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Choose two positions independently and uniformly from $\{1,\dots ,n\}$ and swap them. This Markov chain on the symmetric group holds with probability $1/n$ and assigns probability $2/n^2$ to each unordered transposition. Its stationary distribution is uniform, and its ordinary spectral gap is $2/n$ for $n\ge 2$. The latter follows from the Aldous spectral gap theorem: the single-label continuous-time Markov chain has rate $2/n^2$ between every pair of positions, whose mean-zero eigenvalues are all $-2/n$.

The orientation-preserving symmetries of a cube are the $3\times 3$ signed permutation matrices with determinant one. There are $3!2^2=24$ of them. Their group action on the four body diagonals identifies this group with the symmetric group $S_4$. Indeed choose diagonal directions $(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)$, whose sum is zero and whose only linear relation has all coefficients equal. A rotation fixing every diagonal sends each direction to itself or its negative; the relation forces all four signs equal. All negative would give $-I$, which has determinant $-1$, so the kernel of a group homomorphism is trivial. Both groups have order 24, giving the identification. Its face permutations have cycle types $1^6$, $1^{2}4$, $1^2{2}^2$, $2^3$ and $3^2$, occurring $1,6,3,6,8$ times respectively.

The commuting actions of the symmetric group and the general linear group on a tensor power are mutual commutants. The displayed sum ranges over partitions with at most $\dim V$ rows. The Specht modules and Schur modules are the simple factors for the two actions. In particular a primitive Young symmetrizer selects one copy of the matching Schur module.

A regular tetrahedron has a proper rotational symmetry group $T$ of order 12, isomorphic to the alternating group $A_4$ acting on its four vertices. Its full spatial symmetry group $T_d$ has order 24 and is isomorphic to the symmetric group $S_4$; the additional operations reverse orientation. Three-fold vertex/face-axis rotations and half-turns about opposite-edge midpoints generate the proper group. The distinction matters when a field symmetry uses only proper rotations, but a scalar density also admits reflections.

Associate a transposition to each edge of a finite connected graph. For a simple path $v_0,\dots ,v_{\ell }$, let $t_j=(v_{j-1}{v}_j)$. Then $t_1\cdots t_{\ell -1}{t}_{\ell }{t}_{\ell -1}\cdots t_1=(v_0{v}_{\ell })$, using rightmost-first composition. Thus edge transpositions generate every transposition, and hence the full symmetric group.

The symmetric group acts on multivariate polynomials by replacing each variable $x_i$ by $x_{\sigma (i)}$. With $(\sigma \tau )(i)=\sigma (\tau (i))$, substitution gives $\sigma \cdot (\tau \cdot f)=(\sigma \tau )\cdot f$. Thus it is a left group action. A polynomial encoding a combinatorial partition has a group orbit encoding all relabelings and a stabiliser subgroup encoding its label symmetries.

For a Young tableau $t$, let $r_t$ sum its row permutations and $c_t$ be the signed sum of its column permutations in the group algebra of the symmetric group. The product $h_t=c_t{r}_t$ is a Young symmetrizer. The other order is another conventional realization. It satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the hook product of a partition, so division by that scalar gives a primitive idempotent in characteristic zero. Acting on a tensor power constructs a Schur module.

The standard-tableau right-ideal decomposition of $\mathbb{C}{S}_n$, tensored over that algebra with $V^{\otimes n}$, gives this direct sum of general linear group modules. With $e_t=h_t/H_{\lambda }$, the map $e_t\mathbb{C}{S}_n\otimes T\to e_{t}T$ is an isomorphism. Individual summands generally need not be symmetric group submodules; instead each is a Schur module for the commuting action.