Statement
edit
Suppose
H
{\displaystyle H}
subgroup of
G
{\displaystyle G}
S
{\displaystyle S}
G
=
⟨
S
⟩
{\displaystyle G=\langle S\rangle }
Let
R
{\displaystyle R}
transversal of
H
{\displaystyle H}
G
{\displaystyle G}
R
{\displaystyle R}
section of the quotient map
G
→
H
∖
G
{\displaystyle G\to H\backslash G}
H
∖
G
{\displaystyle H\backslash G}
right cosets of
H
{\displaystyle H}
G
{\displaystyle G}
The definition is made given that
g
∈
G
{\displaystyle g\in G}
g
¯
{\displaystyle {\overline {g}}}
R
{\displaystyle R}
H
g
{\displaystyle Hg}
g
∈
H
g
¯
.
{\displaystyle g\in H{\overline {g}}.}
Then
H
{\displaystyle H}
{
r
s
(
r
s
¯
)
−
1
|
r
∈
R
,
s
∈
S
}
.
{\displaystyle \{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.
The group Z 3 = Z /3Z is cyclic. Via Cayley's theorem , Z 3 is a subgroup of the symmetric group S 3 . Now,
Z
3
=
{
e
,
(
1
2
3
)
,
(
1
3
2
)
}
{\displaystyle \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
)
}
{\displaystyle S_{3}=\{e,(1\ 2),(1\ 3),(2\ 3),(1\ 2\ 3),(1\ 3\ 2)\}}
where
e
{\displaystyle e}
S 3 =
⟨
{\displaystyle \scriptstyle \langle }
s 1 =(1 2), s 2 = (1 2 3) }
⟩
{\displaystyle \scriptstyle \rangle }
Z 3 has just two cosets, Z 3 and S 3 \ Z 3 , so we select the transversal { t 1 = e , t 2 =(1 2) }, and we have
t
1
s
1
=
(
1
2
)
,
so
t
1
s
1
¯
=
(
1
2
)
t
1
s
2
=
(
1
2
3
)
,
so
t
1
s
2
¯
=
e
t
2
s
1
=
e
,
so
t
2
s
1
¯
=
e
t
2
s
2
=
(
2
3
)
,
so
t
2
s
2
¯
=
(
1
2
)
.
{\displaystyle {\begin{matrix}t_{1}s_{1}=(1\ 2),&\quad {\text{so}}\quad &{\overline {t_{1}s_{1}}}=(1\ 2)\\t_{1}s_{2}=(1\ 2\ 3),&\quad {\text{so}}\quad &{\overline {t_{1}s_{2}}}=e\\t_{2}s_{1}=e,&\quad {\text{so}}\quad &{\overline {t_{2}s_{1}}}=e\\t_{2}s_{2}=(2\ 3),&\quad {\text{so}}\quad &{\overline {t_{2}s_{2}}}=(1\ 2).\\\end{matrix}}}
Finally,
t
1
s
1
t
1
s
1
¯
−
1
=
e
{\displaystyle t_{1}s_{1}{\overline {t_{1}s_{1}}}^{-1}=e}
t
1
s
2
t
1
s
2
¯
−
1
=
(
1
2
3
)
{\displaystyle t_{1}s_{2}{\overline {t_{1}s_{2}}}^{-1}=(1\ 2\ 3)}
t
2
s
1
t
2
s
1
¯
−
1
=
e
{\displaystyle t_{2}s_{1}{\overline {t_{2}s_{1}}}^{-1}=e}
t
2
s
2
t
2
s
2
¯
−
1
=
(
1
2
3
)
.
{\displaystyle t_{2}s_{2}{\overline {t_{2}s_{2}}}^{-1}=(1\ 2\ 3).}
Thus, by Schreier's subgroup lemma, { e, (1 2 3) } generates Z 3 , but having the identity in the generating set is redundant, so it can be removed to obtain another generating set for Z 3 , { (1 2 3) } (as expected).
References
edit
Seress, A. Permutation Group Algorithms. Cambridge University Press, 2002. 
