Exercises on Group actions

Let σ^S_n and with σ^A_n the conjugate classes of the permutation σ of A_n respectevely in S_n and A_n. Prove that σ^S_n coincides with σ^A_n if there exists an odd permutation of S_n commuting with σ, otherwise

σ^S_n = σ^A_n ∪ {(1,2) σ (1,2)⁻¹}^A_n
Determine the number of conjugacy classes of A₅ and the cardinality of each class.
For every permutation σ ∈ S₃ calculate the stabilizer with respect to the conjugacy action of S₃ onto itself. Verify that the order of S₃ is equal to |St_σ| ⋅ |O(σ)|. Repeat the exercise with S_n.
Determine the center of the group D₄ of square symmetries. Write the class equation for this group.
How many words of six letters can be formed by permutation of the letters of the word MATTER? Use the concepts of group action, orbit, stabilizer.
How many distinct bracelets can be made with four beads of different colors?
Repeat the previous exercies for a bracelet with two green and two yellow beads.
A factory makes cups with p vertical stripes with p prime, of r different colors. How many different cups can be produced? (assume equal-size stripes that can each get any color at all, independent of all the other colors chosen.)
Use Burnside’s lemma to anwer this question: How many striped flags are there having six stripes (of equal width) each of which can be colored red, white, or blue?
Calculate the centralizers of (12) and of (1234) of S₄. Calculate the centralizers of (12) and of (12345) of S₅.
Generalizer the previous point by calculating the r-cycles of S_n.

Solutions

We always have σ^A_n ⊆ σ^S_n so we only need to show the other inclusion. Let σ ∈ A_n. Suppose that σ commutes with odd permutation δ ∈ S_n (so that δσδ⁻¹ = σδδ⁻¹ = σ, used later on). Let τ ∈ σ^S_n, τ = ασα⁻¹, for some α ∈ S_n. If α ∈ A_n, then τ ∈ σ^A_n; if α is odd then δα ∈ A_n, and τ = ασα⁻¹ = αδσδ⁻¹α⁻¹ = (αδ)σ(αδ)⁻¹, from which again τ ∈ σ^A_n; Thus σ^S_n ⊆ σ^A_n hence σ^S_n = σ^A_n.

Suppose now that σ does not commute with any odd permutation. The the centralizer C_{S_n}(σ) of σ in S_n coincides with the centralizer C_{A_n}(σ) of σ in A_n. Owing to Corollary 7.15.7 and applying Lagrange Theorem: n = i * m where n = |A_n| and m = |C_{(A_n)}(σ)|:

|σ^A_n| = |A_n|/|C_{A_n}(σ)| = 1/2 ⋅ |S_n| / |C_{A_n}(σ)| = 1/2 ⋅ |S_n| / |C_{S_n}(σ)| = 1/2 |σ^S_n|

Notice that any odd permutation has the form (1,2)τ for some permutation τ. Thus the conjugate class ((1,2)σ(1,2)⁻¹)^A_n of (1,2)σ(1,2)⁻¹ in A_n coincides with {δσδ⁻¹ | δ odd}. Note in particular that the fact that conjugacy in S_n is determined by cycle type means that if σ ∈ A_n then all of its conjugates in S_n also lie in A_n. Thus

σ^S_n = {τστ⁻¹, τ even} ∪ {τστ⁻¹, τ odd} = σ^A_n ∪ ((1,2)σ(1,2)⁻¹)^A_n

Note that (1 2) is an odd permutation and (1 2)⁻¹ = (1 2). ■
In S_n, every permutation is conjugate to every other permutation of the same cycle type. Let's start to count the different cyclic structures of S₅. The partition of the integer 5 are: 5, 4+1, 3+2, 2+2+1, 2+1+1+1, 1+1+1+1+1, 3+1+1. Note that the partitions 3 + 2 and 2 +3 are considered equal.
- 2+2+1 represents permutations like (a b)(c d) one element is left fixed, that is an even permutation.
- 2+1+1+1, corresponds to (a b) an odd permutation.
- 3+2 represents permutations like (a b c)(d e) nothing is fixed; (a b c) can be decomposed in terms of transpositions (a b c) = (a c) (a b); so we've and odd permutation.
- 3+1+1, corresponds to (a b c) which can be decompend in terms of transpositions (a b c) = (a c) (a b) so it is even.
- 4+1 which decomposed in traspositions (a b c d) = (a d) (a c) (a b), so it's odd.
- 5 corresponds to a 5-cycle which is an even permutation (a b c d e) = (a e) (a d) (a c) (a b).
- 1+1+1+1+1 is the identity which is even.
S₅ has 7 conjugacy classes. There 4 conjugacy classes of even permutations in S₅. This means that there are not fewer than 4 classes in A₅. Some of these conjugacy classes can split. In A₅, 5-cycle splits in two conjugacy classes in e.g:

(4 5)⁻¹ (1 2 3 5 4) (4 5) = (1 2 3 4 5)

but (4 5) is odd; We have a total of five conjugacy classes in A₅ namely those having as representatives: id, (123), (12)(34), (12345) and (1345). The cardinalies are calculated as in section 7.5 and are repsectevely: 1, 20, 15, 12, 12. Where for the 5-cycle classes we have splitted as found in the previous exercise

|σ^A_n| = 1/2 |σ^S_n|

so Binom(5 5) ⋅ 4!/2 = 12. ■
We have St_id = S₃, St₍₁₂₎ = {id, (12)}, St₍₁₃₎ = {id, (13)}, St₍₂₃₎ = {id, (23)}, St₍₁₂₃₎ = {id, (123), (132)}, St₍₁₃₂₎ = {id, (123), (132)}. |O(σ)| = [g], where [g] is the conjugacy class of g ∈ S₃. The class equation holds e.g. |O(123)| ⋅ |St₍₁₂₃₎| = |S₃| = 3 ⋅ 2 = 6. ■
The group D₄ has order 8 and is generated by the rotation r and reflection σ; D₄ = ⟨r,s⟩. The eight symmetries in D₄ can be uniquely written as r^jsⁱ where j = 0, 1, 2, 3 and i = 0, 1. The group is governed by the relations

r⁴ = 1, s² = 1, rsr = s

Fist the centre Z(D₄) has to be non-trivial. Otherwise the class equation would say that

8 = 1 + a sum of even integers

which cannot be. Consider the reflection sr^j and take the square so that sr^jsr^j = 1, then multiply by s⁻¹ to get r^jsr^j = s for all j (s⁻¹ = s), and then

r^js = r^−j

In particular, r²s = rs⁻². But r⁻² = r², since r⁴ = 1. So, r²s = sr² Since r² with the generators s and s of D₄, it must lie in Z(D₄).

Let's compute the conjugates of r. Since conjugate elements all have the same order and since r, r³ are the only elements in D₄ of order 4, these two rotations form a complete conjugacy class. They are also not in Z(D₄), because central elements only have themselves as conjugates. The group relations yield

rsr⁻¹ = r²s

so the reflections s and r²s are conjugate. This also implies they are not in Z(D₄).

We check now that rs and r³s are conjugates. By the group relations, s(rs)s⁻¹ = rs = r³s. Hence these two elements are conjugate, and from this we see that they are not central either.

To conclude we need to check that s, r²s form a separate conjugacy class from that of rs, r³s. It suffices to check that s and rs are not conjugates. We could conjugate s by all eight elements of D₄ and see that we never get rs, more wisely notice that j = 0, 1, 2, 3 and i = 0, 1

r^jsⁱ s (r^jsⁱ)⁻¹ = r^jsⁱ s s⁻ⁱr^−j = r^j r^−j s = r^2js.

which never equals rs.

We have seens that 1 and r² make up the center of D₄. And there are three conjugacy classes each of order 2. Namely,

{r², r³}, {s, r²s}, {rs, r³s}

The class equation for D₄ looks like:

|G| = |Z(G)| + ∑ |G|/|C(x)|
8 = 2 + 2 + 2 + 2

Remark. Another way to see that s and rs are not conjugates is to think of the vertices of the square as the letters 1,2,3,4 respectevely. From this point of view, r is the cycle (1,2,3,4) and s the transposition (2,4) and D₄ is the subgroup of S₄ generated by r and s. Then

rs = (1,2,3,4) (2,4) = (1,2)(3,4)

Since s and rs have different cyclic structures this cannot be. ■
The group S₆ acts on the set X made by all words with the 6 letters of the word MATTER, by permuting the letters, e.g,

(135)(26)MATTER = ERMTTA

Then |X| = 6! = 720. The anagrams of x = MATTER are ins its orbit O(x). The stabilizer of x is made by all permutations σ ∈ S₆ such that

σ(MATTER) = MATTER

it coincides with all permutations that exchange the third and fourth letters. Thus the stabilizer has 2! elements, then,

|O(x)| = |S₆|/|St_x| = 6!/2 = 360 ■
We must consider the action of the group D₄ of the set of possible permutations of five colors say blue, white, pink, and green, which are 4! = 24. We consider D₄ because you can flip the bracelet over obtaining the same one. The symmetries act this way

Using the Burnisde theorem we see that only the identity leaves each configuration fixed, then it results

s = 1/8 ⋅ (24 + 0 + 0 + 0) + (0 + 0 + 0 + 0) = 24/8 = 3 ■

fixed by rotations fixed by reflections
In this case the possible permutations are 4!/2!*2! = 6. The identity fixes all 6 permutations, the rotation of 180° fixes 2 permutations, the reflections 2 permutations each

s = 1/8 ⋅ (6 + 2 + 0 + 0) + (2 + 2 + 2 + 2) = 16/8 = 3 ■

fixed by rotations fixed by reflections
Let A₄ act on itself by conjugation. We calculate the conjugacy classes of A₄ in Example 7.5.11. The stabilizer is such that

St(x) = {g ∈ A₄ : gxg⁻¹ = x}

(123) and (132) are in different orbits (they are not conjugate in A₄) but they each have stabilizer {(1),(123),(132)}, The same feature is true ofgand g⁻¹ for every 3-cycle g ∈ A₄. Applying |O(x)|*|St(x)| = |A₄| to the elements of A₄ we have
- |O(123)| * |St(123)| = 4 * 3 = 12.
- |O(id)| * |St(id)| = 12 * 1 = 12.
- |O((12)(34))| * |St(id)| = 3 * 4 = 12.
Where St((12)(34)) = {id, (12)(34), (13)(24), (14)(23)}.

Note. If instead of conjugacy we used composition as the group action we'll have |O((123))| = 12 and |St((123))| = 1. ■
We examine first the problem from a combinatorics perspective. Consider the r = 5, p = 5, case where we represent a cup as "12345" each number represents a different color; if you rotate in front of you the cup so to have a different color you obtain "51234", "45123", "34512", "23451", so each configuration, excepts the 5 cups with beads of same color has 5 twins, so

(5⁵ − 5)/5 + 5 = 629

we subtracted the 5 cups with same color and then added them at the end of the computation.

All "cup rotations" of a given stripe coloring are still the same coloring, so the group doing the acting here is cyclic group Z_n = ⟨g⟩. Let X be the set of all possible cups. Clearly |X| = r^p. Let a₁, a₂, ..., a_n a cup. Clearly the action of any fixed element aⁱ ∈ G on a given cup yields the same cup. So the number of orbits is the same as the number of different cups. We now compute X_g for each g ∈ G. First

X_e = {x ∈ X| e * x = x} = X

So |X_e| = r^p. Let g ≠ e. Since G is cyclic of prime order, g generates G. Then

X_g = {x ∈ X| g * x = x}
= {x ∈ X | gⁱ * x = x, ∀i}
= {x ∈ X | aⁱ * x = x, ∀a ∈ G}

Therefore, X_g consists of all those cups which are unchanged by any permutation and these are precisely those which are of one color. Hence |X_g| = r ∀g ∈ G, g ≠ e. The Burnside Theorem then gives the number of different necklaces as
1/p

1/p [ r^p + r + ... + r ] = [ r^p + r(p − 1)]

(p − 1 times)

In particular, since the number is necessarily an integer, we get another proof of Fermat's little theorem that a^p − a is always a multiple of p.

p divides r^p + (p − 1)r thus p|r^p − r, that is r^p ≡ r (mod p). ■
There are 3⁶ possible combinations. But the resulting flags are not all indistinguishable. The flags

(b, w, r, b, w, r) and (r, w, b, r, w, b)

are clearly the same just turned over.

Let X be the set of all 6-tuples; if x ∈ X, then

x = (c₁, c₂, c₃, c₄, c₅, c₆)

where each c_i denotes either blue, red, or white. Let τ be the permutation that reverse all the indices:
$τ = (\begin{array}{cccccc} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 4 & 3 & 2 & 1 \end{array}) = (16) (25) (34)$
(thus, τ “turns over” each 6-tuple x of colored stripes) The cyclic group G = <τ> acts on X; for a reflection τ² = e so |G| = 2; Flags with 6 stripes are directly cups with p = 2, a = 3³, i.e. G = Z₂, as the previous exercise; the orbit of any 6-tuple x consists of either 1 or 2 elements: Either τ fixes x or it does not. Since a flag is unchanged by turning it over, it is reasonable to identify a flag with an orbit of a 6-tuple. For example, the orbit consisting of the 6-tuples

(b, w, r, b, w, r) and (r, w, b, r, w, b)

The number of flags is thus the number N of orbits; by Burnside’s lemma, N = 1/2 [Fix((1)) + Fix(τ)]. The identity permutation (1) fixes every x ∈ X, and so Fix((1)) = 3⁶ (there are 3 colors). Now τ fixes a 6-tuple x if it is a “palindrome,” that is, if the colors in x read the same forward as backward. For example,

x = (w, w, r, r, w, w)

is fixed by τ. Conversely, if

x = (c₁, c₂, c₃, c₄, c₅, c₆)

is fixed, by τ = (16)(25)(34), then c₁ = c₆, c₂ = c₅, and c₃ = c₄; that is, x is a palindrome. It follows that Fix(τ) = 3³, for there are 3 choices for each of c₁, c₂ and c₃; if you double a 3 string flag you get a 6-stripe e.g rrg + grr = rrggrr. The number of flags are

N = 1/2(3⁶ + 3³) = 378. ■
The centralizer in S₄ of C(12) = {id, (12), (34), (12)(34)}; C(1234) = {id, (1234), (13)(24), (1432)}. ■
Remember that "cyclic permutations" of the symbols in a single cycle give you gack the same element of S_n: (1 2 3 4) = (2 3 4 1) = (3 4 1 2) = (4 1 2 3).

In S_n C((12)) = {τ, (12) τ | τ permutations that fix 1 and 2}. τ is disjoint from (12) and so τ commutes with (1 2) ⇐⇒ the conjugate (1 2)⁻¹ τ (1 2) is equal to τ. Generally if s and t are any two elements of S_n where neither one moves any sybmol moved by the other, then st = ts .

C((123 ... n)) = {(123 ... n)ⁱ, i = 0,..., n − 1} e.g (123)² (123) (123)⁻² = (123)² (123) (123)⁻¹ (123)⁻¹ = (123)² id (123)⁻¹ = (123). ■

«Actions on gorups Index Sulfides»

s = 1/8 ⋅	(24 + 0 + 0 + 0) +	(0 + 0 + 0 + 0)	= 24/8 = 3	■
	fixed by rotations	fixed by reflections

s = 1/8 ⋅	(6 + 2 + 0 + 0) +	(2 + 2 + 2 + 2)	= 16/8 = 3	■
	fixed by rotations	fixed by reflections

1/p	[	r^p +	r + ... + r	]	=	[	r^p + r(p − 1)]
			(p − 1 times)