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.
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 $1 = c \star e \star f$.