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For a countable discrete group, the Reiter condition requires a sequence of probability measures in ${\ell }^1(G)$ asymptotically invariant under every left translation. It is equivalent to the existence of an invariant mean and a Følner sequence. Separation by bounded functions shows that amenability gives approximate invariance simultaneously on each finite set of translations; countability makes the successive finite tests into a sequence.

For a based Serre fibration with fibre $F$ over the basepoint, lift a based cube in $B$ from its constant bottom face, keeping the side faces fixed. Its final face lies in $F$, giving surjectivity of the relative map. Homotopies in $B$ lift; two lifts of the same cube are relatively homotopic by lifting with both side liftings prescribed. This gives injectivity. For $n\ge 2$ this is a group isomorphism; at $n=1$ it is initially a bijection of pointed sets, unless the relative set is endowed with the group law transported through the fibration.

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.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.

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.

Reflections in the sides of a spherical triangle of angles $\pi /p,\pi /q,\pi /r$ generate a finite group when its images tessellate the sphere. The triangle area is $\pi (1/p+1/q+1/r-1)$ by the spherical excess formula. Label its three vertex types and continue those labels across each reflected side; they are consistent around every vertex and, because the sphere is simply connected, globally. The generated group preserves these labels, so a chamber stabilizer fixes all three vertices and is trivial. Dividing sphere area by chamber area gives its order. Up to permutation the possibilities are $(2,2,n)$ for $n\ge 2$, and $(2,3,3),(2,3,4),(2,3,5)$, of orders $4n,24,48,120$.

A field $k$ is a splitting field for a finite group $G$ if every simple module over its group algebra $kG$ is absolutely simple: it remains simple over every extension field. Equivalently, the quotient by the Jacobson radical is a product of full matrix algebras over $k$. In characteristic $p$, containing all $m$th roots of unity, where $|G|=p^{a}m$ and $p\nmid m$, is sufficient. This is the modular splitting-field theorem; it is substantially stronger than merely having eigenvalues for one chosen group element.

For odd $p$, parity of valuation and the square class of the unit residue give a group $C_2\times C_2$, represented by $1,u,p,pu$ for a unit $u$ with nonsquare residue. For $p=2$, valuation parity and unit residue modulo eight give $C_2^3$, with generators $-1,5,2$. The unit criteria follow from the Hensel lemma, using its strong derivative version for $p=2$.

For a group action of a group $G$ on a set $X$, the stabilizer subgroup of $x\in X$ consists of the transformations that leave $x$ fixed:It is a subgroup: the identity element fixes $x$, products of transformations fixing $x$ fix $x$, and the inverse element of a transformation fixing $x$ also fixes $x$. A stabilizer subgroup is a subset of $G$, whereas the orbit of a group action is a subset of $X$. See Ackerman's algebra notes, definition 0.4.1.2.

For the square group actions defined under group action, direct calculation givesThe reflection $s$ fixes $A$ and $C$ individually. The reflection $rs$ preserves the unoriented side of a polygon $AB$ by exchanging its two vertices of a polygon. The half-turn rotation $r^2$ exchanges the two ends of the unoriented diagonal of a polygon $AC$ while preserving that diagonal of a polygon. Thus fixing a side of a polygon or diagonal of a polygon as an object does not mean fixing every point on it; the latter condition uses a pointwise stabilizer. The four stabilizer subgroups have group orders $2,2,4,8$, matching group orbit sizes $4,4,2,1$ through the orbit-stabilizer theorem.

A stabilizer subgroup records the transformations still available after imposing a constraint. Fixing a vertex of a polygon of the square leaves two choices; fixing its centroid leaves all eight. In a faithful group action, a stabilizer subgroup can still contain a nonidentity transformation, as $s\in (D_4{)}_A$ shows: a faithful group action requires that only the identity element fix every point of $X$ simultaneously, not that every individual stabilizer subgroup be trivial. The two-point cyclic group example under group action has both stabilizer subgroups equal to the group kernel $\{0,2\}$ of its associated group homomorphism.