Unit 3: Symmetric Groups, Alternating Groups & Group Actions
Symmetric group S_n, cycle notation, cycle decomposition theorem, transpositions, the sign homomorphism sgn(ฯ), alternating groups A_n, Dihedral groups D_{2n}, group actions on sets, orbits, stabilizers, Orbit-Stabilizer theorem, and Burnside's Lemma (Frobenius counting formula).
ยง3.1 Symmetric Groups S_n, Permutations & Disjoint Cycle Decomposition
1. The Symmetric Group $S_n$
Let $X = \{1, 2, \dots, n\}$ be a finite set of $n$ symbols.
Definition 3.1 (Permutation & Symmetric Group): A permutation of $X$ is a bijection $\sigma: X \to X$. The set of all permutations of $X$, equipped with functional composition $\circ$, forms the Symmetric Group of Degree $n$, denoted by $S_n$. The order of $S_n$ is:
Two-Line Notation:
A permutation $\sigma \in S_n$ is traditionally displayed as:
2. Cycle Notation and Disjoint Cycles
Definition 3.2 ($k$-Cycle): A permutation $\sigma \in S_n$ is called a cycle of length $k$ (or $k$-cycle), denoted by $(a_1 \; a_2 \; \dots \; a_k)$, if:
and $\sigma(x) = x$ for all $x \notin \{a_1, a_2, \dots, a_k\}$. A 2-cycle $(a \; b)$ is called a transposition.
Definition 3.3 (Disjoint Cycles):
Two cycles $(a_1 \; \dots \; a_k)$ and $(b_1 \; \dots \; b_m)$ are disjoint if they have no elements in common:
Proposition 3.1 (Commutativity of Disjoint Cycles):
Disjoint cycles in $S_n$ commute:
Proof: If $x$ is moved by $\sigma$, it is fixed by $\tau$, so $\tau(\sigma(x)) = \sigma(x) = \sigma(\tau(x))$. If $x$ is fixed by both, both sides leave $x$ unchanged. $\blacksquare$
3. The Disjoint Cycle Decomposition Theorem
Theorem 3.1 (Cycle Decomposition Theorem): Every permutation $\sigma \in S_n$ can be written uniquely (up to the order of factors) as a product of pairwise disjoint cycles.
Theorem 3.2 (Order of a Permutation):
The order of a permutation $\sigma \in S_n$ written in disjoint cycle form $\sigma = c_1 c_2 \dots c_r$ is the least common multiple of the lengths of its cycles:
ยง3.2 Transpositions, Parity Homomorphism & Alternating Groups A_n
1. Transposition Factorization
Any $k$-cycle can be expressed as a product of 2-cycles (transpositions):
Notice that a $k$-cycle decomposes into $(k - 1)$ transpositions.
Corollary 3.1: Every permutation $\sigma \in S_n$ can be expressed as a product of transpositions. (The symmetric group $S_n$ is generated by transpositions!)
2. The Parity Theorem & The Sign Function $\text{sgn}(\sigma)$
While the decomposition into transpositions is not unique (e.g. $(1 \; 2) = (1 \; 2)(1 \; 3)(1 \; 3)$), the parity of the number of transpositions is invariant!
Theorem 3.3 (Parity Invariance Theorem): If a permutation $\sigma \in S_n$ can be written as a product of $r$ transpositions and also as a product of $s$ transpositions:
then $r$ and $s$ have the same parity:
Definition 3.4 (Even and Odd Permutations):
- A permutation $\sigma \in S_n$ is Even if it can be written as a product of an even number of transpositions.
- A permutation $\sigma \in S_n$ is Odd if it can be written as a product of an odd number of transpositions.
The Sign (Signature) Homomorphism $\text{sgn}: S_n \to (\{+1, -1\}, \cdot)$ is:
satisfying $\text{sgn}(\sigma \tau) = \text{sgn}(\sigma) \cdot \text{sgn}(\tau)$.
3. The Alternating Group $A_n$
Definition 3.5 (Alternating Group): The set of all even permutations of $S_n$ forms a normal subgroup called the Alternating Group of Degree $n$, denoted by $A_n$:
Theorem 3.4 (Order of $A_n$):
For $n \ge 2$:
Proof: The map $\text{sgn}: S_n \to \{\pm 1\}$ is a surjective group homomorphism. By the First Isomorphism Theorem:
Theorem 3.5 (Simplicity of $A_n$): The alternating group $A_n$ is a simple group (contains no non-trivial proper normal subgroups) for all $n \ge 5$. (This theorem is the Galois-theoretic cornerstone establishing the insolvability of the quintic by radicals!)
ยง3.3 Dihedral Groups D_2n, Geometric Symmetries & Group Presentations
1. Geometric Symmetries of Regular Polygons
The Dihedral Group $D_{2n}$ (also denoted $D_n$) is the group of all rigid planar symmetries (rotations and reflections) of a regular $n$-gon. The group has order:
consisting of:
- $n$ planar rotations: $R_k = \frac{2\pi k}{n}$ for $k = 0, 1, \dots, n-1$.
- $n$ planar reflections across symmetry axes: $s, sr, sr^2, \dots, sr^{n-1}$.
2. Group Presentation of $D_{2n}$
The group $D_{2n}$ is completely generated by two elements:
- $r$: Counterclockwise rotation by $2\pi/n$ (order $n$).
- $s$: Reflection across an axis passing through a vertex (order 2).
Theorem 3.6 (Presentation of $D_{2n}$):
The fundamental relation $s r = r^{-1} s = r^{n-1} s$ allows any word in $\{r, s\}$ to be reduced into canonical form:
ยง3.4 Group Actions on Sets, Orbit-Stabilizer Theorem & Burnside's Lemma
1. Formal Definition of a Group Action
Definition 3.6 (Group Action): Let $G$ be a group and $X$ a non-empty set. A group action of $G$ on $X$ is a map $\cdot: G \times X \to X$ satisfying:
- Identity: $e \cdot x = x, \quad \forall x \in X$.
- Compatibility / Associativity: $(g_1 g_2) \cdot x = g_1 \cdot (g_2 \cdot x), \quad \forall g_1, g_2 \in G, \; x \in X$.
We write $G \curvearrowright X$ to denote $G$ acting on $X$.
2. Orbits and Stabilizers
Let $G \curvearrowright X$ and $x \in X$.
Definition 3.7 (Orbit):
The orbit of $x$, denoted by $\text{Orb}(x)$ or $G \cdot x$, is the set of all elements in $X$ to which $x$ can be moved by elements of $G$:
The orbits form an equivalence partition of the set $X$: $X = \coprod \text{Orb}(x)$.
Definition 3.8 (Stabilizer / Isotropy Subgroup):
The stabilizer of $x$, denoted by $\text{Stab}(x)$ or $G_x$, is the set of elements in $G$ that fix $x$:
Proposition: $\text{Stab}(x)$ is a subgroup of $G$ for every $x \in X$.
3. The Orbit-Stabilizer Theorem
Theorem 3.7 (The Orbit-Stabilizer Theorem): Let $G$ be a finite group acting on a set $X$. For every $x \in X$:
Equivalently:
Complete Proof:
Let $H = \text{Stab}(x) \le G$. Let $G / H$ denote the set of left cosets $\{g H : g \in G\}$. Define the map $f: G / H \to \text{Orb}(x)$ by:
1. Well-Definedness and Injectivity:
This chain of equivalences proves that $f$ is both well-defined (left to right) and injective (right to left).
2. Surjectivity:
Any element in $\text{Orb}(x)$ has the form $y = g \cdot x = f(g H)$, so $f$ is surjective. Thus $f$ is a bijection between $G/H$ and $\text{Orb}(x)$. Therefore $|\text{Orb}(x)| = |G/H| = [G : H] = \frac{|G|}{|H|} = \frac{|G|}{|\text{Stab}(x)|}$. Multiplying by $|\text{Stab}(x)|$ yields $|G| = |\text{Orb}(x)| \cdot |\text{Stab}(x)|$. $\blacksquare$
4. Burnside's Lemma (Frobenius Counting Formula)
Let $X^g = \{ x \in X : g \cdot x = x \}$ denote the set of elements fixed by $g \in G$.
Theorem 3.8 (Burnside's Lemma / Cauchy-Frobenius Theorem): The number of distinct orbits $N$ of a finite group $G$ acting on a finite set $X$ is the average number of fixed points:
Step-by-step rigorous derivations with unskipped proofs, categorized into Foundational Concepts, Advanced Structural Analysis, and Honors / Proof Challenge tiers.
Consider the permutations in $S_8$:
- Write $\sigma$ as a product of disjoint cycles. 2. Compute the order $|\sigma|$ and parity $\text{sgn}(\sigma)$. 3. Compute the product $\sigma \tau$ in disjoint cycle form, and determine if $\sigma \tau \in A_8$.
Step 1: Disjoint Cycle Decomposition of $\sigma$ Trace the orbits:
- Start with $1$: $1 \to 3 \to 7 \to 4 \to 8 \to 2 \to 5 \to 1$.
This forms a cycle of length 7: $(1 \; 3 \; 7 \; 4 \; 8 \; 2 \; 5)$.
- Check remaining element $6$: $\sigma(6) = 6$ (fixed point).
Thus, the disjoint cycle decomposition is:
Step 2: Order and Parity of $\sigma$
- Order: $|\sigma| = 7$ (length of the single disjoint cycle).
- Parity: A cycle of length $k$ decomposes into $k - 1$ transpositions:
Number of transpositions = $7 - 1 = 6$ (even number). Therefore:
Step 3: Compute $\sigma \tau$ Given $\tau = (1 \; 4 \; 7)(2 \; 8)(3 \; 6 \; 5)$. Compose $\sigma \circ \tau$ from right to left:
- $1 \stackrel{\tau}{\to} 4 \stackrel{\sigma}{\to} 8$
- $8 \stackrel{\tau}{\to} 2 \stackrel{\sigma}{\to} 5$
- $5 \stackrel{\tau}{\to} 3 \stackrel{\sigma}{\to} 7$
- $7 \stackrel{\tau}{\to} 1 \stackrel{\sigma}{\to} 3$
- $3 \stackrel{\tau}{\to} 6 \stackrel{\sigma}{\to} 6$
- $6 \stackrel{\tau}{\to} 5 \stackrel{\sigma}{\to} 1$
This closes the cycle: $(1 \; 8 \; 5 \; 7 \; 3 \; 6)$. Now check the remaining elements $\{2, 4\}$:
- $2 \stackrel{\tau}{\to} 8 \stackrel{\sigma}{\to} 2$ (fixed point)
- $4 \stackrel{\tau}{\to} 7 \stackrel{\sigma}{\to} 4$ (fixed point)
Therefore:
Cycle length is 6. Parity of a 6-cycle: $6 - 1 = 5$ transpositions (odd). Therefore $\text{sgn}(\sigma \tau) = -1$, which means:
A necklace is made of 6 beads arranged in a circle. Each bead can be colored using one of 3 colors (Red, Blue, Green). Two necklaces are considered identical if one can be obtained from the other by a rotation in the plane. 1. Identify the group $G$ acting on the set of $3^6 = 729$ colorings. 2. For each element $g \in G$, compute the number of fixed colorings $|X^g|$. 3. Apply Burnside's Lemma to calculate the exact number of distinct necklaces.
Step 1: Group Action Formulation Let $X$ be the set of all colored bead configurations:
The rotational symmetry group of the necklace is the cyclic group $G = C_6 = \{R_0, R_1, R_2, R_3, R_4, R_5\}$, where $R_k$ represents a clockwise rotation by $k \times 60^\circ$. $|G| = 6$.
Step 2: Fixed Point Analysis for Each $g \in G$ For a coloring to be fixed by a rotation $R_k$, all beads in each disjoint cycle of $R_k$ must share the same color. If $R_k$ decomposes into $c(R_k)$ cycles, the number of fixed colorings is:
1. $R_0$ (Identity, $0^\circ$ rotation):
Disjoint cycles: $(1)(2)(3)(4)(5)(6) \implies c(R_0) = 6$.
2. $R_1$ ($60^\circ$ rotation):
Full cycle of length 6: $(1 \; 2 \; 3 \; 4 \; 5 \; 6) \implies c(R_1) = 1$.
3. $R_2$ ($120^\circ$ rotation):
Two cycles of length 3: $(1 \; 3 \; 5)(2 \; 4 \; 6) \implies c(R_2) = 2$.
4. $R_3$ ($180^\circ$ rotation):
Three cycles of length 2: $(1 \; 4)(2 \; 5)(3 \; 6) \implies c(R_3) = 3$.
5. $R_4$ ($240^\circ$ rotation):
Two cycles of length 3: $(1 \; 5 \; 3)(2 \; 6 \; 4) \implies c(R_4) = 2$.
6. $R_5$ ($300^\circ$ rotation):
Full cycle of length 6: $(1 \; 6 \; 5 \; 4 \; 3 \; 2) \implies c(R_5) = 1$.
Step 3: Burnside's Lemma Summation Sum of fixed points:
Applying Burnside's Lemma:
There are exactly 130 distinct non-equivalent 3-colored necklaces.
- For any action of a finite group $G$ on a finite set $X$, prove that the orbits form an equivalence partition of $X$. 2. Prove that the map $\psi: G/\text{Stab}(x) \to \text{Orb}(x)$ given by $\psi(g \text{Stab}(x)) = g \cdot x$ is a well-defined bijection, establishing $|G| = |\text{Orb}(x)| \cdot |\text{Stab}(x)|$. 3. Using double-counting of the set $S = \{(g, x) \in G \times X : g \cdot x = x\}$, derive Burnside's Lemma:
Part 1: Proof that Orbits Partition $X$ Define the relation $\sim$ on $X$ by:
- Reflexivity: $e \cdot x = x \implies x \sim x$.
- Symmetry: If $x \sim y$, then $g \cdot x = y \implies x = g^{-1} \cdot y \implies y \sim x$ (since $g^{-1} \in G$).
- Transitivity: If $x \sim y$ and $y \sim z$, then $g_1 \cdot x = y$ and $g_2 \cdot y = z$.
Then $z = g_2 \cdot (g_1 \cdot x) = (g_2 g_1) \cdot x$. Since $g_2 g_1 \in G$, $x \sim z$. Hence $\sim$ is an equivalence relation. The equivalence classes are the orbits $\text{Orb}(x)$, which partition $X$. $\blacksquare$
Part 2: Proof of the Orbit-Stabilizer Bijection Let $H = \text{Stab}(x) = \{ h \in G : h \cdot x = x \}$. Define $\psi: G/H \to \text{Orb}(x)$ by $\psi(g H) = g \cdot x$.
- Well-defined and Injective:
This proves simultaneously that $\psi$ is well-defined ($\implies$) and injective ($\impliedby$).
- Surjective: For any $y \in \text{Orb}(x)$, $y = g \cdot x = \psi(g H)$ for some $g \in G$.
Hence $\psi$ is a bijection, establishing:
Part 3: Derivation of Burnside's Lemma via Double Counting Define the incidence set:
Count the cardinality of $S$ in two ways:
1. Summing over $g \in G$:
2. Summing over $x \in X$:
Equating the two expressions:
By the Orbit-Stabilizer theorem, $|\text{Stab}(x)| = \frac{|G|}{|\text{Orb}(x)|}$. Substituting:
Since the orbits partition $X$, group the sum over distinct orbits $\mathcal{O} \in X/G$:
Therefore: