ping-pong lemma

ping-pong lemma

Theorem

(Ping Pong Lemma).

Let

k

≥

2

and let

G

be a group acting on a space

X

. Suppose
we are given a class

M

=

{

A

1

,

A

2

,

…

,

A

k

,

B

1

,

B

2

,

…

,

B

k

}

of

2

⁢

k

pairwise disjoint

subsets of

X

and suppose

y

1

,

y

2

,

…

,

y

k

are elements of

G

such that

B

i

c

⊆

y

i

⁢

(

A

i

)

i

=

1

,

2

,

…

,

k

(

B

i

c

is the

complement

of

B

i

in

X

). Then, the

subgroup

of

G

generated by

y

1

,

y

2

,

…

,

y

k

is free.

Before turning to prove the lemma let’s state three
simple facts:

Fact 1

.

For all

i

=

1

,

…

,

k

we have

y

i

⁢

(

A

i

c

)

⊆

B

i

and

y

i

-

1

⁢

(

B

i

c

)

⊆

A

i

Proof.

B

i

c

⊆

y

i

⁢

(

A

i

)

⟹

B

i

⊇

y

i

⁢

(

A

i

)

c

=

y

i

⁢

(

A

i

c

)

∎

Fact 2

.

If

i

≠

j

then

A

j

∪

B

j

⊆

A

i

c

∩

B

i

c

.

Proof.

A

i

and

A

j

are

disjoint

therefore

A

j

⊆

A

i

c

.
Similarly,

A

j

⊆

B

i

c

so

A

j

⊆

A

i

c

∩

B

i

c

.
In the same way,

B

j

⊆

A

i

c

∩

B

i

c

so

A

j

∪

B

j

⊆

A

i

c

∩

B

i

c

.
∎

Fact 3

.

If

R

,

S

∈

M

then

R

c

⊈

S

Proof.

Assume by

contradiction

that

R

c

⊆

S

. Then,

X

=

R

∪

S

and therefore any element of

M

intersects

with either

R

or

S

. However, the elements of

M

are pairwise disjoint and there
are at least 4

elements

in

M

so this is a contradiction.
∎

Using the above 3 facts, we now turn to the proof of the Ping Pong Lemma:

Proof.

Suppose we are given

w

=

z

n

ϵ

n

⁢

⋯

⁢

z

2

ϵ

2

⁢

z

1

ϵ

1

such that

z

ℓ

∈

{

y

1

,

y

2

,

…

,

y

k

}

and

ϵ

ℓ

∈

{

-

1

,

+

1

}

.
and suppose further that

w

is freely

reduced

, namely, if

z

i

=

z

i

+

1

then

ϵ

i

=

ϵ

i

+

1

. We want to show
that

w

≠

1

in

G

. Assume by contradiction that

w

=

1

. We get a
contradiction by giving

R

,

S

∈

ℳ

such that

w

⁢

(

S

c

)

⊆

R

and therefore contradicting Fact

3

above since

S

c

=

w

⁢

(

S

c

)

⊆

R

.

The set

S

is chosen as follows. Assume that

z

1

=

y

i

then:

S

=

{

A

i

if

⁢

ϵ

1

=

1

B

i

if

⁢

ϵ

1

=

-

1

Define the following subsets

P

0

,

P

1

,

…

,

P

n

of

X

:

P

0

=

S

c

;

P

1

=

z

1

ϵ

1

⁢

(

P

0

)

,

…

,

P

n

=

z

n

ϵ

n

⁢

(

P

n

-

1

)

=

w

⁢

(

S

c

)

To

complete

the proof we show by

induction

that for

ℓ

=

1

,

2

,

…

,

n

if

z

ℓ

=

y

i

then:

1.

if

ϵ

ℓ

=

1

then

P

ℓ

⊆

B

i

.

2.

if

ϵ

ℓ

=

-

1

then

P

ℓ

⊆

A

i

.

For

ℓ

=

1

the above follows from Fact

1

and the
specific choice of

P

0

. Assume it is true for

ℓ

-

1

and
assume that

z

ℓ

=

y

i

. We have two cases to check:

1.

z

ℓ

-

1

≠

z

ℓ

: by the

induction hypothesis

P

ℓ

-

1

is a subset of

A

j

∪

B

j

for some

j

≠

i

.
Therefore, by Fact

2

we get that

P

ℓ

-

1

is a subset of

A

i

c

∩

B

i

c

.
Consequently, we get the following:

P

ℓ

=

z

ℓ

ϵ

ℓ

⁢

(

P

ℓ

-

1

)

=

y

i

ϵ

ℓ

⁢

(

P

ℓ

-

1

)

⊆

y

i

ϵ

ℓ

⁢

(

A

i

c

∩

B

i

c

)

Hence, if

ϵ

ℓ

=

1

then:

P

ℓ

⊆

y

i

⁢

(

A

i

c

∩

B

i

c

)

⊆

y

i

⁢

(

A

i

c

)

⊆

B

i

And if

ϵ

ℓ

=

-

1

then:

P

ℓ

⊆

y

i

-

1

⁢

(

A

i

c

∩

B

i

c

)

⊆

y

i

-

1

⁢

(

B

i

c

)

⊆

A

i

2.

z

ℓ

-

1

=

z

ℓ

: by the fact that

w

is freely reduced we
get an equality between

ϵ

ℓ

-

1

and

ϵ

ℓ

. Hence, if

ϵ

ℓ

=

1

then

P

ℓ

-

1

⊆

B

i

⊆

A

i

c

and
therefore:

P

ℓ

=

z

ℓ

ϵ

ℓ

⁢

(

P

ℓ

-

1

)

=

y

i

⁢

(

P

ℓ

-

1

)

⊆

y

i

⁢

(

A

i

c

)

⊆

B

i

Similiarly, if

ϵ

ℓ

=

-

1

then

P

ℓ

-

1

⊆

A

i

⊆

B

i

c

and therefore:

P

ℓ

=

z

ℓ

ϵ

ℓ

⁢

(

P

ℓ

-

1

)

=

y

i

-

1

⁢

(

P

ℓ

-

1

)

⊆

y

i

-

1

⁢

(

B

i

c

)

⊆

A

i

∎

Title

ping-pong lemma

Canonical

name

PingpongLemma

Date of creation

2013-03-22 17:11:21

Last modified on

2013-03-22 17:11:21

Owner

uriw (288)

Last modified by

uriw (288)

Numerical id

8

Author

uriw (288)

Entry type

Theorem

Classification

msc 20F65

Synonym

table-tennis lemma

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