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Artin fixed-field theorem says that for a finite group $G\le \mathrm{Aut}L$, $L/L^G$ is Galois with group $G$ and degree $|G|$. Let any finite $G$ act faithfully by left translation on variables $X_g$ in $L=\mathbb{C}(X_g:g\in G)$. Artin's theorem then realizes $G$ as $\mathrm{Gal}(L/L^G)$.

$s_i$ is the sum of all square-free degree-$i$ monomials in the $X_j$. The symmetric group $S_n$ permutes variables and fixes exactly $K=\mathbb{C}(s_1,\dots ,s_n)$, so Artin gives $[L:K]=n!$. The Vandermonde $f={\prod }_{i<j}(X_i-X_j)$ is not symmetric, but $f^2$ is symmetric and hence belongs to $K$.

The relations are ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$, so the generated group is $D_{2n}$. The functions $XY$ and $X^n+Y^n$ are fixed. Conversely $X^n,Y^n$ are roots of $T^2-(X^n+Y^n)T+(XY{)}^n$, and adjoining $X$ then $Y$ gives degree at most $2n$. Artin gives degree exactly $|G|=2n$, proving $M^G=\mathbb{C}(X^n+Y^n,XY)$.

If a group $K$ acts on a finite set $X$, it acts on the direct product of groups $H^X$ by permuting its factors. The resulting semidirect product is the permutation wreath product. For $K=S_m$ acting on $m$ two-point blocks and $H=C_2$, it describes independent within-block swaps combined with a permutation of blocks, and has order $2^{m}m!$. The action on $X$ is part of the definition; it need not be the regular action used in a regular wreath product.

Suppose nontrivial groups $G_1,G_2$ act on a set and there are disjoint nonempty subsets $X_1,X_2$ such that every nonidentity element of $G_1$ maps $X_2$ into $X_1$, while every nonidentity element of $G_2$ maps $X_1$ into $X_2$. Assume at least one factor has at least three elements. Then the subgroup generated by $G_1$ and $G_2$ is their free product. The cardinality hypothesis excludes the counterexample in which $G_1=G_2=C_2$ acts by swapping $X_1$ and $X_2$. Versions allowing both factors to have order two need an additional separation hypothesis.

In an accepted padded convolution of words, a loop after one tape ends can be deleted or repeated while keeping that entire tape fixed. The altered other tape still represents the same required neighbouring group element. A shortest representative has a suffix no longer than the fixed number of automaton states; an accepted longer suffix gives infinitely many representatives of the same element.

A reduced word contains no adjacent generator and inverse-generator pair. Cancel such pairs to obtain the unique reduced representative in a free group. One proof constructs the free group itself from reduced words, with multiplication given by concatenation followed by cancellation; its action by these operations on words distinguishes distinct reduced representatives.

Representation theory describes abstract symmetries by linear maps on vector spaces. A group representation assigns an invertible linear map to each group element, respecting multiplication. The payoff is that a symmetry problem becomes a problem about matrices, invariant subspaces, and eigenvalues. Decomposing the action can reveal combinations of variables that a symmetric system cannot mix.

Take the cyclic group $C_3=\{e,g,g^2\}$, with $g^3=e$. On three real coordinates, let $g$ send $(a,b,c)$ to $(c,a,b)$:This permutation representation does more than relabel a list: it acts linearly on every possible assignment of values to the three positions. The vector $(1,1,1)$ stays fixed. The vector subspace $W=\{(a,b,c):a+b+c=0\}$ is also preserved, since a permutation does not change the sum.

Every vector separates uniquely into these two types:For example, $(4,1,1)=2(1,1,1)+(2,-1,-1)$. The first term belongs to a trivial representation; the second is a contrast with zero coordinate sum. This is a direct sum decomposition ${\mathbb{R}}^3=\mathbb{R}(1,1,1)\oplus W$. The symmetry preserves the mean and rotates the contrasts.

The rotation is literal after choosing the orthonormal basis $u=(2,-1,-1)/\sqrt{6}$, $v=(0,1,-1)/\sqrt{2}$ of $W$. In this basis,It is a rotation through $120$ degrees, and three applications return to the starting point. There is no nonzero invariant real line, because this rotation has no real eigenvector. Thus the two-dimensional action is an irreducible representation over the real numbers. Over the complex numbers it instead splits into two eigenlines with eigenvalues $e^{2\pi i/3}$ and $e^{-2\pi i/3}$. The coefficient field is part of an irreducibility claim.

Why would this decomposition matter for a physical calculation? Consider three equal masses with scalar displacements $x_i$ and the symmetric quadratic coupling energyThe equations of motion are $m\ddot{x}=-kLx$, whereis a Graph Laplacian and $J$ is the all-ones matrix. The constant vector has eigenvalue zero: translating all coordinates equally costs no coupling energy. On $W$, $J$ vanishes and $L=3I$, so both independent contrast directions have angular frequency $\sqrt{3k/m}$. A three-coordinate coupled problem reduces to a free mean and two identical harmonic oscillators.

For initial displacement $x(0)=(4,1,1)$ and initial velocity zero, the exact solution is $x(t)=2(1,1,1)+\cos (\sqrt{3k/m}t)(2,-1,-1)$. The fixed mean and oscillating contrast can be read directly from the decomposition. More generally, a linear operator commuting with the group action carries its fixed-vector subspace into itself; with an invariant inner product, it preserves the corresponding orthogonal complement too.

For a finite group, averaging over the action constructs an invariant inner product over the real numbers or complex numbers. The resulting invariant complements explain Maschke's theorem: a finite-dimensional group representation decomposes into irreducible representations. Characters compress the action into traces and help identify those pieces. The assumptions matter; in modular representation theory the coefficient field can have characteristic dividing the group order, and such decompositions can fail. Chapter 4 of Pavel Etingof and coauthors' notes develops these mechanisms. ^[ref].

The restricted direct sum of a family of groups $G_i$ is the subgroup of their direct product of groups consisting of tuples with only finitely many nonidentity coordinates. Multiplication is coordinatewise. If every factor is a finite group, each subgroup that is a finitely generated group lies in a finite subproduct, so the restricted direct sum is a locally finite group. This gives concrete infinite torsion groups without a uniform bound on element orders.

For a scalar function on a group, use $(R_{a}f)(x)=f(xa)$ as the right-translation convention. For Fourier analysis on a finite group this gives $\widehat{R_{a}f}(\rho )=\hat{f}(\rho )\rho (a{)}^{\ast }$. Specifying whether $a$ or $a^{-1}$ appears in the definition avoids sign and multiplication-order ambiguity.

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.

For a group $G$, a subgroup $H$, and a generating set $S$, the Schreier graph has the left cosets $G/H$ as vertices and an edge from $gH$ to $sgH$ for each $s\in S$. Its radius-$m$ ball records the cosets met by $S^m$.

The sign homomorphism sends an even permutation to $1$ and an odd permutation to $-1$; its kernel is the alternating group.

A symmetric group is the group of all bijections from a set to itself, with function composition as its operation.

For inclusion of a point $i:\{x\}\to X$, this is the direct-image sheaf of a set. It is right adjoint to the stalk functor. It uses the singleton terminal set on opens omitting $x$, rather than the zero group used for abelian-group skyscraper sheaves.

The six-element dihedral group generated by the displayed rotations already forces every invariant second-rank tensor to have the same form as an axially invariant second-rank tensor. The axial threefold rotation eliminates mixed entries and leaves only a planar scalar plus a planar antisymmetric part; the transverse half-turn eliminates the latter. No smaller finite subgroup suffices: cyclic rotation groups retain an axial antisymmetric tensor, while a four-element noncyclic rotation group retains arbitrary diagonal tensors along its three mutually perpendicular half-turn axes.