AXIOMATIC DUALITY BETWEEN ASSOCIATIVITY AND INVERTIBILITY 2
Recall that the conventional axiomatisation of groups is that of a binary law that
is associative, admits a unit, and inverses. What remains to be checked therefore
is associativity and invertibility.
Let therefore x,yand zbe three items of our structure S. As it is assumed that
interprets a binary law ◦, we may define uto be u:
=x◦yand vto be v:
=y◦z.
The first axiom can therefore be rewritten as follows:
x◦y=u
y◦z=v
x◦v=w
u◦z=w
The premises being valid by construction, this proves the existence of a wsuch
that x◦v=wand such that u◦z=w. Which proves associativity.
Let now γbe any item of our structure X. We have assumed the existence
of a unit e. We may therefore specialise the second axiom with the following
assignments: x→γ,z→γ,u→e,v→eand w→γ. Which yields:
γ◦e=γ
e◦γ=γ
γ◦y=e
y◦γ=e
As eis known to be a unit, the premises hold. Which implies the existence of
aysuch that the conclusions holds. Which shows that yis an inverse to γand
therefore that all items of Xadmit inverses.
Hence Sis a group, and any structure satisfying these two axioms under the
assumption that interprets a binary operation admitting a unit is a group.
Conversely, suppose that Gis a group with a neutral element eand a composition
law ◦. We then interpret to be a ternary relation where ab cmeans a◦b=c.
We want to show that the two axioms are verified as that would prove our theorem.
We now check the first axiom, and therefore suppose that xy uand yz v
hold. Which means that x◦y=uand y◦z=v. We then define w0and w1as
w0:
=x·vand w1:
=u·z. As Gis a assumed to be a group, the composition law
is associative, and we then know that w0=w1, and we call wthis item. We then
have xv wand uz w. Which proves that the first axiom holds.
We now check the second axiom, and therefore suppose that xv wand uz w.
Which means that x◦v=w=u◦z. As Gis assumed to be a group, it admits
inverses, and we then have x◦v◦z−1=uand v=x−1◦u◦z. Defining y0and y1
as y0:
=v◦z−1and y1:
=x−1◦u, we then get x◦y0=uand y1◦z=v. In other
words, xy0 uand y1z vhold.
But we also know that the following also holds:
y0=v◦z−1=x−1◦x◦v◦z−1=x−1◦u◦z◦z−1=x−1◦u=y1
We may therefore define yto be that element y0=y1, and we then know that
xy uand yz vhold. Therefore the second axiom holds.
The converse is therefore proven: all groups satisfy the two axioms. This shows
that groups are therefore axiomatised by these two axioms under the assumption
that they interpret a composition law admitting a unit. Which proves the theorem.