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Symmetric Group — Definition, Formula & Examples

The symmetric group SnS_n is the set of all possible permutations (rearrangements) of nn distinct objects, together with the operation of composing those permutations. It serves as one of the most fundamental examples of a group in abstract algebra.

For a positive integer nn, the symmetric group SnS_n is the group whose elements are all bijections from the set {1,2,…,n}\{1, 2, \ldots, n\} to itself, with the group operation being function composition. The identity element is the identity permutation, and the inverse of each permutation is its functional inverse.

Key Formula

∣Sn∣=n!|S_n| = n!
Where:
  • ∣Sn∣|S_n| = The order (number of elements) of the symmetric group on n elements
  • n!n! = n factorial, equal to n × (n−1) × ⋯ × 2 × 1

How It Works

Each element of SnS_n is a permutation that reassigns every element of {1,2,…,n}\{1, 2, \ldots, n\} to a (possibly different) position. You compose two permutations by applying one after the other. The result is always another permutation in SnS_n, so the group is closed under composition. Composition is associative, the identity permutation leaves everything fixed, and every permutation can be undone. Note that for n≥3n \geq 3, composition is not commutative — the order in which you compose matters — so SnS_n is a non-abelian group.

Worked Example

Problem: List all elements of the symmetric group S3S_3 and compute the composition (1 2 3)∘(1 2)(1\ 2\ 3) \circ (1\ 2).
Step 1: Find the order of S3S_3. Since n=3n = 3, the group has 3!=63! = 6 elements.
∣S3∣=3!=6|S_3| = 3! = 6
Step 2: List all six permutations in cycle notation: the identity ee, three transpositions (1 2)(1\ 2), (1 3)(1\ 3), (2 3)(2\ 3), and two 3-cycles (1 2 3)(1\ 2\ 3) and (1 3 2)(1\ 3\ 2).
S3={e,  (1 2),  (1 3),  (2 3),  (1 2 3),  (1 3 2)}S_3 = \{e,\;(1\ 2),\;(1\ 3),\;(2\ 3),\;(1\ 2\ 3),\;(1\ 3\ 2)\}
Step 3: Compose (1 2 3)∘(1 2)(1\ 2\ 3) \circ (1\ 2) by applying (1 2)(1\ 2) first, then (1 2 3)(1\ 2\ 3). Under (1 2)(1\ 2): 1→21 \to 2, then (1 2 3)(1\ 2\ 3): 2→32 \to 3, so 1→31 \to 3. Under (1 2)(1\ 2): 2→12 \to 1, then (1 2 3)(1\ 2\ 3): 1→21 \to 2, so 2→22 \to 2. Under (1 2)(1\ 2): 3→33 \to 3, then (1 2 3)(1\ 2\ 3): 3→13 \to 1, so 3→13 \to 1.
(1 2 3)∘(1 2)=(1 3)( 1\ 2\ 3) \circ (1\ 2) = (1\ 3)
Answer: The composition (1 2 3)∘(1 2)=(1 3)(1\ 2\ 3) \circ (1\ 2) = (1\ 3), which is the transposition swapping 1 and 3.

Why It Matters

Cayley's theorem states that every finite group is isomorphic to a subgroup of some symmetric group, making SnS_n a universal building block in group theory. Symmetric groups arise directly in combinatorics, physics (particle statistics), and cryptography whenever the structure of rearrangements matters.

Common Mistakes

Mistake: Confusing the order of composition — applying the left permutation first instead of the right one.
Correction: In most algebra textbooks, σ∘τ\sigma \circ \tau means apply τ\tau first, then σ\sigma. Always check your textbook's convention, and trace individual elements through step by step.