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"?(adsbygoogle = window.adsbygoogle || []).push({});

The pair [itex]\big(G,(x,y)\mapsto xy\big)[/itex] 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^{-1}x=xx^{-1}=e can be simplified to ex=x and x^{-1}x=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^{-1}x is thesameleft 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

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.

- 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
^{-1}x.- If x
^{-1}x 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.

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

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# Left identity, left inverse.

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