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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 determinant-one matrix over ${\mathbb{F}}_2$ permutes the three nonzero vectors of ${\mathbb{F}}_2^2$. This group action is faithful because fixing two basis vectors forces the matrix to be the identity. There are $(4-1)(4-2)=6$ invertible matrices, all of determinant one since the only nonzero field element is one, so the action identifies the group with the symmetric group $S_3$. Reduction of integer determinant-one matrices modulo two is onto: $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ induce two distinct transpositions, which generate $S_3$. Its kernel consists of matrices congruent to the identity modulo two, the level-two principal congruence subgroup.

The full symmetry group of a regular tetrahedron is the symmetric group $S_4$ acting on its vertices. Its subgroups with at most six elements have eight types up to conjugation, with the following representative subgroups and counts:A transposition is a reflection through the plane containing the two fixed vertices and the midpoint of the exchanged edge. A double transposition is a half-turn about the axis joining the midpoints of the two exchanged edges. A three-cycle is a one-third-turn about a vertex-to-opposite-face axis. A four-cycle is a quarter-turn composed with a reflection perpendicular to the opposite-edge midpoint axis; it is an improper orthogonal transformation.

The unique Klein four-group containing only double transpositions consists of the three opposite-edge half-turns and the identity element. The other three Klein four-groups consist of two perpendicular mirror reflections, their half-turn product, and the identity element. Each $S_3$ is a stabilizer subgroup of a vertex, including the three rotations and three mirror reflections preserving that vertex.

For completeness, a Klein four-group containing $(12)$ must contain the commuting involution $(34)$ or $(12)(34)$, yielding the displayed subgroup; one containing no transpositions must contain all three double transpositions. An $S_3$ subgroup has a unique cyclic subgroup of cardinality three, whose unique fixed vertex must therefore be preserved by the entire $S_3$. These observations and the disjoint cycle decomposition exhaust the types allowed by groups of order at most six. The counts sum to $25$.

A standard Young tableau of shape $\lambda ⊢n$ is a Young tableau that uses $1,\dots ,n$ exactly once and strictly increases left to right along every row and top to bottom down every column. Formally, a bijection $T:Y(\lambda )\to \{1,\dots ,n\}$ is standard whenwhenever the indicated cells belong to the Young diagram. We write $\mathrm{SYT}(\lambda )$ for the set of such fillings and $f^{\lambda }=|\mathrm{SYT}(\lambda )|$ for its cardinality.

For $(3,2)$, the five standard Young tableaux are listed below, with each pair giving the top row followed by the bottom row:There are no others: the upper-left entry must be $1$, and choosing the two bottom-row entries fixes both increasing row orders. Of the six choices of two entries from $\{2,3,4,5\}$, only bottom row $(2,3)$ fails, because its $3$ would lie below $4$. Hence $f^{(3,2)}=5$ by direct enumeration, independently of the hook-length formula.

The cells occupied by $1,\dots ,r$ form a Young diagram: every cell above or to the left of an occupied cell has a smaller label. Consequently a standard Young tableau is equivalently a path from the empty shape to $\lambda$ in the Young branching graph, adjoining one addable node of a Young diagram at each step. For $T_2$, the successive shapes areConversely, number the cell added at step $r$ by $r$ to recover the standard Young tableau. This path interpretation also makes a standard Young tableau a linear extension of a partially ordered set defined by the row and column requirements; see Stanley's Algebraic Combinatorics, Chapter 8, on tableaux and paths in Young's lattice.

The usefulness goes beyond counting: the standard polytabloid basis of a Specht module $S^{\lambda }$ over the complex numbers is indexed by $\mathrm{SYT}(\lambda )$, so ${\dim }_{\mathbb{C}}{S}^{\lambda }=f^{\lambda }$. Each of the five pictured standard Young tableaux therefore labels one basis element of the irreducible representation $S^{(3,2)}$ of the symmetric group $S_5$, whose dimension over the complex numbers is five. The filled diagrams index the basis; the actual vectors are polytabloids. The hook-length formula computes this dimension without listing all the standard Young tableaux.

The $g$th symmetric product of a space $X$ is ${\mathrm{Sym}}^g(X)=X^g/S_g$, where the symmetric group permutes the factors. Its points are unordered $g$-tuples, with repetitions allowed. If $X$ is a Riemann surface, elementary symmetric functions in local complex coordinates make its symmetric product a smooth complex manifold, including along the locus of repeated points. A collection of disjoint circles in $X$ gives a product torus in this symmetric product.

Take the four-point setthe vertices of a polygon of a rectangle whose side lengths are four and two. Three choices of structure give three different symmetry groups on these same points.

As an unstructured set, $R$ has all $4!=24$ permutations as symmetries, forming the symmetric group $S_4$. If Euclidean distance is part of the data, the allowed maps are only the identity, a half-turn rotation, and the reflections across the two coordinate axes. They form a Klein four-group of group order four. To see that nothing is missing, fix the image of one vertex of a polygon: its three other distances, $2$, $4$, and $\sqrt{20}$, are distinct, so they uniquely determine the images of the remaining vertices.

Preserving affine structure permits more. The affine mapcycles all four points of $R$. It sends a horizontal side of length four to a vertical side of length two, so it is not an isometry. Yet it preserves the rectangle's vertex set and all affine combinations among those points.

Let $A=\mathrm{diag}(2,1)$ and let $r(x,y)=(-y,x)$ be the square's quarter-turn rotation. ThenConjugating every square symmetry by $A$ gives eight affine maps preserving $R$. This group isomorphism makes the affine symmetry group a dihedral group, although some of its maps do not preserve the original Euclidean distances. There are no further affine symmetries: an affine map preserves the common midpoint of the diagonals, so it must preserve opposite pairs. The image of one vertex has four choices and the image of an adjacent vertex has two; those choices determine the map through three noncollinear points.

ThusAn arbitrary permutation need not be an affine map. For example, exchanging the two top vertices while fixing the two bottom vertices sends the common diagonal midpoint to two different points, which no single affine map can do. The phrase "the symmetry of these points" becomes unambiguous only after stating which structure matters.

For even n, the even-weight subspace of $F_2^n$ has radical $⟨1⟩$ for the dot product. Its quotient is a nondegenerate alternating bilinear form space of dimension n−2. The symmetric group acts by coordinate permutations. At n=6 this action is faithful and gives $S_6\cong S{p}_4(2)$.

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.

A group action on a nonempty set $X$ is transitive when every two points lie in the same orbit of a group action, equivalently when for all $x,y\in X$ there is a group element $g$ with $g\cdot x=y$. There is exactly one group orbit: every position is reachable from every other. Requiring $X\ne \varnothing$ avoids calling the group action on the empty set transitive merely because the quantified statement has no points to test.^[ref]

Here a left group action of a group $G$ means a function $G\times X\to X$, written $(g,x)\mapsto g\cdot x$, satisfyingwhere $e$ is the identity element. Products move the point by $h$ first, then by $g$. Transitivity says that an orbit map $G\to X$, $g\mapsto g\cdot x$, is surjective for every $x\in X$; it suffices to check one chosen point.^[ref][ref]

For example, the symmetric group $S_3$ acts on $X=\{1,2,3\}$ by $\sigma \cdot i=\sigma (i)$. This is a transitive group action, but it is not a free group action: the stabilizer subgroup of $1$ is $\{e,(23)\}$. Both the transposition $(12)$ and the cyclic permutation $(123)$ send $1$ to $2$. In contrast, the cyclic group $C_3=\mathbb{Z}/3\mathbb{Z}$ acts on itself by $k\cdot x=k+x(\mathrm{mod}3)$, using modular arithmetic. The group element sending $x$ to $y$ is uniquely $k=y-x(\mathrm{mod}3)$. These two group actions reach the same number of points, but only the second has unique transport between them.

The stabilizer subgroup $G_x=\{h\in G:h\cdot x=x\}$ measures the ambiguity of transport. If $g_0\cdot x=y$, thenThe first set is a left coset, and the second identity says that stabilizer subgroups in one group orbit are conjugate subgroups. Briefly, $g\cdot x=g_0\cdot x$ exactly when $g_0^{-1}g\in G_x$. Thus, for a finite group in a transitive group action, the orbit-stabilizer theorem gives $|G|=|X||G_x|$, and there are exactly $|G_x|$ transports for each pair of points. The example counts are $6=3\cdot 2$ for $S_3$ and $3=3\cdot 1$ for $C_3$. This separates the number of reachable positions from the number of ways to reach each one.^[ref]

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 trefoil is the torus knot $T_{2,3}$. It is the simplest nontrivial knot by crossing number of a knot: $c(T_{2,3})=3$. A standard three-crossing alternating knot diagram is reduced, so the Tait crossing-number theorem establishes minimality. The four-crossing knot diagram shown below also represents the trefoil knot, illustrating that a particular projection need not achieve that minimum. Its Seifert genus is one.

Here is a small algebraic certificate that a trefoil knot cannot be deformed into an unknot. One Seifert matrix for a trefoil knot isTaking the determinant gives the Alexander polynomial of a knotwhere $\doteq$ allows multiplication by a Laurent unit $\pm t^n$ in the Laurent polynomial ring. No such multiplication turns this three-term Alexander polynomial of a knot into the unknot's value $1$. Since the Alexander polynomial of a knot is a knot invariant, this computation rules out every possible ambient isotopy between the two knots; it does not merely rule out the deformations we happened to try.

The knot group supplies another certificate. The presentation $⟨a,b\mid a^2=b^3⟩$ maps onto the symmetric group $S_3$ by $a\mapsto (12)$ and $b\mapsto (123)$, since both sides of the relation map to the identity. These images generate a non-abelian group, so the knot group cannot be the unknot's infinite cyclic group. However, a trefoil knot and its mirror of a link have isomorphic knot groups: the abstract knot group alone does not distinguish all knots.

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.

A Young tableau is a filling of a Young diagram with entries, typically positive integers. The shape is the partition of an integer giving its row lengths. We draw Young tableaux in the English convention of Young diagram, with left-aligned rows arranged from top to bottom. A filling need not obey increasing conditions unless these are specified: standard Young tableaux use each of $1,\dots ,n$ exactly once and strictly increase along rows and down columns; Semistandard Young tableaux permit repetitions and require weakly increasing rows and strictly increasing columns.

In the constructions for a symmetric group involving Young symmetrizers, tabloids, and polytabloids, an unqualified numbered Young tableau means a bijective filling by $1,\dots ,n$, with no increasing condition imposed. This narrower convention is compatible with the broader use needed for Semistandard Young tableaux. Historically, Young's work connected these fillings to representation theory of the symmetric group; in particular, the equality between standard Young tableau counts and irreducible representation degrees over the complex numbers predates the modern hook-length formula. See Stanley's Algebraic Combinatorics, historical notes to Chapter 8.