The universal property of the free group basically tells us that "the definition of the free group doesn't depend (up to isomorphism) on the exact details of the set $X$ we picked; only on its cardinality", which is morally a very useful thing to know. You may skip down to the next subheading if you might be scared of category theory, but the property itself doesn't need category theory and is helpful.

The universal property is the technical category-theoretic fact that the free-group functor is left adjoint to the forgetful functor, and it is not so immediately useful as the other more concrete properties on this page, but it is exceedingly important in category theory as a very natural example of a pair of adjoint functors and as an example for the general_adjoint_functor_theorem.

Statement and explanation

The universal property which characterises the free group is:

The free group $FX$ on the set $X$ is the group, unique up to isomorphism, such that for any group $G$ and any function of sets $f:X\to G$ ^[1], there is a unique group homomorphism $\overline{f}:FX\to G$ such that $\overline{f}({\rho }_{a_1}{\rho }_{a_2}\dots {\rho }_{a_n})=f\left(a_1\right)\cdot f\left(a_2\right)\cdot \cdots \cdot f\left(a_n\right)$.

This looks very opaque at first sight, but what it says is that $FX$ is the unique group such that:

Given any target group $G$, we can extend any map $f:X\to G$ to a unique homomorphism $FX\to G$, in the sense that whenever we're given the image of each generator (that is, member of $X$) by $f$, the laws of a group homomorphism force exactly where every other element of $FX$ must go. That is, we can specify homomorphisms from $FX$ by specifying where the generators go, and moreover, every possible such specification does indeed correspond to a homomorphism.

Why is this a non-trivial property?

Consider the cyclic group $C_3$ with three elements; say $\{e,a,b\}$ with $e$ the identity and $a+a=b$, $a+b=e=b+a$, and $b+b=a$. Then this group is generated by the element $a$, because $a=a$, $a+a=b$, and $a+a+a=e$. Let us pick $G=(Z,+)$. We'll try and define a map $f:C_3\to Z$ by $a\mapsto 1$.

If $C_3$ had the universal property of the free group on $\{e,a,b\}$, then we would be able to find a homomorphism $\overline{f}:C_3\to Z$, such that $\overline{f}\left(a\right)=1$ (that is, mimicking the action of the set-function $f$). But in fact, no such homomorphism can exist, because if $\overline{f}$ were such a homomorphism, then $\overline{f}\left(e\right)=\overline{f}(a+a+a)=1+1+1=3$ so $\overline{f}\left(e\right)=3$, which contradicts that the image of the identity under a group homomorphism is the identity.

In essence, $C_3$ "has extra relations" (namely that $a+a+a=e$) which the free group doesn't have, and which can thwart the attempt to define $\overline{f}$; this is reflected in the fact that $C_3$ fails to have the universal property.

A proof of the universal property may be found elsewhere.

1. ^{^︎}

   Here we're slightly abusing notation: we've written $G$ for the underlying set of the group $G$ here.