Motivating Question
Definitions
Constructive Membership Testing
Bases and Strong Generating Sets
Daniel Rogers
June 13, 2014
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
Motivating Question
Given (finite) groups G = hX i ≤ H and h ∈ H, how can we decide
algorithmically whether h ∈ G ?
Algorithms which solve this problem and, if they are successful, write h as
a word in X , the generators of G , are known as constructive membership
tests.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Bases
Let G be a group acting on a set Ω.
Daniel Rogers
Bases and Strong Generating Sets
Constructive Membership Testing
Motivating Question
Definitions
Constructive Membership Testing
Bases
Let G be a group acting on a set Ω.
Definition
A base for G is a sequence B = [b1 , ..., bm ] ⊂ Ω such that the only
element of G which stabilizes each bi is the identity.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
Strong Generating Sets
Let G be a group acting on a set Ω and B = [b1 , ..., bm ] ⊂ Ω a base.
Definition
The basic stabilizer chain associated with a base B is a chain of
pointwise stabilizers of the form
G = G (0) ≥ G (1) ≥ ... ≥ G (m) = 1
where G := G (0) and G (i) := G(b1 ,...bi ) is the group of all elements of G
which stabilize the first i base points.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
Strong Generating Sets
Let G be a group acting on a set Ω and B = [b1 , ..., bm ] ⊂ Ω a base.
Definition
The basic stabilizer chain associated with a base B is a chain of
pointwise stabilizers of the form
G = G (0) ≥ G (1) ≥ ... ≥ G (m) = 1
where G := G (0) and G (i) := G(b1 ,...bi ) is the group of all elements of G
which stabilize the first i base points.
Definition
A strong generating set for G is a subset S ⊂ G such that S contains
generators for each of the groups G (i) for 0 ≤ i ≤ m.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
Strong Generating Sets
Let G be a group acting on a set Ω and B = [b1 , ..., bm ] ⊂ Ω a base.
Definition
The basic stabilizer chain associated with a base B is a chain of
pointwise stabilizers of the form
G = G (0) ≥ G (1) ≥ ... ≥ G (m) = 1
where G := G (0) and G (i) := G(b1 ,...bi ) is the group of all elements of G
which stabilize the first i base points.
Definition
A strong generating set for G is a subset S ⊂ G such that S contains
generators for each of the groups G (i) for 0 ≤ i ≤ m.
Definition
A BSGS is a pair (B, S) of a base and a strong generating set.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
Constructive Membership Testing
We return to the motivating question: suppose we have groups G ≤ H
(both acting on the same set Ω) and h ∈ H, and suppose additionally
that B = [b1 , ..., bm ] and S are a BSGS for G (so in particular G = hSi ).
We describe a simple method for performing constructive membership
testing in this case.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
I
t
Compute α := bi i−1 .
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
t
I
Compute α := bi i−1 .
I
(in other words, whether there is an
Determine whether α ∈ biG
element of G which stabilizes the first i − 1 base points and also
maps bi to α).
Daniel Rogers
Bases and Strong Generating Sets
(i−1)
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
t
I
Compute α := bi i−1 .
I
(in other words, whether there is an
Determine whether α ∈ biG
element of G which stabilizes the first i − 1 base points and also
maps bi to α).
(i−1)
I
If α ∈
/ biG
Daniel Rogers
Bases and Strong Generating Sets
(i−1)
, then ti−1 ∈
/ G (i−1) , and so h ∈
/ G.
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
t
I
Compute α := bi i−1 .
I
(in other words, whether there is an
Determine whether α ∈ biG
element of G which stabilizes the first i − 1 base points and also
maps bi to α).
(i−1)
(i−1)
I
I
If α ∈
/ biG
, then ti−1 ∈
/ G (i−1) , and so h ∈
/ G.
(i−1)
G
(i−1)
If α ∈ bi
, then find gi ∈ G
such that bigi = α. (It is easy to
find such a gi as a word in our strong generating set S).
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
t
I
Compute α := bi i−1 .
I
(in other words, whether there is an
Determine whether α ∈ biG
element of G which stabilizes the first i − 1 base points and also
maps bi to α).
(i−1)
(i−1)
I
I
If α ∈
/ biG
, then ti−1 ∈
/ G (i−1) , and so h ∈
/ G.
(i−1)
G
(i−1)
If α ∈ bi
, then find gi ∈ G
such that bigi = α. (It is easy to
find such a gi as a word in our strong generating set S). Define
ti := ti−1 gi−1 ; then ti stabilizes b1 , ..., bi .
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
Begin by defining t0 := h. We proceed through each base point bi
(i ≥ 1) in turn, assuming at each stage that ti−1 = hg for some g ∈ G ,
and ti−1 stabilizes the first i − 1 base points.
t
I
Compute α := bi i−1 .
I
(in other words, whether there is an
Determine whether α ∈ biG
element of G which stabilizes the first i − 1 base points and also
maps bi to α).
(i−1)
(i−1)
I
I
I
If α ∈
/ biG
, then ti−1 ∈
/ G (i−1) , and so h ∈
/ G.
(i−1)
G
(i−1)
If α ∈ bi
, then find gi ∈ G
such that bigi = α. (It is easy to
find such a gi as a word in our strong generating set S). Define
ti := ti−1 gi−1 ; then ti stabilizes b1 , ..., bi .
Iterate.
Daniel Rogers
Bases and Strong Generating Sets
Motivating Question
Definitions
Constructive Membership Testing
The Algorithm
If we complete this process for all base points, then we have
h ∈ G ⇐⇒ tm = 1, and moreover in this case we have h = gm gm−1 ...g1
has been expressed as a product of words in S.
Daniel Rogers
Bases and Strong Generating Sets