Schreier's Lemma - Example

Example

Let us establish the evident fact that the group Z3 = Z/3Z is indeed cyclic. Via Cayley's theorem, Z3 is a subgroup of the symmetric group S3. Now,

where is the identity permutation. Note S3 = <{ s1=(1 2), s2 = (1 2 3) }>.

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,

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

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