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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?

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Also, if a structure has a left and a right identity then they are necessarily the same element and it has an identity... – user1729 May 28 '12 at 9:50
Which is not what we have here, because $0-x=-x\not=x$ in $\mathbb{Z}_4$. – Gregor Bruns May 28 '12 at 10:05
up vote 4 down vote accepted

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:

enter image description here

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+1, nice diagram! – lhf May 28 '12 at 11:00
@lhf Note that the diagram is from the Wikipedia magma page linked above. – Bill Dubuque May 28 '12 at 16:01

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 $g=1$...

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I think it is not so hard to imagine "subtraction mod $4$" to mean subtract the two numbers and reduce mod $4$. I see no definedness problems. Also, the OP distinctly did not claim that there were both left and right identities. – mixedmath May 28 '12 at 11:17
@mixedmath: I know, I was just making this clear. Also, I wasn't entirely sure what the OP meant by a double identity. I have realised what he means now, so I'll get rid of the paragraph! – user1729 May 28 '12 at 19:58
Thanks for pointing out the terms "opposite group" and "quasigroup" as well for the detailed explanation of their meanings. – Dávid Tóth May 28 '12 at 20:34

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