A modification of a cyclic group that seems to break being a group. What is it?

Question

Background: I came up with this trying to answer an actually silly question of "when can $1+1=3$ be true" ?

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Consider a set $\mathcal S = \{0,\cdots, N-1\}$ coupled with an operation "+" that takes any pair of elements $a,b\in\mathcal S$

$$a+b \to \text{mod}(N-\text{mod}(a+b,N),N)$$

In other words like a cyclic group but after the rotation a reflection is carried out.

My question is: can we classify what I have constructed?

If we analyze the group axioms

1. Closure is trivial to prove.
2. Inverse element is easy to assign, for $s$ just pick $N-s$, the rotation takes us to $0$ and the subsequent reflection does not move $0$ anywhere.
3. Identity element is much trickier. $0+0 = 0$, but for any $s+0$, this will not be $s$.

However, applied twice, it will act as identity $s+0+0 = s$ as we will reflect back again.

So in the language of abstract algebra, what have I done!?

Answer 1

I will let $a\oplus b $ denote $N-\text{mod}(a+b,N)$. This operation is not even associative, as $$(s\oplus0)\oplus0=s\ne N-s=s\oplus0=s\oplus(0\oplus0)$$ Actually, it is a commutative quasigroup, which is a magma with the Latin square property. Let's check the axioms:

1. Closure and commutativity: trivial
2. For each $a,b\in S$, take $x=y=a\oplus b=N-\text{mod}(a+b,N)$, then $$a\oplus x=a\oplus (a\oplus b)=a\oplus (\text{mod}(N-a-b,N))=N-\text{mod}(N-b,N)=b$$ $$y\oplus a=(a\oplus b) \oplus a =a\oplus (a\oplus b)=b$$

By the way, $N-x$ is not the inverse of $x$, because $0$ is not an identity.

Answer 2

I see a few problems with your construction : first of all, it doesn't seem to be well defined over S, you would have to include N (else, if you compute 1*(N-1), you get N).

If you would like to compare this operation with that coming from a group law, it seems to me that the first property that you should look for is associativity (before anything else). It doesn't seem to ever hold here ; an explicit counter-example when N=4 is given by (I will use the symbol * to denote your operation)

(1*1)*2 = 2*2 = 4 but 1*(1*2) = 1*1 = 2.

It does seem like a fun operation and perhaps it would be possible to 'measure' how far this operation is from being associative (for instance, what happens if you compute x*(y*z)-(x*y)*z ?). Hope this helps !