# Using one-sided axioms to show <G,*> is a group

1. Sep 26, 2007

### eq1

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

Prove that a set G, together with a binary operation * on G, satisfying the
following three axioms:
A1) The binary operation * on G is associative
A2) There exists a left identity element e in G such that e*x=x for all x in G
A3) For each a in G, there exists a left inverse a' in G such that a'*a=e
is a group

2. Relevant equations

Our definition of a group:

A group is a set G, and a closed binary operation * on G, such that
the following axioms are satisfied:
G1) * is associative on G
G2) There is an element e in G such that for all x in G
x*e = e*x = x (existence of an identity element e for *)
G3) Corresponding to each a in G there is an element a' in G such that
a*a' = a'*a = e (existence of inverse a' of a)

3. The attempt at a solution

G1, is given by A1

By A2 we know e*x=x for all x in G.
We must show x*e=x for all x in G to finish proving G2.

x*e=x, what is to be proved
x'*(x*e)=x'*x, we know x' exists from A3, existence of inverse
(x'*x)*e=e, using A1, associativity
e*e=e, using A3, existence of inverse
e=e, therefore x*e=x for x in G proving G2 holds.

By A3 we know there exists x' in G such that x'*x=e for all x in G.
We must show x*x'=e for all x in G to finish proving G3.

x*x'=e, what is to be proved
(x'*x)*x'=x'*e, using A1, associativity
e*x'=x', using the above theorem
x'=x', therefore x*x'=e for x in G proving G3 hold.

Therefore the set G, together with a binary operation * on G, satisfying
A1, A2 and A3 is a group.

4. My questions

Does ending with the e=e statement or x'=x' statement show the original statement holds? Since it ends up with LHS=RHS I think it does but maybe I should try to string statements in the form e*x = ... = x*e although I am not sure why that would be better. I guess it is just what I am used to seeing.

Also, I skipped the part about the binary operation being closed for a group. Kind of important, I know. I think (hope) it was indirectly mentioned in the axioms when it was stated * in associative -on- G. Is this correct or is it something I will also need to prove? If so, any ideas?

Thank you for taking the time to read all this.

2. Sep 26, 2007

### mcampbell

"Does ending with the e=e statement or x'=x' statement show the original statement holds?"

No, you've only shown that e=e and x'=x'...profound statements but not what you set out to show. Here's a hint: x' is in G so x' also has a left inverse...what is it?

As for the closure, I am unsure. I don't see how one could show the operation is closed without more information about the nature of the group.

3. Sep 26, 2007

### HallsofIvy

Staff Emeritus
Doesn't the phrase "binary operation on G" (as opposed to "binary operation from G to H") imply the operation always gives a value in G? In other words, can't you take closure as "given" by the hypotheses?

4. Sep 26, 2007

### eq1

mcampbell,
I think understand what you are trying to tell me and unfortunately I don't think I even showed e=e and x'=x' since I started from unproved statements. I'll work with your suggestion now and see what I can come up with.

HallsofIvy,
I was thinking that too but I wasn't positive. Specifically I wasn't sure if on and onto were equivalent. I kind of suspected they were as my text seems to use it that way, but it doesn't explicity state it.

Thank you guys for the replies.

5. Sep 26, 2007

### quasar987

It's not hard, but you gotta start at the right place or you're gonna try for a long god damn time. The right place to start is first with the inverse business. You're gonna try to show that the left inverse of each element in also a right inverse. And for this, you're gonna start with the equation

a*a'=a*e*a'

Then showing the left identity is also a right identity is easy.

6. Sep 27, 2007

### eq1

Wow quasar987, you were not kidding about the long god damn time. I spun on this for almost two hours. But when I combined your approach and mcampbell's approach I think I have a proof that works.

Proof of right inverse

As x' is in G by A3 there exists an inverse, say x'', such that x''*x'=e

Thus x'*x = e, and x''*x'=e
setting them equal to eachother
x' * x = x'' * x'
x' * x = x'' * e * x'
x' * x = x'' * x' * x * x'
x' * x = e * x * x', by associativity and left identity
x' * x = x * x' = e

And the right identity follows easily.
Thank you guys so much. This problem is going to give me nightmares. So simple but so hard at the same time.