# Equivalence of definitions of finitely generated group

This article gives a proof/explanation of the equivalence of multiple definitions for the term finitely generated group

View a complete list of pages giving proofs of equivalence of definitions

## Contents

## Statement

The following are equivalent for a group :

- It has a finite generating set.
- Every generating set of the group has a subset that is finite and is also a generating set.
- The group has at least one minimal generating set and every minimal generating set of the group is finite.
- The minimum size of generating set of the group is finite.
- The group is a join of finitely many cyclic subgroups.

## Proof

### (1) implies (2)

**Given**: A group with a finite generating set and a generating set (not necessarily finite).

**To prove**: There exists a finite subset of that is also a generating set for .

**Proof**:

Step no. | Assertion/construction | Facts used | Given data used | Previous steps used | Explantion |
---|---|---|---|---|---|

1 | For each element , there exists a finite subset of such that . | is a generating set for , | Since generates , and , we can write as a word in terms of the generating set . The word has finite length, hence can use only finitely many of the elements of . Denote by the finite subset of comprising the elements used in such a word. Note that depends on the choice of word, and as such is not unique, but this is not a problem for us. | ||

2 | Let where is the subset defined using Step (1). Then, is a finite subset of . | is finite. | Step (1) | By Step (1), each is finite. Since is finite, is a union of finitely many finite sets, hence is finite. | |

3 | is a finite subset of that is a generating set for | is a generating set for . | Step (1) | By Step (1), for each . Since , for each . Thus, , so . We already know that , so . This forces that . |

### (2) implies (3)

**Given**: A group with the property that every generating set for contains a finite subset that is also a generating set.

**To prove**: has at least one minimal generating set, and every minimal generating set of the group is finite.

**Proof**:

**Proof of the first part**: Start with as a generating set for itself. By the given, there exists a subset that is finite and is a generating set for . Let be the collection of subsets of that generate . is nonempty since . Since it is finite, it must have a minimal element. Thus, there exists a minimal generating set for .

**Proof of the second part**: Suppose is a generating set for . If is infinite, then it cannot be minimal, since by assumption, it contains a finite (and hence strictly smaller) generating set for . Thus, must be finite.

### (3) implies (4)

This is direct.

### (4) implies (1)

This is direct.

### (1) implies (5)

For any finite generating set, the group is the union of the cyclic subgroups generated by the individual generators.

### (5) implies (1)

If the group is a union of finitely many cyclic subgroups, pick one (cyclic) generator for each cyclic subgroup. The set comprising these generators is a finite generating set for the group.