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Prove that we cannot define an binary operation $*$ on the set of integers Z satisfy all of the three properties below simultaneously:

For any $A∈Z,B∈Z,C∈Z:$

1.$A*B=-(B*A)$

2.$(A*B)*C=A*(B*C)$ (Associative Law)

3.For every $A\in Z$ there exist $B∈Z,C∈Z$ such that $A=B*C$

I have got stucked for three days on this questions.Anyway,I will show some result and idea I had:

1.for any X∈Z,we have $X*X=-(X*X)$.So we have $X*X=0$

2.for any X∈Z,we have $X*0=-(0*X)=X*(X*X)=(X*X)*X=0*X$.So we have $X*0=0*X=0$

3.Now For any $X∈Z$($X≠0$).We define the orbit of X--$Ox$ to be the set $Ox$={S|$∃Y∈Z$ such that $X*Y=S$},and the stabilizer of X--$Fx$ to be the set $Fx$={T|$T*X=X*T=0$}.

My goal is to prove that actually $Ox=Fx$.And therefore since $X∈Fx$,so $X∈Ox$,and we reach a contradiction since $X∉Ox$(otherwise if $∃Y∈Z$ such that $X*Y=X$,then $(X*Y)*Y=X*(Y*Y)=X*0=X*Y=X=0$)

It is easy to see that $Ox⊆Fx$,since for any $S∈Ox$,we have $X*S=X*(X*Y)=(X*X)*Y=0*Y=0$,so $S∈Fx$

However,for the other direction,I cannot deduce out,which I need help.

I think the backgroud of this question is the orbit&stabilizer theorem in the course abstract algebra.So I have a strong intuition that I am on the right track.

Andrew
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    Not sure this is clear. Does $AB$ mean $A\star B$? If so, how can you deduce, say, that $X\star X=-X\star X\implies X\star X=0$? All we know is that $X\star X=(-1\star X)\star X$. If $AB$ doesn't mean $A\star B$ I think you need to clarify when you are using the new operation and when you are using ordinary multiplication. – lulu Dec 25 '18 at 14:00
  • @AndrewArmstrong So you mean they are all supposed to be the new operaton? That there is no ordinary multiplication involved in any of these laws? – Bram28 Dec 25 '18 at 14:07
  • In that case, do you agree with me that it is not obvious that $X\star X=-X\star X\implies X\star X=0$? At least that requires a proper proof. here, by the way, is a good tutorial on formatting for this site. I am using "\star" to get $\star$). – lulu Dec 25 '18 at 14:07
  • To be clear: I am guessing that "anti-commutativity" is meant to mean $A\star B=-1\star A\star B$ where the placement of the parenthesis is irrelevant (thanks to associativity), though of course I could be guessing wrong. But, really, I think the whole thing needs a lot of clarification. – lulu Dec 25 '18 at 14:09
  • @AndrewArmstrong Sure! Can you please edit the rest of your post likewise? Thanks! I also took out the 'typical' from the title of your question, by the way ... that was not a good use of the word here. Did you mean to say that it is a 'strange' or 'difficult' question? – Bram28 Dec 25 '18 at 14:13
  • @AndrewArmstrong Hmm, but can you a least place $ signs around all your mathematical expressions? That will do a lot already! – Bram28 Dec 25 '18 at 14:30
  • @lulu Why can't we conclude $X\star X=0$ if it equals its negative? $X\star X$ is an integer and negation hasn't changed. – John Douma Dec 25 '18 at 14:31
  • @JohnDouma Well, you are guessing that the OP means that $-1$ is multiplying through in the ordinary way while I was guessing that the OP meant to use $\star$ there as well. But guessing is a waste of time...the thing should be clarified. – lulu Dec 25 '18 at 14:37
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    @lulu In his first requirement he has parentheses around the product so $X\star X=-(X\star X)$. In ordinary integer arithmetic I interpret $-X$ as the additive inverse of $X$ and it is a thing to prove that $-X=(-1)*X$,.i.e. the additive inverse of an integer is the additive inverse of $1$ times the integer. – John Douma Dec 25 '18 at 14:41
  • @lulu I believe you are correct about this because even if $X=-X$ we cannot conclude that $X=0$ because we don't know if the integers are still an integral domain under this product. – John Douma Dec 25 '18 at 14:51
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    @JohnDouma if $-A$ is intended to denote the additive inverse of the integer $A$ in the usual sense, then $A=-A$ does indeed imply that $A=0$. And it is entirely possible that this is what is intended. – lulu Dec 25 '18 at 14:55

1 Answers1

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Note the associativity condition implies that parentheses are redundant, so something like $a \star b \star c \star d$ is uniquely-defined without parentheses.

Observe that, by axiom $3$, $1 = c \star d$ for some $c,d \in \mathbb{Z}$. A repeated application of axiom $3$ gives us that $d= e \star f$ for some integers $e,f$.

Hence we get that $1 = c \star e \star f$. Observe now that $-1 = f \star c \star e$ by the first axiom. A repeated application of the first axiom gives that $1 = e \star f \star c$. A third application gives $-1 = c \star e \star f$. But this is a contradiction, since we know that $1 = c \star e \star f$.

MathematicsStudent1122
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  • More generally, in any abstract algebra with binary $ \star $ and unary $ - $ satisfying the 3 properties in the question, $ - $ must be the identity. This can be proven by supposing $x ≠ -x$ for some $x$ and deriving a contradiction as above. – Hans Lub Dec 26 '18 at 10:14