Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

2 Answers 2

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

share|improve this answer
    
+1, nice diagram! –  lhf May 28 '12 at 11:00
2  
@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 $g=1$...

share|improve this answer
1  
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. –  David Toth May 28 '12 at 20:34

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.