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group [2014/12/18 18:48] nikolaj |
group [2015/04/16 19:16] nikolaj |
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| | @#55EE55: postulate | @#55EE55: $\forall g.\,\exists g^{-1}.\;(g*g^{-1}=g^{-1}*g=e)$ | | | @#55EE55: postulate | @#55EE55: $\forall g.\,\exists g^{-1}.\;(g*g^{-1}=g^{-1}*g=e)$ | | ||
| + | ----- | ||
| === Alternative definitions === | === Alternative definitions === | ||
| - | Let $\langle G,* \rangle $ be a set $G$ with a binary operation. I'll rewrite the group axioms explicitly in the first order language: | + | == Group axioms explicitly in the first order language == |
| + | Let $\langle G,* \rangle $ be a set $G$ with a binary operation. | ||
| 1. $\forall (a,b\in G).\ (a*b\in G)$ | 1. $\forall (a,b\in G).\ (a*b\in G)$ | ||
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| 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. | 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. | ||
| - | ----- | + | == Sharper definitions == |
| + | We could just define left units and left inverses and prove from the group axioms that they are already units and inverses. | ||
| + | ----- | ||
| === Subset of === | === Subset of === | ||
| [[Monoid]], [[Loop]] | [[Monoid]], [[Loop]] | ||
