Show the function f(x)=2x-3 domain M and range R without 0 is isomorphism of groups M,o) and (R without 0, *) M= r wityhout 3/2   xoy=2xy-3x-3y+6

You need to prove that the given function `f(x) = 2x -3`  is a group homomorphism from `(R-{3/2},o) to (R-{0},*), ` hence, you need to check if `f(xoy) = f(x)*f(y)`  such that:

`f(xoy) = 2(xoy) - 3`

Substituting `2xy-3x-3y+6`  for `xoy`  yields:

`f(xoy) = 2(2xy-3x-3y+6) - 3 => f(xoy) = 4xy - 6x - 6y + 12 - 3`

`f(xoy) = 4xy - 6x - 6y + 9`

`f(x)*f(y) = (2x - 3)(2y - 3)`

Opening the brackets, yields:

`f(x)*f(y) = 4xy - 6x - 6y + 9`

Since the left side is equal to the right side,`4xy - 6x - 6y + 9 = 4xy - 6x - 6y + 9` , then `f(xoy) =f(x)*f(y), ` hence, the function `f(x) = 2x -3`  is a group homomorphism from `(R-{3/2},o) to (R-{0},*).`

You need to check if the homomorphism f(x) is a bijection, since if f(x) is a bijection, then f(x) is a group isomorphism.

You need to prove that the function f(x) is a bijection, hence, you need to prove first that the function is injective, such that:

`f(x_1) = f(x_2) = > x_1 = x_2`

`2x_1 - 3 = 2x_2 - 3 => 2x_1 = 2x_2 => x_1 = x_2 => f(x) = 2x - 3`  is injective.

You need to prove that the function is also surjective, such that:

`AA` `y in R-{0}, ` `EE` `x in ` `R - {3/2}`  such that `x = (y+3)/2` , hence, the function `f(x) = 2x - 3`  is surjective

Since the function is injective and surjective, hence, the function `f(x) = 2x - 3`  is a bijection and it is a group isomorphism from `(R-{3/2},o) to (R-{0},*).`

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