Alternative Axioms for Groups
Contents |
Theorem
In the definition of a group, the axioms for the existence of an identity element and for closure under taking inverses can be replaced by the following two axioms:
- Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a left identity;
- For any element $g$ in a group $G$, there exists at least one left inverse of $g$.
Alternatively, we can also replace the aforementioned axioms with the following two:
- Given a group $G$, there exists at least one element $e \in G$ such that $e$ is a right identity;
- For any element $g$ in a group $G$, there exists at least one right inverse of $g$.
Thus we can formulate the group axioms as either of the following:
Group Axioms (Left)
A group is an algebraic structure $\left({G, \cdot}\right)$ which satisfies the following four conditions:
| \((G 0):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a, b \in G:\) | \(\) | \(\displaystyle a \cdot b \in G\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Closure Axiom | |
| \((G 1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a, b, c \in G:\) | \(\) | \(\displaystyle \left({a \cdot b}\right) \cdot c = a \cdot \left({b \cdot c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity Axiom | |
| \((G_L 2):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \exists e \in G:\) | \(\) | \(\displaystyle \forall a \in G: e \cdot a = a\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Left Identity Axiom | |
| \((G_L 3):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a \in G:\) | \(\) | \(\displaystyle \exists b \in G: b \cdot a = e\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Left Inverse Axiom |
Group Axioms (Right)
A group is an algebraic structure $\left({G, \cdot}\right)$ which satisfies the following four conditions:
| \((G 0):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a, b \in G:\) | \(\) | \(\displaystyle a \cdot b \in G\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Closure Axiom | |
| \((G 1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a, b, c \in G:\) | \(\) | \(\displaystyle \left({a \cdot b}\right) \cdot c = a \cdot \left({b \cdot c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity Axiom | |
| \((G_R 2):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \exists e \in G:\) | \(\) | \(\displaystyle \forall a \in G: a \cdot e = a\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Right Identity Axiom | |
| \((G_R 3):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a \in G:\) | \(\) | \(\displaystyle \exists b \in G: a \cdot b = e\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Right Inverse Axiom |
Proof
Suppose we define a group $G$ in the usual way, but make the first pair of axiom replacements listed above:
- the existence of a left identity;
- every element has a left inverse.
Let $e \in G$ be a left identity and $g \in G$.
Then, from Left Inverse for All is Right Inverse, each left inverse is also a right inverse with respect to the left identity.
Also from Left Inverse for All is Right Inverse we have that the left identity is also a right identity.
Also we have that such an Identity is Unique, so this element can rightly be called the identity.
So we have that:
- $G$ has an identity;
- each element of $G$ has an element that is both a left inverse and a right inverse with respect to this identity.
Therefore, the validity of the two axiom replacements is proved.
$\blacksquare$
The proof of the alternate pair of replacements (existence of a right identity and closure under taking right inverses) is similar.
Warning
Suppose we build an algebraic structure with the following axioms:
| \((0):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a, b \in G:\) | \(\) | \(\displaystyle a \cdot b \in G\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Closure Axiom | |
| \((1):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a, b, c \in G:\) | \(\) | \(\displaystyle \left({a \cdot b}\right) \cdot c = a \cdot \left({b \cdot c}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Associativity Axiom | |
| \((2):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \exists e \in G:\) | \(\) | \(\displaystyle \forall a \in G: a \cdot e = a\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Right Identity Axiom | |
| \((3):\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall a \in G:\) | \(\) | \(\displaystyle \exists b \in G: b \cdot a = e\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Left Inverse Axiom |
Then this does not (necessarily) define a group (although clearly a group fulfils those axioms).
Let $\left({S, \circ}\right)$ be the algebraic structure defined as:
- $\forall x, y \in S: x \circ y = x$
That is, $\circ$ is the left operation.
From Left Operation All Elements Right Identities, every element serves as a right identity.
Then given any $a \in S$, we have that $x \circ a = x$ and as $x$ is an identity, axiom $(3)$ is fulfilled as well.
But from the complementary result of More than One Left Identity then No Right Identity, there is no right identity and therefore no identity element, so $\left({S, \circ}\right)$ is not a group.
Sources
- Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $7.12$
- Richard A. Dean: Elements of Abstract Algebra (1966): $\S 1.2$
- Ian D. Macdonald: The Theory of Groups (1968): $\S 1$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 26 \gamma$
