Category: Group Theory

  • Accessibility

    G is a finitely presented group.

    X is its presentation complex (a simplicial 2-complex). Since G is finitely presented, the number of vertices of X is finite. Suppose \( u_1 , \cdots , u_q \) be the vertices of X.

    \( \tilde {X} \) be its universal cover.

    Fix lifts of the vertices of X. Suppose they are \( v_1 , \cdots , v_q \).

    Choose \( \mathcal {A} \) : a preferred set of subgroups of G, stable under conjugation and taking subgroups. We will be working with trees, whose edge stabilizers come from\( \mathcal {A} \)

    Let \( T_1 \leftarrow \cdots  \leftarrow T_{k+1} \leftarrow T_{k} \leftarrow \cdots \) be refinements of \( \mathcal {A} \) -trees.

    This means there is a collapse map from \( T_{k+1} \to T_k \) for each k.

    Collapse Map

    A collapse map (is a G-equivariant map) that maps some edges to vertices. On the remaining tree, it is a graph isomorphism.

    Pre-image of midpoint of every edge \(e \) of \( T_k \) is the midpoint of an edge e’ in \( T_{k+1} \) that maps isomorphically onto e.

    We will construct equivariant maps \( f_k \) from \( \tilde {X} \) to \( T_k \) for all k inductively. 

    Here is the recipe:

    • Map the (favorite) vertices \( v_i \) of \( \tilde {X} \) to vertices of \( T_1 \). 
    • Extend equivariantly to all vertices of \( \tilde {X} \). The map is now defined on 0- skeleton.
      That is if \( v_i \to f_1 (v_i) \) then \( gv_i \to f_1 (gv_i) = g f(v_i)  \) 
    • In order to extend the map to 1-skeleton, first note that we know where the endpoints of each edge are going. 
    • \( \tilde {e} \) be an edge of \( \tilde {X} \). Let v, w be its endpoints. Since \( T_1 \) is a tree, there is an unique edge path connecting \( f_1 (v) \) to \( f_1 (w) \) 
    • Map \( \tilde {e} \) to that path. 
    • For \( T_2 \), first send v to any vertex \( t_v \)  in the set \( p_1^{-1} ( f_1 (v) ) \). Similarly w is mapped to any vertex \(t_w \)  in \( p_1^{-1} (f_1 (w) ) \) 
    • Restrict \( p_1 \) to the segment \( t_v t_w \). It is a collapse map. 
    • In particular, the pre-image of the midpoint of an edge \( e_k \) of \( T_k \) is the midpoint of the edge of \(T_{k+1}\) mapping onto \( e_k \). 
    • Since \( f_1 \) is constant or injective on \( \tilde {e}\) this allows us to define \( f_{2} \) on \( \tilde {e}\) as a map which is either constant or injective.
    • Extend \( f_{2} \) equivariantly on the 1-skeleton.
    • Finally, we will extend \( f_2 \) to the two simplices. Whenever it is not constant, the image is tripod. Pre-images of midpoints of edges of  \(T_2\) are straight arcs joining two sides of the triangle. 

    Pattern of Dunwoody

    The pre-images of midpoints of \( T_k \) is \( \tilde {\tau_k } \subset \tilde {X} \)

     

    Dunwoody Pattern

    Also note that \( \tilde {\tau_k} \subset \tilde {\tau_{k+1}} \) 

    We will project the Dunwoody pattern to X (the presentation complex we started off with). This is a finite graph. 

    \( S_k \) be the tree dual to the pattern \( \tilde {\tau_k} \in \tilde {X} \). There is a natural induced map that sends \( S_k \to T_k \), sending edge to edge. Hence edge stabilizers of \( S_k \) are in \( \mathcal{A} \). 

    Edge Stabilizers of \( S_k \) are finitely generated

  • The Alexander Trick

    Here is the original paper:

    J. W. Alexander, On the deformation of an n-cell

    (A 2-page paper that influenced a remarkable amount of later work).

  • Ends

    Motivation


    Start with a locally finite simplicial complex X.

    Locally finite: Each vertex is attached to only finitely many simplices.

    Why locally finite: To make sure it is a CW complex. Notice that the closure-finite criteria require each cell of a CW complex to meet only finitely many other cells.

    Hence we do not have a situation like this:

    Screen Shot 2018-03-07 at 6.40.14 AM

    Suppose K is any finite subcomplex of X.

    We are interested in the number of infinite connected components of X – K. Call that number n(K).

    Example: 

    Screen Shot 2018-03-07 at 6.45.26 AM

    Consider the Real line (with usual triangulation). If we remove one vertex (shown in the picture), then n(I) = 2 (as there are two open rays or two infinite connected components).

    From \[\mathbb{R}^n, n \ge 2 \] onward, if you remove a finite subcomplex K, that can be enclosed within a (large enough) closed ball. Hence there can only one infinite connected component of \[\mathbb{R}^n – K , n \ge 2 \]

    This number n(K) can be different for different K.  Take for example the infinite binary tree.

    binary

    Removing the top vertex, the number of infinite connected components are 2. But removing up to level 2, we will have 4 infinite connected components. And as we go down the levels, n(K) will grow.

    So, we try with all possible such finite subcomplexes (possibly infinitely many of them) and check each time how many infinite connected components of X – K is there.

    This is precisely what leads us to ends of X.

    Ends of (X) =  E(X) = sup n(K)

    Example: \[E(\mathbb{R}) = 2, E(\mathbb{R^2}) = 1, E(T) = \infty, \textrm{T = infinite binary tree} \]

    Star of K: If K is a (finite) subcomplex of a simplicial complex X, then st(K) or star of K is defined to be the interior of all simplices which meet the vertices in K.

    Since K is locally finite st(K) is an open finite subset of K.

    If two point \[v_1, v_2 \] in X – st(K) can be connected by a path in X – st(K) then they can be connected via an edge path in X – st(K). Hence they are in a path component of X – K. Hence it is sufficient to look into the 1-skeleton of X – K to find out n(K).

    Cohomology 


    Recall that given a simplicial complex, one can create Chain complexes and Co-Chain Complexes.

    A chain complex, \[C_0 (X) \] of vertices is simply the set of formal sums of vertices in X. In other words, if \[v_1, v_2, v_3 \] are vertices in X then \[v_1 + v_2 – v_3 \] is an element of \[C_0 (X) \]. In fact it has all such finite formal sums.

    Once the chain complex is defined, one may quickly define co-chain complexes. They are formal sums of homomorphisms from \[C_0(X) \to G \] where G is a suitable group. Usually \[G = \mathbb{Z}_2, \textrm {or} G = \mathbb{Z} \]. It depends on what we are trying to achieve.

    One may think of each member of 0th cochain complex as an assignment of values to each vertex in X.

    Screen Shot 2018-03-07 at 7.14.56 AM

    In the above picture, we have assigned 0’s and 1’s to each vertex of the finite simplicial complex. This is one member of \[C^0(X) \]. Each such assignment is nothing but a function from \[C_0(X) \to \mathbb{Z}_2 \].

    We collect all such functions and their formal sum. They constitute \[C^0(X) \].

    In the present context, we may think of \[\mathbb{Z}_2 \] as {off, on}. Whenever we pick a finite subcomplex K in the simplicial complex X, we are indirectly assigning 1 to each vertex in K and 0 to all the vertices in X – K. That assignment is indeed a function from the set of vertices to {0, 1} hence a member of the cochain complex.

    Formally we say that the function has finite support; that is it is non-zero at only finitely many points and zero everywhere else.

    Thus the member of \[C^0(X) \] which have finite support, are our algebraic tools for ‘picking’ finite subcomplexes K from simplicial complex X. In fact, each such cochain corresponds to one such ‘pick’.

    Let \[C_f^0 (X) \] denote the subset of \[C^0 (X) \] with finite support.

    Notice that similarly we can define \[C_f^1 (X) \]. They are maps from (formal sums of) edges of X to \[\mathbb{Z}_2 \] which have 1 assigned to finitely many edges (and 0 elsewhere).

    \[C_f^0 (X) \] maps inside \[C_f^1 (X) \] under the coboundary map.

    Why? Suppose \[\phi \in C_f^0 (X) \] . Then \[\delta \phi (v_1, v_2) = \phi (v_2) – \phi (v_1) \] (Recall that the coboundary map assigns ‘difference’ of values at the vertices. This can be intuited as a change in height function: value at each edge is the difference of height at its vertices).

    Clearly \[\phi (v_2) – \phi (v_1) \] can be non zero if and only if one of them is 0 and the other one is 1. But as \[\phi \in C_f^0(X) \] has finite support, it gives nonzero output only at finitely many vertices hence \[\phi (v_2) – \phi (v_1) \] can be non zero only at finitely many edges.

    ‘Ends’ is the dimension of Cohomology


    Recall that cohomology groups of a space X are the homology groups of Cochain Complex.

    In more simpler terms, consider the following Cochain complex:

    \[… \overset{\delta_2} \Leftarrow C^1(X) \overset{\delta_1} \Leftarrow C^0(X) \overset{\delta_0} \Leftarrow 0 \]

    Then \[H^0 (X) = \frac{ker (\delta_1)}{im (\delta_0)} = ker (\delta_1) \]

    Intuition: Image of the lower level difference map contains lower level ‘difference’ data. By quotienting that out, we may focus only on higher level differences.

    We previously constructed \[C_f^0 (X) \] that singled out the finite subcomplexes. (K’s). Next, we wish to turn our attention toward X – K. Hence it makes sense to ‘quotient out’ \[C_f^0 (X) \] from \[C^0 (X) \]. Then we will be left out with those functions which assign 1 to infinitely many vertices.

    Define \[C_e^0 (X) = \frac{C^0 (X)}{C_f^0 (X)}  \]

    Similarly define \[C_e^1 (X) = \frac{C^1 (X)}{C_f^1 (X)}  \]

    Intuition: Imagine as ends. Functions in \[C_e^0 (X) \] flow 1’s upto ends (as 1’s appear infinitely many times at the vertices).

    Claim: \[dim_{\mathbb{Z}_2} H_e^0 (X) = E(X)  \]

    It is not hard to see why this could be true. Afterall what is $latex H_e^0 (X) $ ? It is the collection of all those \[\phi \in C_e^0 (X) \] that maps to 0 of $latex C_e^1 (X) $.

    More explicitly, these are 0-1assignments on vertices, which have 1 at infinitely many vertices, but non-zero differences at finitely many edges.

    Example: 

    Screen Shot 2018-03-07 at 8.27.28 AM

    Again consider the real line. Assign 1 at all vertices leftward from U. Assign 0 W onward (to the right).

    Note that this is a member of (an equivalence class of) \[C^0_e (X) \] as it has 1 at infinitely many vertices. Call it \[\phi \] .

    But after applying the coboundary map \[\delta \] to \[\phi \], we get 0 at all edges except on UV.  Clearly \[\delta \phi \] has finite support, hence is a member of \[C^1_f (X) \]. Therefore it is in the equivalence class of 0 in \[C^1_e (X) \] (as we have quotiented out all edge-functions with finite support).

    Note that U and V are the crossover points in the above example. When we talk about the \[ker \delta_1 \], we are seeking vertex-functions in \[C^0_e (X) \] we have finitely many crossover points (but infinitely many points with 1 assigned).

    Why? Because if we cut at the crossover, we have an infinite component at least at one side of it (after all the infinitely many 1’s need to accommodated somewhere).

    The recipe for finding the finite crossover but infinitely flowing vertex functions:

    • Look at all vertex-functions that map to edge-functions with finite support (\[\delta^{-1} {C^1_f(X) } \])
    • Quotient out those which are 1 (on) at finitely many vertices (they do not flow to the ends)
    • Final result \[\frac{\delta^{-1} {C^1_f(X) }}{C^0_f(X)} \]

    The rest is easy! 

    Each path component of X – K must bear the same value at each vertex. Hence we can create linearly independent vertex functions corresponding to each component of X – K.

    But we have taken care of all possible K as we considered all cochains.

    Hence the number of linearly independent finite crossover but infinitely flowing vertex maps provide the Ends of X.

    Good News: This data is cohomological. Hence it is independent of the triangulation of the space. (It is a deep theorem of algebraic topology that cohomology groups are independent of triangulation; infact can be achieved without triangulation).

  • A SAGE experiment on Nested Approximate Subgroups

    (this is a continuing write-up mostly for personal records)

    Spoiler

    Before you read on, here is what I have found curious so far (this section is constantly changing):

    • A little computation reveals that though group action conjugation preserves most elements of a subgroups in majority of cases, there exists some subgroups of \[S_4 \] which are ill mannered.
    • Ill -Mannered Subgroups when conjugated with non members, have exactly 33% of the elements common with the conjugated set
      • example: [(), (1,3), (1,3,4), (3,4), (1,4,3), (1,4)] is an ill mannered subgroup.

    Section 0

    This concerns a computational experiment (in SAGE platform) inspired by an article by Terence Tao on Nested Approximate Subgroup. 

    To learn more about SAGE please visit this link.

    Pre-requisites

    • A preliminary knowledge on group theory (ref: Topics in Algebra by Herstein)
    • Basics of mathematical programming (it is useful to have an exposure to Python or C).
    • Tao’s article

    Section 1

    Conjugation and Preservation

    The basic question is very simple. Suppose A is a subgroup of a group B (both finite). Say the index of A in B i.e. [B:A] = K. Hence we will get K disjoint cosets of A that partitions B.

    Suppose \[s_1 A, s_2 A ,… , s_k A\] be the K cosets. Let S = {\[s_1 , … , s_k \] } Then we may write B = SA.

    What happens if an arbitrary element s acts (via conjugation) on subgroup A. That is if A = \[\{ a_1 , a_2 , … , a_n \} \] and \[A^s = sAs^{-1} = \{ s a_1 s^{-1} , s a_2 s^{-1} , … , s a_n a^{-1} \} \]  then how different is  A and \[A^s \]?

    If A is normal in B then ofcourse \[A = A^s \]  (by definition of normal subgroup)

    However if A is not normal then? Turns out that the group action ‘approximately’ preserves A (I am deliberately using of some Terence Tao’s remarks here).

    Here we take a pause and create the first set of computation. We examine \[S_4 \] that is 24 elements, compute it’s subgroups and identify normal subgroups. Next we focus on the non normal subgroups and make elements of \[S_4 \] We  conjugation on them. Finally we will compare how different the conjugated sets are from actual subgroups.

    We use this little program to:

    1. Compute all subgroups of \[S_4 \]
    2. Compute their conjugated cosets (with each element of the main group)
    3. Compute intersection of the coset with the corresponding subgroup.
    4. Find the ratio of length of this intersection with the length of the subgroup

    Program

    Screen Shot 2015-06-06 at 10.31.16 PMG=SymmetricGroup(4)
    G1=G.list()
    mag=[]
    sub=[]
    sub.append(G1)
    for k in range(0,23):
    for r in range (0,23):
    if (k<r): H=G.subgroup([G[k], G[r]]) H1=H.list() z=set(H1) length=[] for x in range(0,len(sub)): y=set(sub[x]) m = y.symmetric_difference(z) n= list(set(m)) t = len(n) length.append(t) lengthmin = min(length) if(lengthmin>0):
    sub.append(H1)
    print “The subgroup H is”
    print H1
    print “\n”
    coset=[]
    for p in G1:
    b=p
    d=p**(-1)
    for a in H1:
    e=(b*a)*d
    coset.append(e)
    print “The coset %s H %s is” %(b,d)
    print coset
    print “\n”
    intcoset = list(set(H1) & set(coset))
    num = float(float(len(intcoset))/float(len(H1)))
    print num
    mag.append(num)
    coset = []
    print len(sub)
    print mag

    Result:

    [1.0, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.25, 1.0, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.5, 1.0, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 0.25, 0.25, 1.0, 1.0, 1.0, 0.25, 0.25, 1.0, 1.0, 0.25, 1.0, 0.25, 0.25, 0.25, 0.25, 1.0, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 1.0, 0.5, 0.5, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 1.0, 0.3333333333333333, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 0.5, 0.5, 0.5, 1.0, 1.0, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 1.0, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 1.0, 0.3333333333333333, 1.0, 1.0, 0.25, 1.0, 0.25, 0.25, 0.25, 0.25, 1.0, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.5, 1.0, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 1.0, 0.5, 0.5, 0.5, 0.5, 0.5, 1.0, 1.0, 1.0, 0.5, 0.5, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 1.0, 0.25, 1.0, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 1.0, 0.3333333333333333, 1.0, 0.3333333333333333, 0.3333333333333333, 0.3333333333333333, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 1.0, 0.25, 0.25, 0.25, 0.25, 0.25, 1.0, 1.0, 1.0, 0.25, 0.25]

    Remark:

    The value ‘1.0’ says the conjugated set is exactly same as corresponding subgroup. 0.25 indicates that one fourth of the values of the coset match the original subgroup.

    As far as \[S_4 \] is concerned, a glance at the values reveal \[[A : A \cap A^s ] \ge 3 \] for large majority of values.

    Notice apart from the trivial subgroup consisting of identity element, we have considered all remaining 29 subgroups (including the three other normal subgroups). For each of those 29 subgroups we have computed 24 conjugated subgroups, with respect to each of the 24 elements of \[S_4 \]. Thus we found \[29 \times 24 = 696 \] conjugated subgroups.

    Next we found the intersection set of each of 24 conjugations with corresponding subgroups, and found the ratio of order of intersection with order of subgroup.

    Among the 696 values thus found, maximum value is obviously 1 (when, conjugated subgroup is same as original subgroup). Since \[A_4, V_4 , S_4 \] are normal in $S_4$, they will generate 72 one. Among remaining 624 values, it turns out that 240 values are less than or equal to 0.33.

    So for about 60% of the cases (of non -normal subgroups), \[A \] and \[A^s \] have exactly half or all the elements in common.

    One may be interested in those subgroups which are different that is whose conjugations have less than 33% element in common. Question is are those 0.33 or 0.25 values distributed all over the values generated? Actually not. They come in chunks. Then definitely there are some subgroups for which the values falls below 33%

    Section 2

    The Ill mannered Conjugates

    So are there some subgroups whose cojugate subgroups are ill mannered? Turns out that for \[S_4 \] there are. We revised the code found in our program to find out how each subgroup is behaving. Here is the revised code.

    Screen Shot 2015-06-07 at 5.56.14 PM

    Screen Shot 2015-06-07 at 5.56.42 PM

    The result is interesting:

    • Some subgroups ill-mannered. That is they have either at most 33% common or they are exactly equal with ALL their conjugated subgroups.
    • Ill mannered Subgroups are also consistent in their behavior. They have exactly two behaviors
      • Exactly 33% common with their conjugated subgroup (mostly this)
      • Or 100% common (only with elements belonging to this subgroup)

    Examples of Ill Mannered Subgroups of \[S_4 \]

    [(), (1,3,4), (1,4,3)]

    [(), (1,3,4), (1,4,3)]

    [(), (1,4,2), (1,2,4)]

    [(), (1,3,2), (1,2,3)]

    [(), (1,2), (1,4,2), (2,4), (1,2,4), (1,4)]

    [(), (1,2), (1,3,2), (1,3), (1,2,3), (2,3)]

    [(), (1,3), (1,3,4), (3,4), (1,4,3), (1,4)]

    [(), (3,4), (2,3,4), (2,4,3), (2,4), (2,3)]