SOLUTION: show that the binary operation is a group and determine if it is Abelian H = {(a,b)belong to R^(2): a can not equal 0} with product (a,b)(c,d) = (ac,ad + b)

Question 26684: show that the binary operation is a group and determine if it is Abelian
H = {(a,b)belong to R^(2): a can not equal 0} with product
(a,b)(c,d) = (ac,ad + b)
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
(a,b)*(c,d) = (ac,ad + b)
TO SHOW THIS OPERATION * IS A GROUP,WE NEED TO PROVE...
1..CLOSOURE...
AC IS ALSO REAL NUMBER SINCE A AND C ARE REAL...AD IS ALSO REAL AND SO IS AD+B.SO (AC,AD+B)IS AN ELEMENT OF R^2
2..ASSOCIATIVE....
TPT...{(A,B)*(C,D)}*(E,F)=(A,B)*{(C,D)}*(E,F)}..
LHS=(AC,AD+B)(E,F)=(ACE,ACF+AD+B)
RHS=(A,B)(CE,CF+D)=(ACE,ACF+AD+B)=LHS
3...EXISTENCE OF IDENTITY....LET IT BE (I1,I2)
WE SHOULD HAVE (A,B)*(I1,I2)=(A,B) AND (I1,I2)*(A,B)=(A,B)
AI1=A...SO...I1=1
AI2+B=B...SO....I2=0....SO...(1,0) IS THE IDENTITY ELEMENT..IT IS AN ELEMENT OF R^2...OK....
LET US CHECK...
(1,0)*(A,B)=(1A,1B+0)=(A,B)...OK...
4...EXISTENCE OF INVERSE....
WE SHOULD FIND (X,Y)SO THAT (X,Y)(A,B)=(1,0)=(A,B)*(X,Y)
XA=1....OR..X=1/A.....
XB+Y=0.....Y=-XB=-B/A.....SO INVERSE IS (1/A,-B/A)..SINCE...A IS NOT ZERO ,WE HAVE THE INVERSE AS AN ELEMENT OF R^2.
LET US CHECK...
(A,B)(1/A,-B/A)=(A*1/A,((A*-B)/A)+B)=(1,0) ....OK....
HENCE * AS A BINARY OPERATION IS A GROUP.
WE FIND THAT
(a,b)*(c,d) = (ac,ad + b)..WHERE AS
(C,D)*(A,B) = (CA,CB+D) WHICH ARE NOT EQUAL.
HENCE THIS IS NOT ABELIAN

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