Schreier's lemma

Constructive statement
Suppose $$H$$ is a subgroup of a group $$G$$, and $$S$$ is a left transversal of $$H$$ in $$G$$ with the property that the representative of the coset $$H$$ itself is the identity element. Suppose $$A$$ is a generating set for $$G$$. Then, define $$B$$ as the set of all elements in $$H$$ that can be written as $$s'^{-1}as$$ where $$s'$$ and $$s$$ are in $$S$$ and $$a$$ is in $$A$$.

Then, $$B$$ is a generating set for $$H$$.

Factual statement
Any subgroup of finite index in a  finitely generated group (see also:  subgroup of finite index in finitely generated group) is finitely generated. This follows because when $$A$$ and $$S$$ are both finite sets, so is $$B$$.

Applications

 * Local finiteness is extension-closed: A locally finite group is a group in which every finitely generated subgroup is finite. If $$G$$ is a group with a locally finite normal subgroup $$N$$ such that $$G/N$$ is also locally finite, then $$G$$ is locally finite. The proof of this uses Schreier's lemma.
 * Freeness is subgroup-closed: Any subgroup of a free group is free, and moreover, if the subgroup has finite index, the number of generators of the subgroup is given as a function of the number of generators of the whole group, and the index of the subgroup. The proof of this is based on a closer analysis of the constructive statement of Schreier's lemma.

Analogues in other algebraic structures

 * Artin-Tate lemma: This states that if $$A \le B$$ are algebras over a commalg:Noetherian ring $$R$$, with $$B$$ finitely generated as a $$R$$-algebra and $$B$$ finitely generated as an $$A$$-module, then $$A$$ is finitely generated as a $$R$$-algebra. Here, being finitely generated as a $$R$$-algebra plays the analogous role to being a finitely generated group, and $$B$$ being finitely generated as an $$A$$-module plays the analogous role to a subgroup of finite index.

Related ideas

 * Schreier coset graph
 * Todd-Coxeter algorithm

Proof idea
The key idea is to rewrite words originally in terms of $$A$$ in terms of $$B$$, with a possible leading term from $$S$$. If the word happens to be in the subgroup, the leading term from $$S$$ must be trivial, and thus, the word originally written in terms of $$A$$ is expressible in terms of $$B$$.

Proof details in the form of a proof by induction
Given: A group $$G$$ with a subgroup $$H$$ having a left transversal $$S$$. $$A$$ is a generating set for $$G$$. $$B$$ is defined as the set of all elements of $$H$$ expressible as $$s'^{-1}as$$ where $$s',s \in S$$ and $$a \in A$$.

To prove: $$\langle B \rangle = H$$.

Proof: Let $$C = A \cup A^{-1}$$ be the set of all elements of $$A$$ and their inverses. Let $$D = B \cup B^{-1}$$ be the set of all elements of $$B$$ and their inverses. We will prove that for any $$n$$:

$$C^n \subseteq SD^n$$,

where $$C^n, D^n$$ denote the elements of $$G$$ expressible as products of length at most $$n$$ from elements of $$C$$, $$D$$ respectively.

We prove this claim by induction on $$n$$. Let's first do the case $$n = 1$$. We need to show that any element of $$C$$ is in $$SD$$. We make two cases:


 * The element is an element $$a \in A$$: In this case, let $$s$$ be the coset representative of $$a$$. Then $$a = s(s^{-1}ae)$$, where $$e$$ is the identity element. Since the identity element is the coset representative of $$H$$, $$s^{-1}ae \in B$$ so $$s^{-1}ae \in D$$ by definition, so $$a \in SD$$.
 * The element is $$a^{-1}$$, with $$a \in A$$: In this case, let $$s$$ be the coset representative of $$a^{-1}$$. Then, $$a^{-1} = s(s^{-1}a^{-1}e)$$. We can write $$s^{-1}a^{-1}e = (eas)^{-1}$$. By assumption, $$as \in H$$, so $$e$$ is its coset representative, so $$(eas) \in B$$, so $$(eas)^{-1} \in D$$, again showing that $$a^{-1} = s(eas)^{-1} \in SD$$.

We now give the induction step.

Suppose $$g \in C^n$$. Then we can either write $$g = ag'$$ or write $$g = a^{-1}g'$$ for $$a \in A, g' \in C^{n-1}$$. We examine both cases.


 * $$g = ag', a \in A$$: By the induction step, write $$g' = sd$$, with $$s \in S, d \in D^{n-1}$$. Then $$g = asd$$. Let $$s'$$ be the coset representative of $$as$$. Then, $$g = s'(s'^{-1}as)d$$. By assumption, $$s'^{-1}as \in B \subseteq D$$, so $$g \in SDD^{n-1} = SD^n$$.
 * $$g = a^{-1}g', a \in A$$: By the induction step, write $$g' = sd$$, with $$s \in S, d \in D^{n-1}$$. Then $$g = a^{-1}sd$$. Suppose $$s'$$ is the coset representative of $$a^{-1}s$$. Then, $$g = s'(s'^{-1}a^{-1}s)d$$. The middle term $$s'^{-1}a^{-1}s$$ is the inverse of $$s^{-1}as'$$, which is by definition in $$B$$. Hence, $$s'^{-1}a^{-1}s \in B^{-1} \subseteq D$$. Thus, $$g \in SDD^{n-1} = SD^n$$.

Thus, we have established that:

$$C^n \subseteq SD^n$$.

Now, any element of $$H$$ is in particular an element of $$G$$, and since $$A$$ generates $$G$$, every element of $$H$$ is in $$C^n$$. Thus, for any element $$h \in H$$, we have:

$$h = sd, s \in S, d \in D^n$$.

But since $$D \subseteq H$$, we have $$D^n \subseteq H$$, so $$d \in H$$. In particular, this yields $$s = hd^{-1} \in H$$. But the coset representative in $$H$$ is by assumption trivial, so we get $$h \in D^n$$. Thus, $$h \in \langle B \rangle$$. Since $$h$$ was arbitrary, this shows that every element of $$H$$ is in the subgroup generated by $$B$$.

As an algorithm
Schreier's lemma can give an algorithm to compute a generating set for a subgroup starting from a generating set of the whole group and a system of coset representatives for the subgroup. The idea is to traverse all tuples of the form $$s'^{-1}as$$ where $$a \in A$$ and $$s,s' \in S$$.

The problem of determining a system of coset representatives is solved by means of a bradth-first search in the Cayley graph of the action of $$G$$ on $$G/H$$ in terms of the specified generating set of $$G$$.