10.1 Groups, Subgroups, and Permutations

10.1 Groups, Subgroups, and Permutations

Group theory forms the bedrock of modern abstract algebra. On the GRE Mathematics Subject Test, questions test structural properties of groups, subgroup criteria, cyclic generators, permutation order and sign calculations in $S_n$ and $A_n$, and geometric symmetry groups such as $D_n$.

---

Group Axioms and Core Properties

A group $(G, )$ is a set $G$ equipped with a binary operation $: G \times G \to G$ satisfying four fundamental axioms:

1. Closure: For all $a, b \in G$, $a * b \in G$.
2. Associativity: For all $a, b, c \in G$, $(a * b) * c = a * (b * c)$.
3. Identity Element: There exists an element $e \in G$ such that for all $a \in G$, $e * a = a * e = a$.
4. Inverse Element: For each $a \in G$, there exists an element $a^{-1} \in G$ such that $a * a^{-1} = a^{-1} * a = e$.

If the operation additionally satisfies commutativity ($a * b = b * a$ for all $a, b \in G$), the group is termed abelian (or commutative).

Elementary Deductions

• Uniqueness: Both the identity element $e$ and the inverse $a^{-1}$ of any element $a \in G$ are unique.
• Shoes and Socks Property: $(a * b)^{-1} = b^{-1} * a^{-1}$.
• Cancellation Laws: Left cancellation ($a * b = a * c \implies b = c$) and right cancellation ($b * a = c * a \implies b = c$) hold universally in any group.
• Order of a Group and Element: The order of a group $|G|$ is its cardinality. The order of an element $g \in G$, denoted $|g|$ or $\operatorname{ord}(g)$, is the smallest positive integer $n$ such that $g^n = e$. If no such positive integer exists, $g$ has infinite order. A critical divisibility property states: $g^k = e \iff |g| \mid k$ Furthermore, for any integer $k$, $|g^k| = \frac{|g|}{\gcd(|g|, k)}$.

---

Subgroups and Subgroup Tests

A subset $H \subseteq G$ is a subgroup (written $H \le G$) if $H$ forms a group under the operation inherited from $G$. Every group $G$ has trivial subgroups ${e}$ and $G$ itself; any other subgroup is proper.

Subgroup Tests

To establish whether a non-empty subset $H \subseteq G$ is a subgroup, full verification of the four axioms is unnecessary:

• Two-Step Subgroup Test: $H \le G$ if and only if:
  1. For all $a, b \in H$, $ab \in H$ (closed under multiplication).
  2. For all $a \in H$, $a^{-1} \in H$ (closed under inverses).
• One-Step Subgroup Test: $H \le G$ if and only if for all $a, b \in H$, $ab^{-1} \in H$.
• Finite Subgroup Test: If $H$ is a finite non-empty subset of a group $G$, then $H \le G$ if and only if $H$ is closed under multiplication ($a, b \in H \implies ab \in H$). Inverses are automatically guaranteed because the sequence of powers $a, a^2, a^3, \dots$ must repeat by the pigeonhole principle, yielding $a^k = e$ and hence $a^{-1} = a^{k-1} \in H$.

Important Canonical Subgroups

• Center of $G$: $Z(G) = {z \in G : zg = gz \text{ for all } g \in G}$. The center is always an abelian subgroup of $G$.
• Centralizer of an element: $C_G(a) = {g \in G : ga = ag}$.
• Normalizer of a subgroup: $N_G(H) = {g \in G : gHg^{-1} = H}$.

---

Cyclic Groups and Their Subgroups

A group $G$ is cyclic if there exists an element $a \in G$ such that $G = \langle a \rangle = {a^k : k \in \mathbb{Z}}$. The element $a$ is called a generator.

Classification of Cyclic Groups

Every infinite cyclic group is isomorphic to $(\mathbb{Z}, +)$, and every finite cyclic group of order $n$ is isomorphic to $(\mathbb{Z}_n, +)$.

Fundamental Theorem of Cyclic Groups

Let $G = \langle a \rangle$ be a cyclic group of order $n$:

1. Subgroup Characterization: Every subgroup of $G$ is cyclic.
2. Divisor Correspondence: For each positive divisor $d$ of $n$, there exists exactly one subgroup of order $d$, specifically generated by $a^{n/d}$.
3. Generators: An element $a^k$ generates $G$ if and only if $\gcd(k, n) = 1$. Consequently, the total number of generators of $\mathbb{Z}_n$ is given by Euler's totient function $\phi(n)$.
4. Elements of Order $d$: For any divisor $d$ of $n$, the number of elements in $G$ of order exactly $d$ is $\phi(d)$.

---

Symmetric Groups $S_n$ and Alternating Groups $A_n$

The symmetric group $S_n$ is the group of all bijections (permutations) from the set ${1, 2, \dots, n}$ to itself under function composition. The order of $S_n$ is $|S_n| = n!$. For $n \ge 3$, $S_n$ is non-abelian.

Disjoint Cycle Decomposition

A $k$-cycle $(a_1\ a_2\ \dots\ a_k)$ maps $a_1 \mapsto a_2, \dots, a_k \mapsto a_1$, leaving other elements fixed. Two cycles are disjoint if they share no moved elements.

• Disjoint cycles commute: If $\sigma$ and $\tau$ are disjoint cycles, $\sigma\tau = \tau\sigma$.
• Canonical Factorization: Every permutation $\sigma \in S_n$ can be factored uniquely (up to the order of factors) as a product of disjoint cycles.
• Order via LCM: The order of a permutation $\sigma$ written in disjoint cycle form is the least common multiple of the lengths of its cycles: $\operatorname{ord}(\sigma) = \operatorname{lcm}(\ell_1, \ell_2, \dots, \ell_m)$

Transpositions, Parity, and $A_n$

A transposition is a 2-cycle $(i\ j)$. Every permutation can be expressed as a product of transpositions.

• While the factors and number of transpositions are not unique, the parity (even or odd) of the number of transpositions is invariant.
• A cycle of length $k$ can be written as the product of $k - 1$ transpositions: $(a_1\ a_2\ \dots\ a_k) = (a_1\ a_k)(a_1\ a_{k-1})\dots(a_1\ a_2)$.
• Sign Homomorphism: $\operatorname{sgn}: S_n \to {+1, -1}$ satisfies $\operatorname{sgn}(k\text{-cycle}) = (-1)^{k-1}$.
  • An odd-length cycle is an even permutation.
  • An even-length cycle is an odd permutation.
• The alternating group $A_n$ is the set of all even permutations in $S_n$. It is a subgroup of order $|A_n| = \frac{n!}{2}$. For $n \ge 5$, $A_n$ is a simple group (having no non-trivial proper normal subgroups).

---

Dihedral Groups and Small Group Classification

The dihedral group $D_n$ represents the group of symmetries of a regular $n$-sided polygon in the plane, containing $n$ rotations and $n$ reflections. Its order is $|D_n| = 2n$.

• Generators and Relations: $D_n = \langle r, s \mid r^n = e,\ s^2 = e,\ srs = r^{-1} \rangle$.
• The $n$ rotations form a cyclic subgroup $\langle r \rangle \cong \mathbb{Z}_n$ of index 2.
• Every reflection has the form $sr^k$ and satisfies $(sr^k)^2 = e$, meaning all $n$ reflections have order 2.

Small Group Landscape ($|G| \le 8$)

| Order $|G|$ | Abelian Groups | Non-Abelian Groups | Key Properties | | :--- | :--- | :--- | :--- | | 1 | ${e}$ | None | Trivial group | | 2 | $\mathbb{Z}_2$ | None | Prime order, cyclic | | 3 | $\mathbb{Z}_3$ | None | Prime order, cyclic | | 4 | $\mathbb{Z}_4, V_4 \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ | None | $V_4$ is Klein four-group; every non-identity has order 2 | | 5 | $\mathbb{Z}_5$ | None | Prime order, cyclic | | 6 | $\mathbb{Z}_6 \cong \mathbb{Z}_2 \times \mathbb{Z}_3$ | $S_3 \cong D_3$ | Smallest non-abelian group; $|S_3| = 6$ | | 7 | $\mathbb{Z}_7$ | None | Prime order, cyclic | | 8 | $\mathbb{Z}_8, \mathbb{Z}_4 \times \mathbb{Z}_2, \mathbb{Z}_2^3$ | $D_4, Q_8$ | $Q_8 = {\pm 1, \pm i, \pm j, \pm k}$ has 6 elements of order 4 |

---

Step-by-Step Worked Problem

Problem: In the symmetric group $S_7$:

1. Determine the maximum possible order of an element.
2. Count the number of elements of order 10 in $S_7$.

Solution:

• Step 1: Partitioning 7 to maximize LCM. The order of any $\sigma \in S_7$ is $\operatorname{lcm}(c_1, \dots, c_k)$ where $\sum c_i \le 7$. Testing candidate integer partitions of integers $\le 7$:
• $7$: order $7$.
• $5 + 2$: order $\operatorname{lcm}(5, 2) = 10$.
• $4 + 3$: order $\operatorname{lcm}(4, 3) = 12$.
• $3 + 2 + 2$: order $\operatorname{lcm}(3, 2, 2) = 6$.

The maximum possible element order in $S_7$ is 12, achieved by any permutation with cycle structure $(3, 4)$.

• Step 2: Counting elements of order 10. To achieve order 10, the cycle lengths must have $\operatorname{lcm} = 10$ and sum $\le 7$. The unique cycle structure is a disjoint product of a 5-cycle and a 2-cycle ($5 + 2 = 7$).
• Choose 5 elements from 7 for the 5-cycle: $\binom{7}{5} = \frac{7 \times 6}{2} = 21$.
• The number of distinct circular arrangements of 5 elements is $(5 - 1)! = 4! = 24$.
• The remaining 2 elements form a unique transposition: $(2 - 1)! = 1$.
• Total count: $21 \times 24 \times 1 = 504$ elements of order 10.

---

GRE Exam Traps & Pitfalls

Trap 1: Cycle Length Parity vs. Permutation Parity A cycle of length $k$ has sign $(-1)^{k-1}$. A 3-cycle is an even permutation ($(-1)^2 = +1$), whereas a 2-cycle (transposition) is odd ($(-1)^1 = -1$). Do not assume an odd-length cycle is an odd permutation.

Trap 2: Adding Cycle Lengths Instead of Taking LCM The order of $(1\ 2)(3\ 4\ 5)$ is $\operatorname{lcm}(2, 3) = 6$, never $2 + 3 = 5$. Sums dictate the ambient index $n$ in $S_n$; LCM dictates element order.

Trap 3: Assuming Non-Cyclic Implies Non-Abelian The Klein four-group $V_4 \cong \mathbb{Z}_2 \times \mathbb{Z}_2$ is abelian, but it has no element of order 4 and is therefore not cyclic.

Trap 4: Subgroup Order Divisibility Converse Lagrange's theorem asserts that if $H \le G$, then $|H| \mid |G|$. However, if $d \mid |G|$, $G$ does not necessarily possess a subgroup of order $d$. The alternating group $A_4$ has order 12, but has no subgroup of order 6.