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My group theory is very rusty. If I want to just start with left inverses and left identities, must I link the axioms, or can I leave them independent?

e.g., is it enough to say "there exists at least one $e \in G$ s.t. $ea=a$ for all $a \in G$" and "for each $a$ in $G$, there exists $a^{-1} \in G$ s.t. $a^{-1}a$ is an identity", or must I say "there exists at least one $e \in G$ s.t. ($ea=a$ for all $a \in G$ AND for each $a \in G$ there exists $a^{-1} \in G$ s.t. $a^{-1}a=e$)".

This seems like it must be a FAQ, but I just can't find it. If I don't assume the linkage, I run into problems where I show (for example) that if $a^{-1}a=e_1$, $b=aa^{-1}$, $b^{-1}b=e_2$, then $b=e_2$. That will get me things like $ae_1=a$, but not the general case.

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3 Answers 3

up vote 5 down vote accepted

Suppose $X$ is any set with at least two elements, and consider the binary operation $*$ on $X$ given by $x*y=y$. Then $*$ is clearly associative, and everything is a left identity. Thus in your weak sense, everything is left inverse to everything. But of course $(X, *)$ is not a group.

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1  
Ugh. Thanks, Chris. I'd even been looking at that example, but just not making the leap from "everything is an identity" to "therefore, everything is an inverse". –  Johann Hibschman Apr 3 '12 at 16:25

It is the linked definition that is formally correct, so it is to be preferred. The problem is that the unlinked definition has become so standardized that we are stuck with it, and the majority of textbooks use it.

In fact, the way it is usually stated admits (at least) two distinct plausible interpretations. Are we assuming the inverse axiom for some $e$ that satisfies the identity axiom, or for all such $e$? Of course it turns out that this makes no difference, but we do not know that at a priori.

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The main difficulty with the first phrasing you have is in the meaning of "identity" in the first formulation. You do need to connect the identity from the inverse clause to the identity from the identity clause.

That is, if you require:

  1. The operation $\cdot$ is associative;
  2. There exists $e\in G$ such that $e\cdot x = x$ for all $x\in G$; and
  3. For every $x\in G$ there exists $y\in G$ such that $yx=e$ (the same $e$ from 2)

then you do have enough to guarantee a group. If you replace $3$ with:

3'. For every $x\in G$ there exists $y\in G$ such that for all $z\in G$, $(yx)z = z$.

then you don't have enough, as the example given by Chris Eagle shows.

With 1, 2, and 3, however, you can establish what you have a group as follows:

Step 1. If $x\in G$ satisfies $xx=x$, then $x=e$.

Proof. Let $y\in G$ be such that $yx=e$. Then $e = yx = y(xx) = (yx)x = ex=x$.

Step 2. Let $x\in G$ and let $y$ be such that $yx=e$. Then $xy=e$

Proof. $(xy)(xy) = x((yx)y) = x(ey) = xy$; hence, by step 1, $xy=e$.

Step 3. $xe=x$ for all $x\in G$.

Proof. Let $y$ be such that $yx=e$; then $xy=e$ by Step 2, so $$xe = x(yx) = (xy)x = ex = x.$$

Thus, $e$ is a two-sided identity, and inverses from 3 are two-sided inverses, giving a group.

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Thanks, Arturo. This is one of those things that I'd been wondering about every now and then, but I never thought of it when I was in the same room as Sarah's copy of Herstein. –  Johann Hibschman Apr 3 '12 at 18:11

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