# Group Theory: Is the following a valid Group?

1. Oct 8, 2015

### sa1988

1. The problem statement, all variables and given/known data

Is the following a valid group?

The values contained in the set of all real numbers ℝ, under an operation ◊ such that x◊y = x+y-1

2. Relevant equations

Axioms of group theory:
Closure
Associativity
There must exist one identity element 'e' such that ex=x for all x
There exists an inverse 'y' for every element such that xy = yx = e

3. The attempt at a solution

Testing of each axiom:

Closure: All x◊y will give a value existing in ℝ. Closure holds.

Associativity: Test: x◊y◊z
= x◊(y◊z) = x◊(y+z-1) = x+y+z-2
= (x◊y)◊z = (x+y-1)◊z = x+y+z-2
Associativity holds.

Identity: e◊x = x
Group definition says e◊x = e+x-1.
Hence, e◊x = e+x-1 = x
This is true for all x when e = 1
There is an identity

Inverses: x◊y = y◊x = e = 1
Meaning, x+y-1 = y+x-1
x,y∈ℝ so the additive property defined by operation ◊ means that all elements will be able to have an inverse in ℝ such that x◊y=e
Inverses exist

Final answer, YES THIS IS A GROUP.

So, that's my attempt at an answer. However I don't know if I've really gone about it the right way, or even if I'm correct. Thoughts?

Thanks.

2. Oct 8, 2015

### Staff: Mentor

I don't see anything wrong with what you did, although I have to admit it's been many years since I did this stuff.

3. Oct 8, 2015

### sa1988

Well, that's a positive enough response for now! Thanks :)

4. Oct 9, 2015

### Fredrik

Staff Emeritus
The part about inverses can be improved a bit. You didn't make it clear how you know that each x has an inverse. The best way is to find a formula for the inverse of an arbitrary x.

5. Oct 9, 2015

### sa1988

Thanks, I've gone over it and written:

x◊y = x+y-1 = e = y+x-1 = y◊x
Meaning
y = e+1-x
which is true and different for all x because it only uses addition to create the inverse

6. Oct 9, 2015

### HallsofIvy

Didn't you just show that e= 1? So you are saying that if y= 1+ 1- x= (1+ 1)- x= 2- x then x◊ y= x+ y- 1= x+ (2- x)- 1= 1.

7. Oct 10, 2015

### sa1988

I wasn't sure if I was 'allowed' to do a proof with e=1, or if I was supposed to keep it as 'e' so it's more general.

As I'm primarily a physics student, I haven't really done much on proofs before.

If the 'e=1' method is fine then I'll certainly use it!

8. Oct 10, 2015

### Fredrik

Staff Emeritus
You can and should use it. The equality $e=1$ means that the symbols $e$ and $1$ represent the same element of the set. If they represent the same element, then the expression $e+1-x$ must represent the same element as the expression $2-x$. In other words, since $e=1$, we have $e+1-x=2-x$.

Since you have completed the solution, I'll show you mine. It's essentially the same as yours.

For all $x,y,z\in\mathbb R$, we have
$$(x\diamond y)\diamond z= (x+y-1)+z-1= x+(y+z-1)-1=x\diamond(y\diamond z).$$ So $\diamond$ is associative. For all $x\in\mathbb R$, we have
$$x\diamond 1=x+1-1=x$$ and
$$1\diamond x=1+x-1=x.$$ So $1$ is an identity of $\diamond$. Let $x\in\mathbb R$. We have
$$x\diamond (2-x)=x+(2-x)-1=1$$ and
$$(2-x)\diamond x=(2-x)+x-1=1.$$ Since $1$ is an identity of $\diamond$, this means that $2-x$ is an inverse of $x$. Since $x$ is an arbitrary element of $\mathbb R$, this means that every element of $\mathbb R$ has an inverse.