Is group associativity equivalent to assuming equivalence of left and right inverses?

Question

Associativity is usually taken as one of the axioms that define a group, however I'm sure I recall reading that it can instead be proved if it is taken as axiomatic that the left and right inverses are the same. In other words, given a set, $G$, and a binary relation, $\cdot$, can we prove that $(a\cdot b)\cdot c = a\cdot(b\cdot c)$ for all $a,b,c \in G$ from the following axioms?

• Closure: $a\cdot b \in G$ for all $a, b \in G$, .
• Identity: There exists a unique element $e \in G$ such that $e\cdot a = a\cdot e = a$ for all $x \in G$.
• Inverses: For all $x \in G$, there exists a unique element $x^{-1} \in G$ such that $x\cdot x^{-1} = x^{-1}\cdot x = e$.

If not, am I misremembering and is there similar axiom that allows associativity to be proved?

Answer 1

No. The natural numbers under the Euclidean metric, $(\mathbb{N}, d)$ where $d(a, b)=|b-a|$, is a counter-example.

Identity is $0$, as $d(n, 0)=|n-0|=n$ and $d(0, n)=|0-n|=n$.

Inverse of $n$ is $n$, as $d(n, n)=|n-n|=0$.

It is non-associative as, for example, $0=d(1, 1)=d(1, d(2,3))\neq d(d(1,2), 3)=d(1, 3)=2$.

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Thanks to lulu for spotting an issue with a previous version!

Answer 2

A small counter-example is the set $G=\{e,a,b\}$ with a binary operation $\,\cdot\,$ where $$ e\cdot x=x\cdot e=x\cdot x=x \;\; \text{ for all }\;\; x\in G$$ while $a\cdot b=b\cdot a=e$.

Verify that it satisfies closure, identity, and inverses axioms. Now consider

$$ a\cdot(b\cdot b)=a\cdot b = e \ne (a\cdot b)\cdot b = e\cdot b = b$$

which implies that the magma $(G,\cdot)$ is not associative but it is idempotent and commutative.

The operation table is $$ \begin{array}{|c|c|c|c|} \hline \cdot & e & a & b \\ \hline e & e & a & b \\ \hline a & a & a & e \\ \hline b & b & e & b \\ \hline \end{array}$$

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[Edit Jun 26, 2003] Actually, more is true. That is, using the same set $G=\{e,a,b\}$ with a closed binary operation $\,\cdot\,$ where $\, e\cdot x = x\cdot e = x\,$ for all $\, x\in G \,$ and either $\, a\cdot a = b\cdot b = e\,$ or$\, a\cdot b = b\cdot a = e,\,$ then the binary operator $\,\cdot\,$ is associative if and only if $\, a\cdot b = b\cdot a = e,$ $a\cdot a = b,\,$ $b\cdot b = a\,$ (cyclic group). In only one case out of eight is the operation associative.

Answer 3

No.

Definition: A loop is a set $L$ with a binary operation $\ast$ such that

• There is an identity element $1_L$ in $L$ with respect to $\ast$.

• For all $a,b\in L$, there exist $x,y\in L$ with $$\begin{align} a\ast x&=b,\\ y\ast a&=b.\end{align}$$

A loop need not be associative. (Consider the octonions.)

You can check whether a loop has left and right inverses (that need not be the same).

Let $a\in L$ be arbitrary and let $b=1_L$ in the above.

For examples of loops with two-sided inverses, see here:

Examples of loops which have two-sided inverses.