Permutations

A permutation written as the list of its images: $\sigma =(231)$ records $\sigma (1)=2$, $\sigma (2)=3$, $\sigma (3)=1$. Position carries the input, entry carries the output, and the ambient $n$ is visible as the length of the list.

This form makes inversions readable directly: scan the pairs and count how many have . For $(231)$ the pairs are $(2,3)$ in order, $(2,1)$ inverted, $(3,1)$ inverted, two inversions, so the sign is $+1$. Nothing has to be traced or reconstructed; the count is a scan of the written symbols.

It is also the form the Leibniz formula indexes: the term for $\sigma$ takes $a_{1,\sigma (1)}{a}_{2,\sigma (2)}\cdots$, reading the entry positions straight off the list, and it is the form a permutation matrix is built from, row $i$ has its 1 in column $\sigma (i)$.

What it hides is the orbit structure. That $(231)$ is a single three-element loop while $(2143)$ is two independent swaps is not apparent from the lists, and neither is the order of the permutation or its decomposition into transpositions. Those require tracing where each element travels, which is the cycle form's business.

The same permutation written as its disjoint orbits: $(123)$ means $1\mapsto 2\mapsto 3\mapsto 1$. Following any element until it returns closes one cycle; starting again from an unused element opens the next. The decomposition is unique up to the order the cycles are written and the starting point within each.

This form makes the sign immediate. A cycle of length $k$ is $k-1$ transpositions, so its parity is the parity of $k-1$, an even-length cycle is odd, an odd-length cycle is even. The whole permutation's sign is the product over its cycles, which for $(2143)=(12)(34)$ is $(-1)(-1)=+1$ without counting a single inversion.

It also exposes what one-line notation buries: the order of the permutation is the least common multiple of the cycle lengths, composition with a disjoint cycle is independent, and the structure of $S_n$ into conjugacy classes is exactly the classification by cycle type.

Two hazards attach. Fixed points are conventionally omitted, so $(13)$ is a permutation of $\{1,2,3\}$ or of $\{1,\dots ,9\}$ depending on an $n$ the notation does not state. And the length of a cycle is not its parity, the most common error in this representation is reading $(1234)$ as even because 4 is even, when it is three transpositions and odd.

What this form loses is direct access to $\sigma (i)$ for a given $i$: answering that means locating $i$ in its cycle and stepping once, and building a matrix or a Leibniz term means converting back.

Each of these is worked by both routes, so the agreement is visible rather than asserted.

The identity, $(1234)$ in $S_4$. No pair is out of order, so $\mathrm{inv}=0$ and the sign is $+1$. As cycles it is four fixed points, contributing no transpositions. It is the one permutation whose matrix is $I$, with $detI=1$ agreeing.

A transposition, $(2134)$. One pair is inverted, the leading $2$ against the $1$, so $\mathrm{inv}=1$ and the sign is $-1$. As cycles, $(12)$ with 3 and 4 fixed: a single swap. This is the smallest odd permutation, and its matrix is the elementary row-exchange matrix whose determinant the determinants unit gave as $-1$.

A 3-cycle, $(231)$ in $S_3$. The inverted pairs are $(2,1)$ and $(3,1)$, so $\mathrm{inv}=2$ and the sign is $+1$. As cycles it is $(123)$, one cycle of length 3, hence $3-1=2$ transpositions. Odd length, even permutation. The relationship that trips people, seen here in the smallest case.

A 4-cycle, $(2341)$ in $S_4$. The inverted pairs are $(2,1)$, $(3,1)$ and $(4,1)$, three, so the sign is $-1$. As cycles it is $(1234)$, one cycle of length 4, hence 3 transpositions. Even length, odd permutation, the mirror of the previous case.

Two disjoint swaps, $(2143)$ in $S_4$. Inverted pairs: $(2,1)$ and $(4,3)$, so $\mathrm{inv}=2$ and the sign is $+1$. As cycles, $(12)(34)$: two transpositions, each contributing $-1$, and $(-1{)}^2=+1$. Signs multiply across disjoint cycles.

The reversal, $(4321)$ in $S_4$. Every one of the $(\genfrac{}{}{0ex}{}{4}{2})=6$ pairs is inverted, so the sign is $(-1{)}^6=+1$. As cycles it is $(14)(23)$, again two transpositions. Reversing four elements is even, although reversing three is odd; the parity depends on $n$ through $(\genfrac{}{}{0ex}{}{n}{2})$ and is not a property of "reversing".

Two of them, the 3-cycle and the double swap, have the same sign by different structures. Two others, the 3-cycle and the 4-cycle, differ in sign by one unit of length. And the count of inversions is never the count of transpositions, $(4321)$ has six inversions and two transpositions, only their parity is shared, which is precisely what the sign records.

Two permutations of the same size, opposite signs.

$\sigma$ displaces more elements than $\tau$ and is nonetheless the even one. Amount of disturbance is not what the sign measures, only the parity of the swap count is.

Same sign, different cycle structure. In $S_4$, both $(123)$ and $(12)(34)$ are even: the first is one 3-cycle, two transpositions; the second is two 2-cycles, also two transpositions. Their one-line forms are $(2314)$ with inversions 2, and $(2143)$ with inversions 2. Equal signs, and no other structural resemblance, sign is a single bit, and it does not determine the permutation.

The reversal, which alternates. The full reversal of $n$ elements has $(\genfrac{}{}{0ex}{}{n}{2})$ inversions:

Reversing four elements is even; reversing three is odd. Nothing about "reversal" fixes a sign. It depends on $n$ through a binomial coefficient, and the pattern has period 4.

Where the contrast bites. Given a matrix whose rows have been reversed, whether $det$ changes sign depends on the size. For a $4\times 4$ the reversal is even, so the determinant is unchanged; for a $3\times 3$ it is odd and the determinant negates. Reasoning from "a reversal is a lot of swapping, so it must flip the sign" gives the wrong answer half the time.

A cycle of even length is an even permutation, so the 4-cycle $(1234)$ has sign $+1$.