# Left identity, left inverse.

1. Sep 4, 2010

### Fredrik

Staff Emeritus
Consider a binary operation on a set G. A an element e of G is said to be a left identity if ex=x for all x. If x is in G, an element y of G is said to be a left inverse of x if yx is a left identity. A right identity and right inverse is defined similarly. Is the following an adequate definition of "group"?

The pair $\big(G,(x,y)\mapsto xy\big)$ is said to be a group if
(1) the binary operation is associative,
(2) every member has a left inverse.

(By my definition of "left inverse", (2) implies that a left identity exists, so no need to mention that in a separate axiom). I have seen the claim that the group axioms that are usually written as ex=xe=x and x-1x=xx-1=e can be simplified to ex=x and x-1x=e without changing the meaning of the word "group", but I don't quite see how that can be sufficient. It's possible that I have weakened the definition too much by not explicitly saying that x-1x is the same left identity for all x. I'll have to add that as part of the definition if it can't be proved from the other axioms. Without including that as an axiom, I'm able to prove that
• There's at most one right identity.
• Every x has at most one right inverse.
• If x-1 is a left inverse of x, then x is a left inverse of x-1.
• If f is a right identity, then for any x, f is a right inverse of x-1x.
• If x-1x has a right inverse, then x is a right inverse of x-1.
• If x is a right inverse of x-1, then x-1 is a right inverse of x.
I'll show the calculations if someone asks for them. I don't see how to finish the proof that a "group" in the sense of the definition above, is actually a group. Looks like I only need to prove that there's a right identity.

This might be one of those times when I see the answer immediately after I post the question.

Last edited: Sep 4, 2010
2. Sep 4, 2010

### Hurkyl

Staff Emeritus
I think there are models that aren't groups.

My idea is to adjust the free group on a set of symbols. We have left inverses so we can cancel off the left, so what if we can't cancel off the right? I want the rightmost symbol to pick out a class of words that are a group, but without the rightmost symbol itself being cancellable.

In the normal free group, the elements are words in the alphabet of generators and inverses of the generators, modulo the rewrite rules
xx' ->
x'x ->​
(the right hand sides are the null strings)

My plan is that you only allow these rewrite rules if they do not occur on the right edge of a word.

If I haven't made any errors, then each of the words of the form xx' and x'x are left identities, for x one of the generators, but all of them are unequal.

3. Sep 4, 2010

### Hurkyl

Staff Emeritus
Reformulation of the same idea:

Let G be a group. Let X = |G| x |G|.

Define a product on X by
(a,b).(c,d) = (abc,d)​

This is associative:
((a,b).(c,d)).(e,f) = (abc,d).(e,f) = (abcde,f) = (a,b).(cde,f) = (a,b).((c,d).(e,f))​
the elements (x',x) are left identities:
(a,a').(b,c) = (aa'b,c) = (b,c)​
and (b',a') is the left inverse of (a,b):
(b',a').(a,b) = (b'a'a,b) = (b',b)​

4. Sep 4, 2010

### Fredrik

Staff Emeritus
Thank you very much. That answers it, and saves me a significant amount of time. There's no right identity in X, since (a,b)=(a,b).(c,d)=(abc,d) implies d=b. So my definition must at least be supplemented by the requirement that x-1x is the same left identity for all x. And it's not hard to see that this is sufficient. x-1x=e implies (xx-1)2=xx-1, which implies xx-1=e. So for arbitrary x, we have x-1x=e=xx-1, and this means that for arbitrary x, we have xe=x(x-1x)=(xx-1)x=ex=x. So the left identity is also a right identity.

By the way, I assume that the notation |G| means "the underlying set of G". Is that notation standard?