Group


Definition

An algebraic structure \((G, \circ)\) is a group if and only if:

(G0) Closure \(\forall x,y \in G : x \circ y \in G\)
(G1) Associativity \(\forall x,y,z \in G : (x \circ y) \circ z = x \circ (y \circ z)\)
(G2) Identity \(\exists 1 \in RE : \forall x \in G : 1 \circ x = x = x \circ 1\)
(G3) Inverse \(\forall x \in G : \exists x^{-1} \in G : x \circ x^{-1} = 1 = x^{-1} \circ x\)

Axioms

        assoc: ?a * ?b * ?c = ?a * (?b * ?c)
 left_neutral: 1 * ?a = ?a
right_neutral: ?a * 1 = ?a
 left_inverse: inverse ?a * ?a = 1
right_inverse: ?a * inverse ?a = 1
THEOREM   

The group inverse is an involution.

$$\left(g^{-1}\right)^{-1} = g$$

Proof available

THEOREM    $$\forall a,b,x \in G : ax = b \Leftrightarrow x = a^{-1}b$$ $$\forall a,b,x \in G : xa = b \Leftrightarrow x = ba^{-1}$$

Proof available

THEOREM   

Let \((G, *)\) be a group. Then \((G, *)\) has the Latin square property.

Proof available

THEOREM   

The identity of a group is unique.

Proof available

THEOREM   

\((\mathbb{N}, +)\) is not a group.

Proof available

THEOREM   

\((\mathbb{N}, *)\) is not a group.

Proof available

THEOREM   

\((\mathbb{Q}, *)\) is not a group.

Proof available

THEOREM   

\((\mathbb{R}, *)\) is not a group.

Proof available

See also

Category: Group Theory - ProofWiki

Parent topics