Follow 1 → σ(1) → σ(σ(1)) → … until you come back, then start a new cycle with the smallest unused number; disjoint cycles commute, and overlapping cycles must be multiplied out.
Two-row notation lists every number, even the ones that do not move. Permutations have a much shorter description: follow each number around until it comes back.
Cycles. The cycle $(a_{1}\;a_{2}\;\dots\;a_{k})$ sends $a_{1} \to a_{2} \to \cdots \to a_{k} \to a_{1}$ and fixes every other number. It has length $k$; a 2-cycle is a transposition. The same cycle can start anywhere: $(1\;4\;3) = (4\;3\;1) = (3\;1\;4)$.
💡 Why it works
Every permutation is a product of disjoint cycles, in exactly one way (apart from the order of the cycles and where each one starts). Algorithm: start at the smallest number, $1$, and follow $1 \to \sigma(1) \to \sigma(\sigma(1)) \to \cdots$. Since there are finitely many numbers and $\sigma$ is one-to-one, you must come back to $1$ (you cannot return to a middle number first, because it already has its preimage in the chain). Close the cycle, then start again at the smallest unused number. Fixed points form cycles of length $1$, which we leave out.
The permutation $1 \mapsto 4$, $2 \mapsto 6$, $3 \mapsto 1$, $4 \mapsto 3$, $5 \mapsto 5$, $6 \mapsto 2$, $7 \mapsto 7$ as arrows; each cycle has its own color. The fixed points $5$ and $7$ are left out of cycle notation.
Smallest unused: $2 \to 6 \to 2$: the cycle $(2\;6)$.
$5$ and $7$ are fixed.
Answer: $\sigma = (1\;4\;3)(2\;6)$.
Disjoint cycles commute: they move different numbers, so it does not matter which acts first. Cycles that share a number usually do not commute, and a product of overlapping cycles must be multiplied out, right to left like any composition.
✏️ Worked example: Multiplying overlapping cycles
Write $(1\;2\;3)(2\;3\;4)$ in $S_{4}$ as a product of disjoint cycles.
Apply $(2\;3\;4)$ first, then $(1\;2\;3)$.
$1$: the right cycle fixes $1$, the left sends $1 \to 2$. So $1 \mapsto 2$.
$2$: right sends $2 \to 3$, left sends $3 \to 1$. So $2 \mapsto 1$.
$3$: right sends $3 \to 4$, left fixes $4$. So $3 \mapsto 4$.
$4$: right sends $4 \to 2$, left sends $2 \to 3$. So $4 \mapsto 3$.
Answer: $(1\;2)(3\;4)$.
✏️ Worked example: Inverting
Find the inverse of $(1\;5\;2\;4)(3\;6)$.
Run each cycle backward: $(1\;5\;2\;4)^{-1} = (4\;2\;5\;1) = (1\;4\;2\;5)$, and a transposition is its own inverse.
The cycles are disjoint, so invert them one at a time.
Answer: $(1\;4\;2\;5)(3\;6)$.
⚠️ Common mistake
Treating overlapping cycles as if they were disjoint. $(1\;2\;3)(2\;3\;4)$ is not "a 3-cycle and a 3-cycle"; multiplied out it is $(1\;2)(3\;4)$. And multiply right to left: done left to right, the same product would give $(1\;3)(2\;4)$. A good check is to trace a single number through the cycles in both orders and see which matches the definition $\sigma\tau = \sigma \circ \tau$.
🔑 Key idea
Every permutation is a product of disjoint cycles, found by following $1 \to \sigma(1) \to \cdots$ until you return. Disjoint cycles commute; overlapping cycles are multiplied right to left; inverses run each cycle backward.
Practice
Pass with 3 right answers, the last 2 in a row. Each right answer earns 2 XP.