Intuition of characteristic property of the free group

Question

Here is a theorem about characteristic property of the free group:

Theorem (Lee TM). Let $S$ be a set. For any group $H$ and any map $f:S\to H$, there exists a unique homomorphism $g:F(S)\to H$ extending $f$.

Here $F(S)$ is free group on $S$.

I know what it says but I don't know why it should be useful. i.e. What is the strategy of such theorems? How it can help to understand $F(S)$?

Can anyone enlighten it by a simple example?

Answer 1

An analogy with vector spaces may be helpful. If $V$ is a vector space with basis $B$, then any set function on $B$ into another vector space $W$ can be extended to a linear map $V \to W$. So $V$ is a free vector space on the basis $B$. In the case of vector spaces this property does not characterise $V$ because every vector space has a basis. This is not true in the case of groups, even if they are abelian. Free groups are special in that they admit "basis" expansions, and group homomorphisms from a free group are determined completely by their action on "basis elements". In the same way that a vector space is determined (upto linear isomorphism) by its vector space dimension, which is the cardinality of any basis for that space, a free group is determined (upto group isomorphism) by the cardinality of its generating set, called its rank.

Answer 2

This characterisation emphasises that free groups are related to homomorphisms. Here is a simple proof which uses this characterisation:

Theorem. If $|X|\geq2$ then $F(X)$ is non-soluble.

Proof. The symmetric group $S_5$ can be generated by two elements. Therefore, by this characterisation of free groups, $F(X)$ subjects onto $S_5$ for all sets $X$ with at least two elements. As $S_5$ is non-soluble, as as solubility is preserved under homomorphic images, the result follows.

(Previously this answer used $S_3$ instead of $S_5$ to prove that free groups are non-nilpotent. The answer is now stronger, as non-soluble implies non-nilpotent.)