Inverse elements in the absence of identities/associativity.

Question

Lets view groups as consisting of a binary operation, a distinguished element $e$, and unary operation $x \mapsto x^{-1}$. Then the group axioms can be stated as follows.

1. $(xy)z=x(yz).$
2. $xe=ex=x.$
3. $xx^{-1}=x^{-1}x=e.$

Supposing we dropped axiom 2 and forgot about $e$, we would need to rephrase 3. One possibility would be

3'. $xx^{-1}y=x^{-1}xy=yxx^{-1}=yx^{-1}x=y.$

So my first question is, can 3' be shortened and simplified?

Now suppose we dropped associativity, as well. Then 3' needs to be further rephrased. One possibility would be

3''. $(xx^{-1})y=(x^{-1}x)y=y(xx^{-1})=y(x^{-1}x)=y,$ and $x(x^{-1}y)=x^{-1}(xy)=(yx)x^{-1}=(yx^{-1})x=y.$

Again, my question is, can 3'' be shortened and simplified?

Answer 1

This isn't exactly an answer to your question, but I thought you might like to know how inverse elements are defined in semigroups (i.e. structures with only your first axiom).

If $S$ is a semigroup, and $a\in S$, then an inverse of $a$ is any element $b\in S$ such that $aba=a$ and $bab=b$. This is equivalent to the usual definition in the case of groups, and doesn't necessitate the existence of an identity (as your 2' ends up doing).

An element can have more than one inverse. If every element of $S$ has a unique inverse, then $S$ is called an inverse semigroup. Inverse semigroups are very nice objects which capture the idea of partial symmetry.

I know very little about non-associative structures, so I'm not sure whether there is a standard definition of inverse there. You might find the section on inverse properties in quasigroups here of interest.

Answer 2

You are assuming (indeed, using) the very axioms you are trying not to introduce or rely upon, even though you clearly are relying on them, which is not at all desirable in mathematics.

We need to state such assumptions explicitly, if we are going to use them, as they also serve as "definitions:" definitions required if 3' and 3'' are to express what you want them to express.

What does $x^{-1}$ mean with respect to $x$ if we do not define it in terms of some element $e$ for which $ex = xe = x$, such that there exists an element $x^{-1}$ with $x^{-1}x = xx^{-1} = e$?

Similarly, with associativity.

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I'm not entirely clear about your motivation. If you want to maintain the integrity of a group structure, we need axioms $1, 2, 3$, though the axioms are equivalent to requiring only left-identity and left-inverse in the second and third properties/axioms:

$(2')$ $\exists e^L, \forall x: \;e^Lx = x$

$(3')$ $\forall x, \exists x^{-1^L}:\; x^{-1^L}x = e^L$

With associativity, one can prove that $e^L = e \;$ as in $(2)$, and $\;x^{-1^L} = x^{-1}\;$ as in $(3)$.

Answer 3

a. You already capture 2. in once with 3'.:

Let $e:=xx^{-1}$ for an arbitrary but fixed $x$, then we have $ey=y=ye$ according to 3'. for all $y$, so $e$ is a unit, so must be unique, alternatively, we can get $yy^{-1}=yy^{-1}xx^{-1}=xx^{-1}$.

b. Again, the first part of 3''. states that $xx^{-1}$ is a unit (and also $x^{-1}x$), so we didn't really get rid of 2., and the associativity is not needed for showing uniqueness of a unit ($e=ef=f$ if $e,f$ are both units).

However, the original group axioms can also be simplified, one only needs to require the existence of a left unit $e$ and a left inverse for all $x$.

Answer 4

Unless I'm missing something obvious, any nonempty set $S$ with axioms 1, 2, and 3' is in fact a group. First, take any $y \in S$. Certainly $yy^{-1} \in S$ because the operations of multiplication and inverse are defined in $S$. Let $z=yy^{-1}$. Then for all $x \in S$, we have $zx=xz=x$. Thus $z$ is an identity. Furthermore, an elementary argument shows $z$ is unique. Voila, $S$ is a group.

But with non-associativity, I am out of my element (no pun intended).....

Hope this helps!

Answer 5

Foreword: I am surprised because I may (I must...) have read this question several times before tonite but only now I thought about adding one definition, not yet mentioned in this thread, that may hopefully be pertinent and interesting as well among the others.

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In associative structures, the inverse element is defined after the identity element. In a generic (non associative) groupoid, an inverse property can be defined without identity element instead.

A groupoid or quasigroup $G$ has the left inverse property if $$ \forall x\in G, \exists\ z\in G : z(xy)=y\ \forall y\in G\tag{1} $$ and has the right inverse property if $$ \forall x\in G, \exists\ t\in G : (yx)t=y\ \forall y\in G.\tag{2} $$ $G$ is also said to have the inverse property if it satisfies both the above equations.

If $z$ as in $(1)$ exists and the groupoid $G$ is a quasigroup, such $z$ is also unique because in a quasigroup the equation $z(xy)=y$ has a unique solution $z$ given $x,y$: therefore we can denote $z$ with $x_l^{-1}$.
If $t$ as in $(2)$ exists and the groupoid $G$ is a quasigroup, such $t$ is also unique because in a quasigroup the equation $(yx)t=y$ has a unique solution $t$ given $x,y$: therefore we can denote $t$ with $x_r^{-1}$.

Let now $G$ be a loop with identity element $e$. Then: $$ \forall x\in G, \exists\,!\,x_l\in G : x_l\,x = e $$ and $$ \forall x\in G, \exists\,!\,x_r\in G : x\,x_r = e. $$ The (existence and) uniqueness of $x_l$ and $x_r$ is again due to the fact that the loop is a quasigroup and hence all linear equations have a unique solution in here. We can call $x_l$ and $x_r$ left inverse of $x$ and right inverse of $x$ respectively.

But beware: in the absence of associativity, the only existence of $x_l$ and $x_r$ does not imply $(1)$ and $(2)$ to be verified, indeed not all the loops have the inverse property.

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