# Uniqueness of Inverse and Identity (Group Theory Abstract Algebra)

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?

• What are the elements of your group? – Hagen von Eitzen Sep 23 '16 at 14:34
• Set of integers, or real numbers. – Avnish Gaur Sep 23 '16 at 14:36
• What is the operation? Your set is a group under addition but not multiplication (why not?). – user1729 Sep 23 '16 at 14:36
• user2277550, Please read the question. Let me know what part you didn't get. I have mentioned everything in my question. – Avnish Gaur Sep 23 '16 at 14:36
• Operation is Congruence modulo N. I have mentioned in the question. – Avnish Gaur Sep 23 '16 at 14:37

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 $,,,\ldots,[n-1]$. 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.