Given the set G=(0,1) and the law of composition x*y=xy/(2xy-x-y+1), Verify if e=1/2 is the neutral element of the group G.
The neutral element for the law of composition given by x*y = xy/ (2xy - x - y +1) is one for which any element x gives x*(1/2) = x
x *(1/2) = (1/2)x/(2(1/2)x - (1/2) - x + 1
=> (x/2)/(x - (1/2) - x + 1)
=> 2*x / 2
Also the law of composition is symmetric with respect to x and y.
As x*(1/2) = x, we have (1/2) as the neutral element for the law of composition x*y = xy/ (2xy - x - y +1).
The property of a neutral element of a group is:
e*g = g*e = g, where g is an element of the set G.
We'll apply the composition law and we'll replace e by 1/2:
x*(1/2) = x
We'll replace x*(1/2) by x/2/(2x/2 - x - 1/2 + 1)
We'll compute the expression to check if it yields x.
x/2/(2x/2 - x - 1/2 + 1) = x/2(x - x - 1/2 + 1)
x/2(1 - 1/2) = x/(2/2) = x/1 = x
We notice that the expression x*(1/2) = x, therefore e = 1/2 represents the neutral element of the given group, whose law of composition is x*y=xy/(2xy-x-y+1).