Schreier's lemma

In mathematics, Schreier's lemma is a theorem in group theory used in the Schreier–Sims algorithm and also for finding a presentation of a subgroup.

Statement
Suppose $$H$$ is a subgroup of $$G$$, which is finitely generated with generating set $$S$$, that is, $$G = \langle S\rangle$$.

Let $$R$$ be a right transversal of $$H$$ in $$G$$. In other words, $$R$$ is (the image of) a section of the quotient map $$G \to H\backslash G$$, where $$H\backslash G$$ denotes the set of right cosets of $$H$$ in $$G$$.

The definition is made given that $$g\in G$$, $$\overline{g}$$ is the chosen representative in the transversal $$R$$ of the coset $$Hg$$, that is,
 * $$g\in H\overline{g}.$$

Then $$H$$ is generated by the set
 * $$\{rs(\overline{rs})^{-1}|r\in R, s\in S\}.$$

Hence, in particular, Schreier's lemma implies that every subgroup of finite index of a finitely generated group is again finitely generated.

Example
The group Z3 = Z/3Z is cyclic. Via Cayley's theorem, Z3 is a subgroup of the symmetric group S3. Now,
 * $$\mathbb{Z}_3=\{ e, (1\ 2\ 3), (1\ 3\ 2) \}$$
 * $$S_3= \{ e, (1\ 2), (1\ 3), (2\ 3), (1\ 2\ 3), (1\ 3\ 2) \}$$

where $$e$$ is the identity permutation. Note S3 = $$\scriptstyle\langle$${ s1=(1 2), s2 = (1 2 3) }$$\scriptstyle\rangle$$.

Z3 has just two cosets, Z3 and S3 \ Z3, so we select the transversal { t1 = e, t2=(1 2) }, and we have
 * $$\begin{matrix}

t_1s_1 = (1\ 2),&\quad\text{so}\quad&\overline{t_1s_1} = (1\ 2)\\ t_1s_2 = (1\ 2\ 3) ,&\quad\text{so}\quad& \overline{t_1s_2} = e\\ t_2s_1 = e        ,&\quad\text{so}\quad& \overline{t_2s_1} = e\\ t_2s_2 = (2\ 3) ,&\quad\text{so}\quad& \overline{t_2s_2} = (1\ 2). \\ \end{matrix}$$

Finally,
 * $$t_1s_1\overline{t_1s_1}^{-1} = e$$
 * $$t_1s_2\overline{t_1s_2}^{-1} = (1\ 2\ 3)$$
 * $$t_2s_1\overline{t_2s_1}^{-1} = e $$
 * $$t_2s_2\overline{t_2s_2}^{-1} = (1\ 2\ 3).$$

Thus, by Schreier's subgroup lemma, { e, (1 2 3) } generates Z3, but having the identity in the generating set is redundant, so it can be removed to obtain another generating set for Z3, { (1 2 3) } (as expected).