| 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 | $\forall g.\,\exists g^{-1}.\;(g*g^{-1}=g^{-1}*g=e)$ |
We could just define left units and left inverses and prove from the group axioms that they are already units and inverses.
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.