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

Statement

The following are equivalent for a group $G$:

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

Proof

(1) implies (2)

Given: A group $G$ with a finite generating set $A$ and a generating set $B$ (not necessarily finite).

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

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explantion
1	For each element $a \in A$, there exists a finite subset $C_a$ of $B$ such that $a \in \langle C_a \rangle$.		$B$ is a generating set for $G$, $A \subseteq G$		Since $B$ generates $G$, and $a \in A \subseteq G$, we can write $a$ as a word in terms of the generating set $B$. The word has finite length, hence can use only finitely many of the elements of $B$. Denote by $C_a$ the finite subset of $B$ comprising the elements used in such a word. Note that $C_a$ depends on the choice of word, and as such is not unique, but this is not a problem for us.
2	Let $C = \bigcup_{a \in A} C_a$ where $C_a$ is the subset defined using Step (1). Then, $C$ is a finite subset of $B$.		$A$ is finite.	Step (1)	By Step (1), each $C_a$ is finite. Since $A$ is finite, $C$ is a union of finitely many finite sets, hence is finite.
3	$C$ is a finite subset of $B$ that is a generating set for $G$		$A$ is a generating set for $G$.	Step (1)	By Step (1), $a \in \langle C_a \rangle$ for each $a \in A$. Since $C_a \subseteq C$, $a \in \langle C \rangle$ for each $a \in A$. Thus, $A \subseteq \langle C \rangle$, so $\langle A \rangle \le \langle C \rangle$. We already know that $G = \langle A \rangle$, so $G \le \langle C \rangle \le G$. This forces that $G = \langle C \rangle$.

(2) implies (3)

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

To prove: $G$ 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 $G$ as a generating set for itself. By the given, there exists a subset $C \subseteq G$ that is finite and is a generating set for $G$. Let $\mathcal{F}$ be the collection of subsets of $C$ that generate $G$. $\mathcal{F}$ is nonempty since $C \in \mathcal{F}$. Since it is finite, it must have a minimal element. Thus, there exists a minimal generating set for $G$.

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