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For an even permutation $g$, its symmetric group class splits into two equal alternating group classes exactly when its symmetric-group centralizer contains no odd permutation. If an odd commuting element exists, an odd conjugator can be multiplied by it to become even; otherwise the index-two subgroup has two conjugation orbits. In cycle notation the splitting types are precisely distinct odd cycle lengths, counting fixed points as length-one cycles: rotating odd cycles is even, whereas an even cycle or the exchange of two equal odd cycles supplies an odd centralizer element.

For the permutation action on $n$ coordinates with $n\ge 2$, every alternating-group invariant is a symmetric polynomial plus the Vandermonde product $\Delta$ times a symmetric polynomial. Thus the invariant ring is $\mathbb{C}[e_1,\dots ,e_n,Z]/(Z^2-D(e_1,\dots ,e_n))$, where $D$ is the discriminant polynomial. The two summands are independent over the symmetric polynomial ring.

For disjoint pairs $A_i,B_i$ such that $A_i\cap B_j\ne \varnothing$ whenever $i\ne j$, the displayed sum is at most one. In a uniformly random permutation, the events that all of $A_i$ precedes all of $B_i$ are disjoint events, and their probabilities are the summands. Taking $B_i$ to be the complement of the members $A_i$ of an antichain recovers the LYM inequality.

Commutation with a fixed-point-free involution preserves its partner relation. A commuting permutation can permute the $m$ pairs and independently swap each pair. This gives the permutation wreath product with order $2^{m}m!$.

A permutation commutes with a transposition exactly when it preserves its two-element support as a set. It may swap those two points and independently permute the remaining points. Thus this centraliser has order $2(n-2)!$.

The cycle type of a permutation is the multiset of lengths in its decomposition into disjoint permutation cycles, including one-cycles for fixed points. Two permutations in the same symmetric group are conjugate exactly when they have the same cycle type.

A cyclic interval of length $r$ is the set of $r$ consecutive positions in a cyclic ordering. For $1\le r<n$, there are $n$ distinct intervals, one ending at each position. Under a uniformly random permutation, a fixed $r$-set appears as an interval with probability $n/\left(\genfrac{}{}{0ex}{}{n}{r}\right)$. This is the counting mechanism of the Katona circle method.

An intersecting family of cyclic intervals of length $r$ in a cyclic order of $n$ positions has at most $r$ members when $1\le r\le n/2$. Rotate one selected interval to end at position $n$. Intervals ending at $r,\dots ,n-r$ miss it. Pair the remaining endpoints, other than $n$, as $(j,j+n-r)$ for $1\le j\le r-1$; each pair represents two disjoint intervals, hence contributes at most one member. This gives $1+(r-1)=r$. Counting such intervals over all permutations proves the Erdős-Ko-Rado theorem.

A cyclic ordering of a finite ground set records its elements around an oriented circle, identifying permutations that differ by rotation. There are $(n-1)!$ such orderings for $n\ge 1$. A uniformly random permutation induces a uniformly random cyclic ordering.

Delete $n$ from the cycle notation of a permutation in $S_n$ to obtain a permutation in $S_{n-1}$. Extend this set map linearly to ${\Pi }_n:\mathbb{C}{S}_n\to \mathbb{C}{S}_{n-1}$. It is an $S_{n-1}$-bimodule map and the identity on $\mathbb{C}{S}_{n-1}$, but is not in general an algebra homomorphism. In particular,For $n\ge 5$ it is the unique bimodule-equivariant set map $S_n\to S_{n-1}$; at $n=4$ another such map exists because $C_{S_3}(S_2)=S_2$.

An antichain in the Boolean lattice has Lubell mass one if and only if it is a full rank level. Equality means every maximal chain in a Boolean lattice meets the family exactly once. Swapping two adjacent entries in the defining permutation then forces the family to contain every set obtained from one member by exchanging an included and an excluded element, hence every set of that rank.

The general linear group over a finite field $G{L}_2({\mathbb{F}}_2)$ acts faithfully on the four vectors of ${\mathbb{F}}_2^2$, since a matrix fixing all vectors fixes the standard basis. The zero vector is fixed by every element; the induced action on the three nonzero vectors identifies this six-element group with $S_3$. Swapping the two coordinates fixes $(0,0)$ and $(1,1)$ and exchanges the other two vectors, so its four-point permutation is odd. This action therefore embeds the group into $S_4$ with nontrivial composite sign homomorphism.

For a permutation $\pi$ of $\{1,\dots ,n\}$, an inversion is a pair $i<j$ with $\pi (i)>\pi (j)$. Its number equals the Coxeter length for adjacent transpositions: each adjacent swap changes the count by one, and swapping adjacent descents sorts the permutation in exactly that many steps.

The commutant of the permutation action on $V^{\otimes n}$ identifies with the symmetric tensors in $\mathrm{End}(V{)}^{\otimes n}$. Polarization spanning of symmetric tensors spans it by $a^{\otimes n}$. Interpolating $(a+tI{)}^{\otimes n}$ at $n+1$ values where $a+tI$ is invertible expresses each such power using invertible ones. Hence the Schur algebra is the linear span of the diagonal general linear action.

For a permutation $f$ of an $N$-element set and a known marked output subset of size $K$, exactly $K$ inputs have marked outputs. The uniform quantum superposition thus has good probability $K/N$, independent of which permutation the oracle represents. This allows a known-angle amplitude amplification schedule without counting unknown preimages.