What is the name of the structure Z4 under subtraction?

Question

If we consider $\mathbb{Z_4}$ under addition, then it forms a cyclic group of order 4. However if we change the binary operation to subtraction on $\mathbb{Z_4}$, we get a different structure $J$ with properties:

• closure
• right identity element, $x*0 = x\:\forall x\in G$
• left "double identity" element, $0*(0*x) = x\:\forall x \in G $
• not associative
• not commutative

If we could reflect the multiplication table of $J$ by the vertical axis crossing its middle we would get a table isomorphic to a corresponding cyclic group table of the same size, namely $C_{\left|{J}\right|}$.

Where can I find more about such mathematical structures? Are they called by some name?

Answer 1

All that I would call this is a magma, or perhaps a quasigroup (but it's not a loop, which requires there to be a general identity element).

There are people who study quasigroups and loops, and books on them. But to be honest, I haven't read them, and I can't give a very good indication of where to start other than the terminology.

You may be interested in this schematic indicating the relationships between 1-binary-operation grouplike structures:

Answer 2

I do not think "changing the operation to subtraction mod $4$" is actually well-defined. This is because $a-b\neq b-a$ unless $a=b$. So, what you need to do is something like, $$h\ast g:=hg^{-1}$$ which is not usually associative as $(h\ast g)\ast k=hg^{-1}k^{-1}$ while $h\ast (g\ast k)=h\ast (gk^{-1})=hkg^{-1}$. However, this structure has a right identity element, and every element has an inverse (which is itself: $g\ast g=gg^{-1}=1$). An alternative construction would be $$h\ast g:=h^{-1}g^{-1}$$ but this is actually an isomorphism (see if you can spot it! It is similar to something called the "opposite" group, which is the group under the operation $$h\ast g:=gh$$ and the opposite group is isomorphic to the group itself (the same is not true for other "opposites", such as opposite rings.))

Anyway, we shall think about the first operation: $h\ast g:=hg^{-1}$. mixedmath has pointed out in his answer that $G$ under this is either a quasigroup or a magma. It is, indeed, a quasigroup, as if $g\ast a=h$ then the only possibility for $a$ is $h^{-1}g$, while if $b\ast g=h$ then the only possibility for $b$ is $hg^{-1}$. (Look up the definition on, say, wikipedia if you are not sure what one is - basically, these $a$ and $b$ must exist and be unique.)

Note that for this operation to be associative you would need $kg^{-1}=g^{-1}k^{-1}$ for all $g, k\in G$ which means every element must have order $2$ (take $g=1$), and so $g^{-1}=g$ so the operation is just the operation of $G$! Note that if every element of $G$ has order $2$ then $G$ is necessarily abelian, so this is a boring case!

Again, this operation is commutative if and only if every element of $G$ has order $2$, as $h\ast g=hg^{-1}$ while $g\ast h=gh^{-1}$ so we need $hg^{-1}=gh^{-1}$ so $hg^{-1}hg^{-1}=1$. Taking $g=1$...