Why is a group injective with regards to its operation?

I have the following group: The set contains the numbers from 0 to 20 and the operation is plus modulo 6. Meaning if we take two numbers 3 and 7, the result is 4.

One can see that the neutral element is 0. However, if I take the number 8, I have two inverse elements: 4 and 10. However, shouldn't there only be one inverse element?

This object is not a group. There is no identity element. If $e$ would be an identity, then $$7+e=7$$ but this is not possible, since $a+b\in \{0,1,2,3,4,5\}$ for all $a,b$, and in particular $7+e$ will never be $7$.

• @WiCK3DPOiSON Yes you're right... Was sloppy – user2520938 Jan 1 '17 at 13:43

Your group is violating three axioms of the four group axioms.

You are violating the Closure axiom, Identity axiom and Inverse axiom.

You have to define a group properly. The above you mentioned is not a group. Your 'group' members have either atleast $4$ identity elements or no identity element and each element has at least $3$ inverses.

• "en.wikipedia.org/wiki/…". On Wikipedia, it says that a group needs to have "an" identity element. Only later do we "prove" that there is only one. – Kevin Wu Jan 1 '17 at 13:20
• @KevinWu Alright tell me what is the identity element of $7$ or tell me for what values of $a$ and $b$ do you get $a.b = 7$. – 8hantanu Jan 1 '17 at 13:24
• yeah, but now we are back at where the question started. Having one identity element, we have multiple possible inverse values for the numbers 7, for example 5 and 11 – Kevin Wu Jan 1 '17 at 13:30
• Identity doesn't exist for $7$, therefore $7$ doesn't belong to the above group. How can you talk about the inverses of $7$. Are you new to the concept of groups or algebra. – 8hantanu Jan 1 '17 at 13:33
• Ok got it. Thanks and yes to the last question – Kevin Wu Jan 1 '17 at 13:37