# 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, f:(2,+infinite)-->(3,+infinite) is isomorphism between groups.

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### 1 Answer

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'.**