Given the groups G=(2,+infinite) and x*y=xy-2x-2y+6 and g'=(3,+infinite), xoy=xy-3x-3y+12prove that the function f(x)=x+a,...

First, we need to determine the identity elements of the
groups G and G'.

One of the four axioms of the group is the
identity.

A group must have an identity element, such
as:

x*e = e*x = x

We'll
determine the identity element for G:

x*e = xe - 2x - 2e +
6

But x*e = x => xe - 2x - 2e + 6 =
x

We'll move all elements that do not contain x, to the
right:

e(x-2) = 2x + x +
6

e(x-2) = 3x - 6

e(x-2) =
3(x-2)

e = 3

We'll determine
the identity element for G':

xoe' =
x

xe' - 3x - 3e' + 12 =
x

We'll move all elements that do not contain x, to the
right:

e'(x-3) = 4x -
12

e'(x-3) = 4(x-3)

e' =
4

If f is isomorphism of groups G and G', then f(3) =
4

f(3) = 3 + a => 3 + a = 4 => a =
1

The function f(x) is f(x) = x +
1.

We'll prove that f is morphism of
groups:

f(x*y) =
f(x)of(y)

f(x*y) = x*y + 1 = xy - 2x - 2y + 7
(1)

f(x)of(y) = (x+1)o(y+1) = (x+1)(y+1) - 3(x+1) - 3(y+1)
+ 12

f(x)of(y) = xy - 2x - 2y + 7
(2)

We notice that (1) = (2) => f(x*y) =
f(x)of(y)

Now, we have to prove that f is bijective, to
verify if f is isomorphism of G and G'.

We'll verify if
f(x) is one to one function:

f(x) = f(y) <=>
x + 1 = y + 1 => x = y => f(x) is one to one function
(3)

We'll verify if f(x) is on to
function:

For any y that belongs to G', there is an x that
belongs to G, such as f(x) = y.

x + 1 = y <=>
x = y - 1

But x>2 => y - 1 > 2
=> y > 3,true => f(x) is on to function
(4)

From (3) and (4) => f(x) is bijective
function.

Since f(x) = x + 1 is a morphism of
G and G' and it is also a bijective function, then f:(2,+inf.)->(3,+inf.) is an
isomorphism of groups G and G'.