TITLE: A generating set of cardinality $n$ in the free group $F_n$ is a free basis.
QUESTION [6 upvotes]: Let $F_n$ be the free group on $n$ letters. Let $S=\{s_1,s_2,\ldots,s_n\}$ be a set of $n$ elements of $F_n$.
 Is there any way to prove that $S$ is in fact a free basis for $F_n$ without using the Nielsen transformations?

REPLY [6 votes]: Poster jgon has presented notions that lie indeed at the heart of the matter. A slight variation in the arguments given above can be presented as follows:

Lemma. Let $G$ be a finitely generated group and $n \in \mathbb{N}^*$ arbitrary. Then the set of subgroups of $G$ of index $n$ is finite.

Proof: We shall write $\Sigma_{n}$ for the symmetric group of degree $n$; for natural numbers $r, s$ the notation $[r, s]$ refers to the interval induced by the natural order on $\mathbb{N}$; for $H \leqslant G$ we will use $(G/H)_{s}$ to denote the quotient of set $G$ by the left congruence modulo $H$.
Set $\mathscr{S}=\{H \leqslant G\ |\ |G:H|=n\}$ and for each $H \in \mathscr{S}$ introduce $\Phi_H=\{\varphi \in \mathrm{Isom}_{\mathrm{Ens}}((G/H)_{s}, [1,n])\ |\ \varphi(H)=1\}$; it is not difficult to see that $\Phi_{H} \neq \varnothing$ for every $H \in \mathscr{S}$, hence $$\prod_{H \in \mathscr{S}} \Phi_{H} \neq \varnothing$$ and we can thus fix a certain (yet arbitrary) $$\theta \in \prod_{H \in \mathscr{S}} \Phi_{H}$$
which is simply put a family of set-isomorphisms (i.e. mere bijections, that is what isomorphisms are in the category of sets) between the set of left congruence classes modulo each $H$ of index $n$ and the interval $[1, n]$, such that the class of $1_{G}$ (which is $H$ itself) is mapped to $1$.
For each $H \in \mathscr{S}$ there is a canonical left action of $G$ on $(G/H)_{s}$, which can be transported via the bijection $\theta_{H}$ to an action of $G$ on $[1, n]$, to be written as $\alpha_{H}$, whose associated morphism of permutation representation we will denote by $\rho_{H} \in \mathrm{Hom}_{\mathrm{Gr}} (G, \Sigma_{n})$. By construction, $\theta_{H}$ is an isomorphism between the left $G$-sets $(G/H)_{s}$ and $[1, n]$ (the latter given by $\alpha_{H}$ of course), hence $$H=\mathrm{Stab}_{\alpha_{H}} 1 \tag{1}$$ From this we infer that the mapping
$$\mathscr{S} \to \mathrm{Hom}_{\mathrm{Gr}}(G, \Sigma_n) \\ H \mapsto \rho_H \tag{2}$$
is injective (if $K, H \in \mathscr{S}$ are such that $\rho_{H}=\rho_{K}$, this is then equivalent to $\alpha_H=\alpha_K$ and relation (1) applies).
Fixing $S \subseteq G$ to be a finite generating set, one easily sees that the restriction mapping
$$\mathrm{Hom}_{\mathrm{Gr}}(G, \Sigma_n) \to \mathrm{Hom}_{\mathrm{Ens}}(S, \Sigma_n) \\ f \mapsto f_{| S} \tag{3}$$
is also injective (two morphisms that agree on a generating system are equal). The injectivity of maps (2) and (3) means that $|\mathscr{S}| \leqslant |\mathrm{Hom}_{\mathrm{Ens}}(S, \Sigma_n)|=\left|\Sigma_n^S\right|=(n!)^{|S|}$; as $S$ is by hypothesis finite (more explicitly, by hypothesis such an $S$ certainly exists), then $(n!)^{|S|} \in \mathbb{N}$ and therefore equally $|\mathscr{S}| \in \mathbb{N}$, $\mathscr{S}$ being thus finite (the implicit set-theoretical conception being that natural numbers are precisely the finite cardinal numbers according to the Bourbaki presentation of Set Theory). $\Box$
For the sake of pure contemplation, may we also present an
Alternative proof: Given arbitrary set A we shall denote its symmetric group by $\Sigma(A)=\mathrm{Aut}_{\mathrm{Ens}}(A)$; for arbitrary sets $A, B$ and bijection $\varphi \in \mathrm{Isom}_{\mathrm{Ens}}(A, B)$ one can introduce the mapping
$$\Sigma(\varphi): \Sigma(A) \to \Sigma(B)\\ \Sigma(\varphi)(\lambda)=\varphi \circ \lambda \circ \varphi^{-1}$$
The correspondence between these objects actually implements a functor (from $\mathrm{Iso(Ens)}$ to $\mathrm{Iso(Gr)}$, that is from the core of the category of sets to the corresponding core of the category of groups), so it is easy to infer that $\Sigma(\varphi) \in \mathrm{Isom}_{\mathrm{Gr}}(\Sigma(A), \Sigma(B))$.
For arbitrary $H \leqslant G$ let us denote the normal core of $H$ by $$H_G=\bigcap_{t \in G}tHt^{-1}$$ For the canonical left action of $G$ on the quotient $(G/H)_s$ we shall denote its associated morphism of permutation representation by $\rho_H \in \mathrm{Hom}_{\mathrm{Gr}}(G, \Sigma((G/H)_s))$ and notice that $\mathrm{Ker}(\rho_H)=H_G$. Employing the same notation as before for $\mathscr{S}$, fix an arbitrary $$\varphi \in \prod_{H \in \mathscr{S}} \mathrm{Isom}_{\mathrm{Ens}}((G/H)_s, [1, n])$$
(since cartesian products of nonempty sets are nonempty); finally, define $\theta_H=\Sigma(\varphi_H) \circ \rho_H$ and notice that $\theta_H \in \mathrm{Hom}_{\mathrm{Gr}}(G, \Sigma_n)$ and also that $\mathrm{Ker} \theta_H=H_G$. Also set $$K=\bigcap_{H \in \mathscr{S}} H_G$$
We denote the canonical projection of index $H$ on the direct product (direct power, even) $\Sigma_n^{\mathscr{S}}$ by $p_H$ and we refer to the universality property of direct products to infer the existence of a unique
$$\psi \in \mathrm{Hom}_{\mathrm{Gr}}\left(G, \Sigma_n^{\mathscr{S}}\right)$$
such that $p_H \circ \psi=\theta_H$ for all $H \in \mathscr{S}$. It will thus be the case that $$\mathrm{Ker} \psi=\bigcap_{H \in \mathscr{S}} \mathrm{Ker} \theta_H=K$$
If we abbreviate $G/K=G'$ and denote the canonical surjection attached to the quotient of $G$ modulo $K$ by $\sigma \in \mathrm{Hom}_{\mathrm{Gr}}(G, G')$, the fundamental (iso)morphism theorem for groups tells us that there exists a unique $\gamma \in \mathrm{Hom}_{\mathrm{Gr}}\left(G', \Sigma_n^{\mathscr{S}}\right)$ such that $\psi=\gamma \circ \sigma$, and this unique morphism is furthermore injective. As $G'$ is the quotient of a finitely generated group, it itself is finitely generated and thus isomorphic (via $\gamma$) to a finitely generated subgroup of $\Sigma_n^{\mathscr{S}}$.
Let us also recall the following version of the correspondence theorem: in general, for an arbitrary surjective morphism $f \in \mathrm{Hom}_{\mathrm{Gr}}(F, F')$, by considering sets $\mathscr{S}=\{E \leqslant F\ |\ E \geqslant \mathrm{Ker}f \wedge |F:E|=n\}$ respectively $\mathscr{T}=\{H \leqslant F'\ |\ |F':H|=n\}$ and mappings
$$\Phi: \mathscr{S} \to \mathscr{T} \\ \Phi(E)=f(E)$$
together with
$$\Psi: \mathscr{T} \to \mathscr{S} \\ \Psi(H)=f^{-1}(H)$$
not only are these maps correctly defined but they are mutual inverses (in fact they are isomorphisms of ordered sets, when ordering both $\mathscr{S}$ and $\mathscr{T}$ with inclusion). In our particular case, considering surjection $\sigma$, any subgroup of index $n$ in $G$ automatically includes the kernel of $\sigma$ by construction, so by the correspondence theorem we can infer that $|\mathscr{S}|=|\mathscr{S'}|$, where $\mathscr{S'}=\{H \leqslant G'\ |\ |G':H|=n\}$.
In other words, we have reduced the claim on subgroups of $G$ to the analogous one for $G'$, and in order to prove this latter claim it will suffice to show that $G'$ is finite. This will be achieved as long as we can establish the next auxiliary result, interesting in and of itself, which we will prove independently. $\Box$

Auxiliary Lemma: let $F$ be a finite group and $\Lambda$ an arbitrary set. Then any finitely generated subgroup $G \leqslant F^{\Lambda}$ is finite.

Proof: For arbitrary set $A$, let us write $\mathscr{Eq}(A)$ for the set of all equivalence relations on $A$ and $\Delta_A$ for the diagonal of $A$ (the relation of equality on $A$); for arbitrary $R \in \mathscr{Eq}(A)$ let us denote the canonical surjection attached to $R$ by $\sigma_R \in \mathrm{Hom}_{\mathrm{Ens}}(A, A/R)$. If $f \in \mathrm{Hom}_{\mathrm{Ens}}(A, B)$ let $\mathrm{E}_f=(f \times f)^{-1}(\Delta_B)$ denote the canonical equivalence induced by $f$.
We shall say that $R$ is a finite equivalence if the quotient set $A/R$ is finite; we go on to remark that if $I$ is a finite and nonempty index set and $R \in \mathscr{Eq}(A)^I$ is a family of finite equivalences, then
$$\bigcap_{i \in I} R_i \in \mathscr{Eq}(A)$$ is also finite (finite nonempty intersections of finite equivalences are again finite equivalences): if $\tau_i=\sigma_{R_i}$ for all $i \in I$ and $$p_i: \prod_{j \in I} A/R_j \to A/R_i$$ is the canonical projection of index $i$ of the cartesian product, then by the universality property there will exist a unique
$$\rho: A \to \prod_{i \in I} A/R_i$$ such that $p_i \circ \rho=\tau_i$ for all $i \in I$. By construction, we have $$\mathrm{E}_{\rho}=\bigcap_{i \in I} \mathrm{E}_{\tau_i}=\bigcap_{i \in I} R_i=S$$
(the last equality serving as the definition of object $S$) and by the fundamental (iso)morphism theorem for sets we infer the existence of a unique map $$\varphi: A/S \to \prod_{i \in I} A/R_i$$ such that $\rho=\varphi \circ \sigma_S$, map which is necessarily injective. Hence, owing to this injectivity we can claim that $$|A/S| \leqslant \left|\prod_{i \in I} A/R_i \right|=\prod_{i \in I} |A/R_i|$$
As finite products of finite cardinals (i.e. natural numbers) are finite cardinals and cardinals less or equal to finite ones are themselves finite, it follows that $S$ is also a finite equivalence, justifying our claim.
Having completed preparations, we move on to the core of the problem and define for each $x \in F^{\Lambda}$ the equivalence $\Gamma_x \in \mathscr{Eq}(\Lambda)$ given by $$\lambda \Gamma_x \mu \Longleftrightarrow x_{\lambda}=x_{\mu}$$
It is immediate to notice that $\Gamma_x$ occurs as the canonical equivalence induced by the map
$$\Lambda \to F \\ \lambda \mapsto x_{\lambda}$$
map which induces an injective quotient map $\Lambda/{\Gamma_x} \to F$; as $F$ is finite by hypothesis, we gather that $\Gamma_x$ is a finite equivalence.
Next, fix a certain (yet arbitrary)  finite nonempty generating system $S \subseteq G$, the existence of which is guaranteed by hypothesis and define $$\Delta=\bigcap_{s \in S} \Gamma_s$$
a finite equivalence as we have explained above. Let us remark the properties
$$\Gamma_x \cap \Gamma_y \subseteq \Gamma_{xy} \tag{1}$$ $$\Gamma_x=\Gamma_{x^{-1}} \tag{2}$$ $$\Gamma_{1_{F^{\Lambda}}}=\Lambda \times \Lambda \tag{3}$$
valid for all $x, y \in F^{\Lambda}$. Upon introducing subset $$E=\{x \in F^{\Lambda}\ |\ \Delta \subseteq \Gamma_x \}$$
then by relation (3) $E$ is nonempty (it contains the unit), by (1) it is multiplicatively stable and by (2) it is closed with respect to inverses, which in other words means that $E \leqslant F^{\Lambda}$.
Let us note that by definition $S \subseteq E$, hence $G=\langle S \rangle \leqslant E$; therefore, to prove our claim it will suffice to show that $E$ is finite. To this end, we fix a certain complete and independent system of representatives for $\Delta$ in $\Lambda$, say $M$; as $|M|=|\Lambda/{\Delta}|$ and $\Delta$ is a finite equivalence, $M$ is a finite set. Let
$$p_M: F^{\Lambda} \to F^{M} \\ p_M(x)=x_{|M}=(x_{\lambda})_{\lambda \in M}$$
denote the morphism of restriction to $M$ (its construction is also justified by the universality property of direct products) and $f=(p_M)_{|E} \in \mathrm{Hom}_{\mathrm{Gr}}\left(E, F^M\right)$ the restriction of this morphism to $E$. As $F$ and $M$ are both finite, it will suffice to establish the injectivity of $f$ in order to conclude the finiteness of $E$.
Let $u \in \mathrm{Ker}f$ and $\lambda \in \Lambda$ be arbitrary; there must exist a (unique!) $\mu \in M$ such that $\lambda \Delta \mu$, and by definition of $E$, as $u \in E$ it follows that $\lambda \Gamma_u \mu$, in other words that $u_{\lambda}=u_{\mu}$; however, since $f(u)=u_{|M}=(1_F)_{\mu \in M}$ we conclude that $u_{\lambda}=1_F$ for any index $\lambda \in \Lambda$, whence $u=1_{F^{\Lambda}}$ and thus $\mathrm{Ker}f$ is trivial. $\Box$
As a remark, this lemma can easily be generalised to monoids: in an arbitrary direct power of a finite monoid, any finitely generated submonoid is finite.

Theorem. If group $G$ is finitely generated and residually finite then it is Hopfian (i.e. any surjective endomorphism is an automorphism).

Proof: Let $f \in \mathrm{End}_{\mathrm{Gr}}(G)$ be surjective and set $$\hat{f}, \check{f}: \mathscr{P}(G) \to \mathscr{P}(G) \\ \hat{f}(X)=f(X)\\ \check{f}(X)=f^{-1}(X)$$ The surjectivity entails that $\hat{f} \circ \check{f}=\mathbf{1}_{\mathscr{P}(G)}$, and thus $\check{f}$ is injective.
It is an elementary fact (universality properties of quotient maps in the category of sets) that for any $H \leqslant G$ $f$ induces a canonical bijection between the quotient sets $(G/f^{-1}(H))_s$ and $(G/H)_{s}$ (this is where the assumption of surjectivity is again indispensable). Therefore, if in continuity with the above notation for arbitrary $n \in \mathbb{N}$ we introduce $\mathscr{S}_n=\{H \leqslant G\ |\ |G:H|=n|\}$ we then have that $\check{f}(\mathscr{S}_n) \subseteq \mathscr{S}_n$; the lemma above tells us that $\mathscr{S}_n$ is finite which together with this inclusion and the injectivity of $\check{f}$ suffice to establish the equality $$\check{f}(\mathscr{S}_n)=\mathscr{S}_n \tag{1}$$
Furthermore, note that for any $H \leqslant G$ we have $\mathrm{Ker}f \leqslant f^{-1}(H)$ so in particular for any $n \in \mathbb{N}$ such that $\mathscr{S}_n \neq \varnothing$ it is the case that $$\mathrm{Ker}f \leqslant \bigcap_{H \in \mathscr{S}_n}\ f^{-1}(H)=\bigcap_{H \in \mathscr{S}_n}\ \check{f}(H)=\bigcap \mathscr{S}_n \tag{2}$$ (the axiomatic system I guide myself by does not allow for empty intersections).
Introducing $M=\{n \in \mathbb{N}\ |\ \mathscr{S}_n \neq \varnothing\}$ (about which we note that $1 \in M$) and $$\mathscr{T}=\bigcup_{n \in \mathbb{N}}\ \mathscr{S}_n=\bigcup_{n \in M} \mathscr{S}_n$$
the residual finiteness property can be expressed as
$$\{1_G\}=\bigcap \mathscr{T}=\bigcap_{n \in M} \bigcap \mathscr{S}_n \tag{3}$$
Relation (2) means that $\mathrm{Ker}f \leqslant \bigcap \mathscr{S}_n$ for any $n \in M$, thus by (3) $\mathrm{Ker}f \leqslant \{1_G\}$ and $f$ is injective. $\Box$