Schreier’s lemma

Schreier’s lemma

Let

G

be a group,

H

a

subgroup

and

T

transversal

of

H

in

G

. For every

g

∈

G

,
we denote by

g

¯

the unique element

t

∈

T

such that

H

⁢

g

=

H

⁢

t

.
We also insist that

1

∈

T

. Under these conditions we can state Schreier’s lemma.

Lemma 1

(Schreier).

Given a group

G

=

⟨

S

⟩

, a subgroup

H

, and transversal

T

for

G

/

H

, then the set

U

=

H

∩

{

t

⁢

s

⁢

(

t

⁢

s

¯

)

-

1

:

s

∈

S

,

t

∈

T

}

generates

H

.

Proof.

[

1

]

We show that

H

is generated by

U

∪

U

-

1

. We already know

U

is contained in

H

and we know every element

h

∈

H

is expressible as
a word over

S

, as

S

generates

G

. So

h

=

s

1

⁢

⋯

⁢

s

k

where

s

i

∈

S

∪

S

-

1

. Note that

h

=

1

⁢

s

1

⁢

⋯

⁢

s

k

which is of
the form

u

1

⁢

⋯

⁢

u

i

⁢

t

i

+

1

⁢

s

i

+

1

⁢

⋯

⁢

s

k

for

u

j

∈

U

∪

U

-

1

,

t

i

+

1

∈

T

and

s

j

∈

S

∪

S

-

1

, where

i

=

0

. We assume for
induction that

h

has this form for some

i

. Now we perform the
sleight-of-hand.

h

=

u

1

⁢

⋯

⁢

u

i

⁢

t

i

+

1

⁢

s

i

+

1

⁢

⋯

⁢

s

k

=

u

1

⁢

⋯

⁢

u

i

⁢

t

i

+

1

⁢

s

i

+

1

⁢

(

t

i

+

1

⁢

s

i

+

1

¯

)

-

1

⁢

(

t

i

+

1

⁢

s

i

+

1

¯

)

⁢

s

i

+

2

⁢

⋯

⁢

s

k

=

u

1

⁢

⋯

⁢

u

i

⁢

(

t

i

+

1

⁢

s

i

+

1

⁢

(

t

i

+

1

⁢

s

i

+

1

¯

)

-

1

)

⁢

t

i

+

1

⁢

s

i

+

1

¯

⁢

s

i

+

2

⁢

⋯

⁢

s

k

=

u

1

⁢

⋯

⁢

u

i

⁢

u

i

+

1

⁢

t

i

+

2

⁢

s

i

+

2

⁢

⋯

⁢

s

k

by letting

u

i

+

1

=

t

i

+

1

⁢

s

i

+

1

⁢

(

t

i

+

1

⁢

s

i

+

1

¯

)

-

1

which is
in

U

∪

U

-

1

, and

t

i

+

2

=

t

i

+

1

⁢

s

i

+

1

¯

∈

T

. At the
end of the

iteration

we have replaced all the

s

i

’s with elements of

U

∪

U

-

1

and the last element

t

k

∈

T

. However,

U

lies in

H

so the

product

u

1

⁢

⋯

⁢

u

k

∈

H

and as

h

∈

H

and

h

=

u

1

⁢

⋯

⁢

u

k

⁢

t

k

we know

t

k

∈

H

. As

H

⁢

t

k

=

H

, and

t

k

∈

T

we now use the fact
that

1

∈

T

to assert that

t

k

=

1

. So

h

is a word in

U

∪

U

-

1

and thus

U

generates

H

.
∎

The lemma was discovered in the course of studying

free groups

as a way to produce

generators

for a subgroup of a free group. The Reidemeister-Schreier
theorem strengthened this result to produce a

presentation

for

H

given one
for

G

and a transversal of

G

/

H

.

The lemma lead to an unexpected use in the 1970’s where it was rediscovered by various authors
to solve certain problems in computational group theory. Sims used the result to find
generators of the

stabilizer

of a point in a

permutation group

. This method could then be
repeated to find generators for each stabilizer in a stabilizer chain of a base of the group.
A naive approach can result in exponentially many generators, so Sims’

algorithm

additionally
removes unnecessary generators along the way. This is now called the Schreier-Sims algorithm.

Independently a version of this algorithm was developed by Frust, Hopcroft, and Luks as a means to
prove the

polynomial-time

complexity

of various problems in

finite group

theory and

graph
theory

. Many improvements followed including versions by Knuth, Jerum, and even a probabilistic
nearly linear time method by Seress.

The lemma still finds use in non-finite groups as Schreier had intended. The associated
algorithms have been repeatedly improved principally by the work of George Havas and Colin Ramsay.

References

1

Seress, Ákos

Permutation group algorithms

,
Cambridge Tracts in Mathematics, vol. 152,
Cambridge University Press, Cambridge, 2003.

Title

Schreier’s lemma

Canonical name

SchreiersLemma

Date of creation

2013-03-22 15:53:55

Last modified on

2013-03-22 15:53:55

Owner

Algeboy (12884)

Last modified by

Algeboy (12884)

Numerical id

17

Author

Algeboy (12884)

Entry type

Theorem

Classification

msc 20K27

Synonym

Schreier-Sims

Related topic

LiftsTransversals

Related topic

FindingTheOrderOfAGroup

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