Question #249909

Check whether R is a group under binary operation

a*b=a+b-ab


1
Expert's answer
2021-10-12T10:21:41-0400

(R,*) is a group, if:

1) a,bR:abR\forall a,b\in R: a*b\in R

2) a,bR:(ab)c=a(bc)\forall a,b\in R: (a*b)*c=a*(b*c)

3) eR\exist e\isin R aR:ae=ea=a\forall a\in R : a*e=e*a=a

4) aR\forall a\isin R a1R:aa1=a1a=e\exists a^{-1}\isin R: a*a^{-1}=a^{-1}*a=e


1) a,bR:(a+bR,abR)abR\forall a,b\in R: (a+b\isin R, ab\in R) \to a*b \in R

2) a,bR:(ab)c=(a+bab)c=a+bab+c(a+bab)c=\forall a,b\in R: (a*b)*c = (a+b-ab)*c = a+b-ab+c - (a+b-ab)c =

=a+bab+cacbc+abc=aabac+abc+(b+cbc)=(a+(b+cbc))a(b+cbc)=a(bc)= a+b-ab + c -ac-bc+abc = a-ab-ac+abc+(b+c-bc) =(a+(b+c-bc)) - a(b+c-bc) = a*(b*c)

3) ae=a+eae=e+aea=eaa*e = a+e-ae=e+a-ea=e*a

a * e = a

a + e - ae = a

e(1 - a) = 0

e = 0

(e=0)R\exist (e = 0)\isin R aR:a0=0a=a\forall a\in R : a*0=0*a=a

4) aa1=ea*a^{-1} = e

a+a1aa1=0a+a^{-1}-aa^{-1} = 0

a+a1(1a)=0a+a^{-1}(1-a) = 0

a1=a1aa^{-1} = -{\frac a {1-a}}

For a=1 there is no a1a^{-1}, so R is not a group under binary operation *.


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