# Changes

## Generating set of a group

, 11:08, 4 July 2008
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{{termrelatedto|combinatorial group theory}}
<section begin=mainbeginner/>
==Definition==
* Every element of the group can be expressed in terms of the elements of this subset by means of the group operations of multiplication and inversion. (note that if the subset is a [[symmetric subset]], i.e., it is closed under taking inverses, then every element of the group must be a product of elements in the subset. Symmetric subsets arise, for instance, when we take a union of subgroups).
* There is no proper subgroup of the group containing this subset<section end=beginner/>
* There is a surjective map from a [[free group]] on that many generators to the given group, that sends the generators of the free group to the elements of this ''generating set''.
<section begin=beginner/>
The elements of the generating set are termed generators.
where for each $a_i$, either $a_i \in S$ or $a_i^{-1} \in S$ (here, the $a_i$s are not necessarily distinct). In the situation where $S$ is a [[symmetric subset]] (i.e. $a_i \in S \implies a_i^{-1} \in S$) we do not need to throw in inverses. This happens, for instance, when $S$ is a union of subgroups of $G$.
* If $H$ is a [[proper subgroup]] of $G$ (i.e. $H$ is a [[subgroup]] of $G$ that is not equal to the whole of $G$), then $H$ cannot contain $S$.<section end=beginner/>
* Consider the natural map from the free group on as many generators as elements of $S$, to the group $G$, which maps the freely generating set to the elements of $S$. This gives a surjective homomorphism from the free group, to $G$.
<section end=main/>
==Constructs==