Abstract AlgebraPermutation Groups

Permutations and Composition

A permutation is a bijection from {1, …, n} to itself; Sₙ is the group of all n! of them under composition, and στ means do τ first, then σ.

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A permutation is a rearrangement, and rearrangements were the first groups ever studied: Lagrange and Galois looked at how the roots of a polynomial can be shuffled. Formally a permutation is a bijection from a set to itself.

Definition. $S_{n}$ is the set of all bijections $\sigma: \{1, \dots, n\} \to \{1, \dots, n\}$, with composition as the operation. It has $n!$ elements. We write $\sigma$ in two-row notation, with each number above its image: $$\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}$$ means $\sigma(1) = 2$, $\sigma(2) = 3$, $\sigma(3) = 1$.

Why it works

$S_{n}$ is a group. A composition of bijections is a bijection (closure). Composition of functions is associative: $(\sigma \circ \tau) \circ \rho$ and $\sigma \circ (\tau \circ \rho)$ both send $x$ to $\sigma(\tau(\rho(x)))$. The identity map is the identity, and the inverse function $\sigma^{-1}$ (read the two rows bottom to top) is the inverse.

Our convention. $\sigma\tau$ means $\sigma \circ \tau$: do $\tau$ first, then $\sigma$, exactly as for functions, $(\sigma\tau)(x) = \sigma(\tau(x))$. (Some books compose left to right; always check.)

Composing right to left: the blue arrows are $\tau$, applied first, and the green arrows are $\sigma$. Following each number through both gives the product.

Worked example: Two products in $S_{3}$

Let $\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{pmatrix}$ and $\tau = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \end{pmatrix}$. Compute $\sigma\tau$ and $\tau\sigma$.
  1. $\sigma\tau$: $1 \xrightarrow{\tau} 2 \xrightarrow{\sigma} 3$, $\ 2 \xrightarrow{\tau} 1 \xrightarrow{\sigma} 2$, $\ 3 \xrightarrow{\tau} 3 \xrightarrow{\sigma} 1$. So $\sigma\tau = \begin{pmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \end{pmatrix}$.
  2. $\tau\sigma$: $1 \xrightarrow{\sigma} 2 \xrightarrow{\tau} 1$, $\ 2 \xrightarrow{\sigma} 3 \xrightarrow{\tau} 3$, $\ 3 \xrightarrow{\sigma} 1 \xrightarrow{\tau} 2$. So $\tau\sigma = \begin{pmatrix} 1 & 2 & 3 \\ 1 & 3 & 2 \end{pmatrix}$.
  3. They differ, so $S_{3}$ is not abelian. (In fact $S_{n}$ is nonabelian for every $n \ge 3$.)

Answer: $\sigma\tau \ne \tau\sigma$.

Worked example: An inverse

Find the inverse of $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix}$.
  1. Swap the rows: $\begin{pmatrix} 3 & 1 & 4 & 2 \\ 1 & 2 & 3 & 4 \end{pmatrix}$.
  2. Sort the columns by the top row: $\sigma^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{pmatrix}$.
  3. Check: $\sigma(\sigma^{-1}(1)) = \sigma(2) = 1$, and so on for each number.

Answer: $\sigma^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 1 & 3 \end{pmatrix}$.

Counting: to build a permutation, choose $\sigma(1)$ ($n$ ways), then $\sigma(2)$ ($n - 1$ ways), and so on, giving $n!$. So $|S_{3}| = 6$, $|S_{4}| = 24$, $|S_{5}| = 120$: these groups grow fast.

Permutation groups matter far beyond shuffling. Later in the course, Cayley's theorem shows that every group can be pictured as a group of permutations, so the habits you build here (follow one input at a time, read right to left) work for every group you will meet.

Common mistake

Composing left to right. With our convention, $\sigma\tau$ applies $\tau$ first. To check a product, follow one input all the way: start at $1$, apply the right-hand factor, then the left-hand one. If your answer for $(\sigma\tau)(1)$ matches $\sigma(\tau(1))$, the order is right.

Key idea

$S_{n}$ is the group of all $n!$ permutations of $\{1, \dots, n\}$ under composition. $\sigma\tau$ means "do $\tau$, then $\sigma$"; inverses come from swapping the two rows; for $n \ge 3$ the group is not abelian.

Practice

Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.

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Sources and further reading

  • Thomas W. Judson, Abstract Algebra: Theory and Applications (open textbook), chapter 5, Permutation Groups (section 5.1, Definitions and Notation)
  • MIT OpenCourseWare 18.703 Modern Algebra (Spring 2013), lecture on permutation groups
  • John A. Beachy and William D. Blair, Abstract Algebra, online study guide (Northern Illinois University), section 2.3, Permutations
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