbb{Z}_n$ cannot exceed $n$. This is exactly what Gerry Myerson says. $\endgroup$ Mar 6, 2018 at 8:22
  • $\begingroup$ Thanks but we haven't yet defined the terms cyclic and order. I googled those terms and while that makes sense, I would like to use another approach. $\endgroup$
    – Derato
    Mar 6, 2018 at 8:26

2 Answers 2

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Suppose $\phi:\mathbb{Z}_{n^2}\to\mathbb{Z}_n\times \mathbb{Z}_n$ is an isomorphism. Notice that $$\phi(x)=\phi(\sum_{i=1}^x1)=\sum_{i=1}^x\phi(1).$$ Now write $\phi(1)=(a,b)$ for certain $a,b$. Then $\sum_{i=1}^x\phi(1)=(\sum_{i=1}^xa,\sum_{i=1}^xb)=x\cdot(a,b)$. It follows that $\phi(1)=(a,b)=(n+1)\cdot(a,b)=\phi(n+1)$. Hence $\phi$ is not injective, a contradiction!

Remark that $a=(n+1)\cdot a$ in $\mathbb{Z}_n$ by definition of $\mathbb{Z}_n$, this explains the equality $(a,b)=(n+1)\cdot(a,b)$ above.

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Z/(n^2Z) is cyclic but Z/(nZ) × Z/(nZ) is non cyclic (as every element of this group has order n). Hence they are not isomorphic.

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  • $\begingroup$ Fine, but OP has already written in the comments that he/she doesn't know what "cyclic" and "order" mean. $\endgroup$ Mar 6, 2018 at 11:41

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