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Consider addition modulo N relation $a \pmod n$ over all the integers for example. We can easily prove that this is a group. Identity is any element which satisfies the following:

$$a \cdot e=a$$

In the given relation $a \pmod n$, we can have $0$ as identity. But $n, 2n$ and so on. These numbers will also satisfy the above condition.

Similarly we can have multiple inverses by just adding $n, 2n$ and so on..

What am I missing here?

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  • $\begingroup$ What are the elements of your group? $\endgroup$ – Hagen von Eitzen Sep 23 '16 at 14:34
  • $\begingroup$ Set of integers, or real numbers. $\endgroup$ – Avnish Gaur Sep 23 '16 at 14:36
  • $\begingroup$ What is the operation? Your set is a group under addition but not multiplication (why not?). $\endgroup$ – user1729 Sep 23 '16 at 14:36
  • $\begingroup$ user2277550, Please read the question. Let me know what part you didn't get. I have mentioned everything in my question. $\endgroup$ – Avnish Gaur Sep 23 '16 at 14:36
  • $\begingroup$ Operation is Congruence modulo N. I have mentioned in the question. $\endgroup$ – Avnish Gaur Sep 23 '16 at 14:37
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What you are missing is that you're thinking of all of $\mathbb{Z}$ when in reality, you don't have that, you have the equivalence classes $[0],[1],[2],\ldots,[n-1]$. For each equivalence class we have $$[i]=\{kn+i:k\in\mathbb{Z}\}$$ Which is a collection fo elements in $\mathbb{Z}$, however what you are dealing with is the quotient group for which each element, the congruence class $[i]$, is a collection of elements in the mother group, in our case $\mathbb{Z}$.

So $0,n,2n,\ldots$ do not represent different elements in your $\mathbb{Z}/n\mathbb{Z}$, but rather is different representation of the same element. Just like with rational numbers $$\frac{p}{q}=\frac{jp}{jq}$$ Different representations, the same element.

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  • $\begingroup$ Alright, that explains it. But what if I don't consider the equivalence classes, and take all the integers separately? I can't do that because they are basically the same numbers, is it so? $\endgroup$ – Avnish Gaur Sep 23 '16 at 14:41
  • $\begingroup$ You can't do that because the group structure is on the equivalence classes as elements, not the individual elements themselves from the mother group. $\endgroup$ – Zelos Malum Sep 23 '16 at 14:44
  • $\begingroup$ What is Podulo in this context? $\endgroup$ – Zelos Malum Sep 23 '16 at 14:53
  • $\begingroup$ I am sorry but I am not understanding what you are trying to say, care to give an example to illustrate your idea? $\endgroup$ – Zelos Malum Sep 23 '16 at 14:55
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Avnish Gaur Sep 23 '16 at 14:57

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