## Groups and neutral elements

This is my first experiment in presenting formalized mathematics in a blog post. Ideally, such presentation should be generated by the theorem proving environment from machine-verified sources. Since the software for converting Isabelle theory files to WordPress postable HTML is not ready yet, the post below is a result of conversion done manually.

theory Group_ZF_1 imports Group_ZF
begin

In a typical textbook a group is defined as a set $G$ with an associative operation such that two conditions hold:

A: there is an element $e\in G$ such that for all $g\in G$ we have $e\cdot g = g$ and $g\cdot e =g$. We call this element a “unit” or a “neutral element” of the group.

B: for every $a\in G$ there exists a $b\in G$ such that $a\cdot b = e$, where $e$ is the element of $G$ whose existence is guaranteed by A.

The validity of this definition is rather dubious to me, as condition A does not define any specific element $e$ that can be referred to in condition B – it merely states that a set of such units is not empty. Of course it does work out in the end as we can prove that the set of units has exactly one element, but still the definition by itself is not valid. You just can’t reference a variable bound by a quantifier outside of the scope of that quantifier.

One way around this is to first use condition A to define the notion of a monoid, then prove the uniqueness of $e$ and then use the condition B to define groups.

Another way is to write conditions A and B together as follows:

$\exists_{e \in G} (\forall_{g \in G} e\cdot g = g \wedge g\cdot e = g) \wedge (\forall_{a\in G}\exists_{b\in G} a\cdot b = e).$

This is rather ugly.

The third way that I want to talk about is an amusing way to define groups directly without any reference to the neutral elements. Namely, we can define a group as a non-empty set $G$ with an associative operation “$\cdot$” such that

C: for every $a,b\in G$ the equations $a\cdot x = b$ and $y\cdot a = b$ can be solved in $G$.

This blog post aims at proving the equivalence of this alternative definition with the usual definition of the group, as formulated in IsarMathLib’s Group_ZF.thy. The informal proofs come from an Aug. 14, 2005 post by buli on the matematyka.org forum.

An alternative definition of a group

We will use the multiplicative notation for the group operation. To do this, we define a context (locale) that tells Isabelle to interpret $a\cdot b$ as the value of function $P$ on the pair $\langle a,b \rangle$.

locale group2 =
fixes P
fixes dot (infixl $\cdot$ 70)
defines dot_def [simp]: $a \cdot b \equiv P\langle a,b \rangle$

The next theorem states that a set $G$ with an associative operation that satisfies condition C is a group, as defined in IsarMathLib’s Group_ZF theory.

theorem (in group2) altgroup_is_group:
assumes A1: $G\neq 0$ and A2: $P \{\text{is associative on}\} G$
and A3: $\forall a\in G.\ \forall b\in G.\ \exists x\in G.\ a\cdot x = b$
and A4: $\forall a\in G.\ \forall b\in G.\ \exists y\in G.\ y\cdot a = b$
shows $\text{IsAgroup}(G,P)$
proof -
from A1 obtain a where D1: $a\in G$
with A3 obtain x where D2: $x\in G$ and D3: $a\cdot x = a$
from D1 A4 obtain y where D4: $y\in G$ and D5: $y\cdot a = a$
have T1: $\forall b\in G.\ b = b\cdot x \wedge b = y\cdot b$
proof
fix b assume A5: $b\in G$
with D1 A4 obtain yb where D6: $y_b \in G$
and D7: $y_b\cdot a = b$
from A5 D1 A3 obtain xb where D8: $x_b\in G$
and D9: $a\cdot x_b = b$
from D7 D3 D9 D5 have
$b = y_b\cdot (a\cdot x)$  and $b = (y\cdot a)\cdot x_b$
moreover from D1 D2 D4 D8 D6 A2 have
$(y\cdot a)\cdot x_b = y\cdot (a\cdot x_b)$ and $y_b\cdot (a\cdot x) = (y_b\cdot a)\cdot x$
using IsAssociative_def
moreover from D7 D9 have
$(y_b\cdot a)\cdot x = b\cdot x$  and  $y\cdot (a\cdot x_b) = y\cdot b$
ultimately show $b = b\cdot x \wedge b = y\cdot b$
qed
moreover have $x = y$
proof -
from D2 T1 have $x = y\cdot x$
also from D4 T1 have $y\cdot x = y$
finally show $x = y$
qed
ultimately have $\forall b\in G.\ b\cdot x = b \wedge x\cdot b = b$
with D2 A2 have $\text{IsAmonoid}(G,P)$ using IsAmonoid_def
with A3 show $\text{IsAgroup}(G,P)$
using monoid0_def monoid0.unit_is_neutral IsAgroup_def
qed

The converse of theorem altgroup_is_group: in every (classically defined) group the condition C holds.
In informal mathematics we can say “Obviously condition C holds in any group.” In formalized mathematics the word “obviously” is not in the language. The next theorem is proven in the context called group0 defined in IsarMathLib’s theory Group_ZF.thy. Similarly to group2 that context defines $a\cdot b$ as $P\langle a,b\rangle$, where $P: G\times G\rightarrow G$ is the group operation. It also defines notation related to the group inverse and adds the assumption that the pair $(G,P)$ is a group to all its theorems. This is why in the next theorem we don’t explicitely assume that $(G,P)$ is a group – this assumption is implicit in the group0 context.

theorem (in group0) group_is_altgroup: shows
$\forall a\in G.\ \forall b\in G.\ \exists x\in G.\ a\cdot x = b$ and $\forall a\in G.\forall b\in G.\ \exists y\in G.\ y\cdot a = b$
proof -
{ fix a b assume A1: $a\in G$  $b\in G$
let x = $a^{-1} \cdot b$
let y = $b\cdot a^{-1}$
from A1 have
$x \in G$  $y \in G$ and $a\cdot x = b$  $y\cdot a = b$
using inverse_in_group group_op_closed inv_cancel_two
hence $\exists x\in G.\ a\cdot x = b$ and $\exists y\in G.\ y\cdot a = b$
} thus
$\forall a\in G.\forall b\in G.\ \exists x\in G.\ a\cdot x = b$ and
$\forall a\in G.\forall b\in G.\ \exists y\in G.\ y\cdot a = b$
qed

end

This blog post has been generated from IsarMathLib’s Group_ZF_1.thy theory file, see the relevant pages of IsarMathLib proof document.