A free generating set $S$ of a free group $F$ has the smallest cardinality of the generating sets.

Question

I'm reading Clara Löh's book "Geometric group theory, an introduction" and i'm going through the free groups section. She stated the following:

Let $F$ be a free group.

1. Let $S\subset F$ be a free generating set of $F$ and let $S' \subset F$ be a generating set. Then $|S'|\geq |S|$.

Here a free generating set is a set such that $F$ satisfies the Universal propertie of free groups. She said that it could be derived from the universal propertie mapping $S$ to $\mathbb{Z}/2\mathbb{Z}$. I tried this and i'm not sure if my argument it's correct.

The cardinality of maps from $S$ to $\mathbb{Z}/2\mathbb{Z}$ is $2^{|S|}=|P(S)|$ (where $P(S)$ is the power set of $S$), and as that cardinality is in bijection to the homomorphisms from $F$ to $\mathbb{Z}/2\mathbb{Z}$ then we can restrict that homomorphism to $S'$, that set is a subset of the functions from $S'$ to $\mathbb{Z}/2\mathbb{Z}$, so $2^{|S'|}\geq 2^{|S|}$ and that implies $|S'|\geq |S|$.

I'm fairly sure that an argument like that is what Clara Löh wanted from her hint, so if it's not correct i would love to see how to correct the argument.

Answer 1

This argument is not correct: $2^{|S'|}\geq 2^{|S|}$ does not necessarily imply $|S'|\geq |S|$ for infinite sets (in fact, this claim is independent of ZFC; see Why continuum function isn't strictly increasing? for instance). The argument does work if $S$ is countable, though.

To handle the uncountable case, note that the subgroup generated by $S'$ has cardinality at most $\max(\aleph_0,|S'|)$. So $|S|\leq\max(\aleph_0,|S'|)$ which implies $|S|\leq |S'|$ if $S$ is uncountable.

Answer 2

This cannot work, since it is consistent with ZFC that there are infinite cardinals $\lambda < \kappa$ with $2^{\lambda} = 2^{\kappa}$. See MO/67473.

Here is a reduction to the case of vector spaces. If $G$ is a group, then $V(G) := G/\langle \langle g^2 : g \in G \rangle \rangle$ is a commutative group, and in fact a vector space over $\mathbb{F}_2$. If $G$ a free group over some set $S$, it follows formally that $V(S)$ is a free $\mathbb{F}_2$-vector space over $S$, since the universal property is transported. If $G$ is generated by $S$, then $V(G)$ is generated by $S$ (as a vector space). (To the category-theoretic minded people: $V : \mathbf{Grp} \to \mathbf{Vect}_{\mathbb{F}_2}$ is left adjoint to the forgetful functor.)

So, if $G$ is free over $S$ and $S'$ is a generating set of $G$, then it follows that $V(S)$ has a basis $S$ and is generated by $S'$, hence $|S'| \geq |S|$.