# Homework Help: Question about a group

1. Feb 14, 2013

### port31

1. The problem statement, all variables and given/known data
So we have this operation x*y=x+2y+4
and then our 2nd one is x*y=x+2y-xy
I need to check if it is commutative,associative, and if it has a identity and an inverse.
3. The attempt at a solution
y*x=y+2x+4 so it is not commutative

x*(y*z)=x+2(y+2z+4)+4=x+2y+4z+12
(x*y)*z=(x*y)+2z+4=x+2y+4+2z+4
Not associative

Now I will solve for the identity
e*x=e+2x+4=x
e=-x-4
x*e=x+2e+4=x
e=-2
Since I have 2 different identity elements this means that one does not exist because
e should be unique.
Since there is no e there is no inverse.

Now for the second one x*y=x+2y-xy
y*x=y+2x-yx Does not commute

(x*y)*z=x+2y-xy+2z-xz-2yz+xyz
x*(y*z)=x+2y+4z-2yz-xy-2xz+xyz
Not associative

x*e=x+2e-xe=x
e=0
e*x=e+2x-ex=x
e(1-x)=-x
The identity element does not seem to be unique so it does not exist.
Just want to know if I am doing this right.

2. Feb 14, 2013

### Staff: Mentor

What do you mean "our 2nd one"? You can't define the * operation on a group in two different ways.

3. Feb 14, 2013

### port31

The second one is just another problem. They are 2 separate problems.

4. Feb 14, 2013

### Staff: Mentor

Then you should identify them as such instead of clumping them together as you did.

5. Feb 14, 2013

### Staff: Mentor

What are you trying to show here? Is the exercise for each to say whether some set with the given operation is a group?

If so, you don't need to check every group axiom for the operation. For example, if the operation isn't associative, then that's enough to say that the set and the operation aren't a group.

Also, just because something isn't unique, that doesn't mean it doesn't exist.

6. Feb 15, 2013

### HallsofIvy

More fundamentally, the identity can't depend upon "x".