Group
Set
| context | $G$ |
| definiendum | $ \langle G,* \rangle \in \mathrm{it}$ |
| inclusion | $\langle G,* \rangle \in \mathrm{monoid}(G)$ |
| let | $e$ |
| such that | $\forall g.\, e*a=a*e=a$ |
| range | $g,g^{-1}\in G$ |
| postulate | $\forall g.\,\exists g^{-1}.\;(g*g^{-1}=g^{-1}*g=e)$ |
Alternative definitions
Sharper definitions
We could just define left units and left inverses and prove from the group axioms that they are already units and inverses.
Group axioms explicitly in the first order language
Let $\langle G,* \rangle $ be a set $G$ with a binary operation.
1. $\forall (a,b\in G).\ (a*b\in G)$
2. $\forall (a,b,c\in G).\ ((a*b)*c=a*(b*c))$
3. $\exists (e\in G).\ \forall (a\in G).\ (a*e=e*a=a) $
4. $\forall (a\in G).\ \exists (a^{-1}\in G).\ (a*a^{-1}=a^{-1}*a=e)$
The first axiom is already implied if “$*$” is a binary operation $*:G\times G\to G$.
For given $G$, the set $\text{group}(G)$ is the set of all pairs $\langle G,* \rangle$, containing $G$ itself, as well a binary operation which fulfills the group axioms. One generally calls $G$ the group, i.e. the set with respect to which the operation “$*$” is defined.
Subset of
