Myujth

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 $Ain Z$ 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 $Ox$=$∃Y∈Z$ such that $X*Y=S$,and the stabilizer of X--$Fx$ to be the set $Fx$=$T*X=X*T=0$.

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 $c,d in mathbbZ$. 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 $1 = c star e star f$.