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The rank of a profinite group is $\mathrm{rk}(G)={\sup }_{K{\le }_{c}G}d(K)$, where $K$ runs through closed subgroups and $d(K)$ is the least number of topological generators. If an open subgroup $H$ has rank $r$, its normal core $C$ is open and has rank at most $r$. Every closed $K\le G$ has a normal subgroup $K\cap C$ generated by at most $r$ elements, while $K/(K\cap C)$ has order at most $[G:C]$. A finite group of order at most $M$ is generated by at most $\lfloor {\log }_{2}M\rfloor$ elements, since each necessary new generator at least doubles the generated subgroup. Lifting generators yields $d(K)\le r+\lfloor {\log }_2[G:C]\rfloor$. Thus finite rank is preserved in both directions under passage to an open subgroup.

For a finite group, the number of irreducible characters which are real-valued characters equals the number of real conjugacy classes. On the complex vector space of class functions, the linear operator $Tf(g)=f(g^{-1})$ permutes the basis of class indicators by inversion and the basis of irreducible characters by pointwise complex conjugation, using the character inversion identity. The trace of a basis permutation is its number of fixed members. Equality of the trace in the two bases proves the theorem. The linear operator $T$ is complex-linear; pointwise complex conjugation on arbitrary class functions is not.

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 chosen inequivalent unitary irreducible representations $\rho ,\sigma$ of a finite group and a $d_{\rho }\times d_{\sigma }$ matrix $M$, the average ${\mathbb{E}}_x\rho (x)M\sigma (x{)}^{\ast }$ is zero when $\rho \ne \sigma$. For $\rho =\sigma$, it is $(\mathrm{tr}M/d_{\rho })I$. This follows from the Schur lemma, since the average intertwines the two group representations. Equivalent representations in different bases require the corresponding intertwiner; the identity-matrix formula assumes literally the same representative.

For inequivalent chosen unitary irreducible representations of a finite group, uniform expectation givesThe Kronecker deltas in the first case refer to entries in the same chosen basis. Averaging an intertwiner and applying the Schur lemma proves the formula; summing diagonal entries gives character orthogonality.

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.

The symmetric cube of a group representation $V$ is the symmetric part of $V^{\otimes 3}$. Over the complex numbers, its character isThe projection onto symmetric tensors averages the six permutations of the three factors. The identity permutation contributes $\chi (g{)}^3$, the three transpositions each contribute $\chi (g)\chi (g^2)$, and the two three-cycles each contribute $\chi (g^3)$. Cubic polynomial invariants are fixed vectors in ${\mathrm{Sym}}^3(V^{\ast })$, whose dimension follows by averaging this character for a finite group.

For a specified finite group action on $A={\mathbb{Z}}^n$, extensions are classified by $e\in H^2(Q,A)$. For an element $q$ of prime order $p$, a lift $g$ satisfies $g^p=a\in A^q$; changing the lift changes $a$ by $N_{q}v=(1+q+\cdots +q^{p-1})v$. Thus the extension is torsion-free precisely when every restriction to a prime-order cyclic subgroup is nonzero in $A^q/N_{q}A$. Any finite-order element has a prime-order power, so this tests all torsion. An arbitrary finite quotient must not be confused with faithful holonomy on a maximal translation lattice.

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]

Suppose distinct Sylow subgroups for $p$ in a finite group have trivial intersections. Conjugation by one of them, $P$, fixes $P$. The stabilizer of any other $Q$ is $H=P\cap N_G(Q)$. Since $H$ normalizes $Q$, the product $HQ$ is a subgroup of $p$-power order; maximality of $Q$ gives $H\le Q$. Thus $H\le P\cap Q=1$. All other orbits have size $|P|$, proving the congruence.

The upper unitriangular group $U{T}_3({\mathbb{F}}_3)$ consists of $M(a,b,c)=\left(\begin{array}{ccc}1& a& c\\ 0& 1& b\\ 0& 0& 1\end{array}\right)$ for $a,b,c\in {\mathbb{F}}_3$. Its multiplication is $M(a,b,c)M(a^{\prime },b^{\prime },c^{\prime })=M(a+a^{\prime },b+b^{\prime },c+c^{\prime }+a{b}^{\prime })$. The elements $M(1,0,0)$ and $M(0,1,0)$ do not commute. Yet every element is $I+N$ with $N^3=0$, so characteristic three gives $(I+N{)}^3=I$. It is a nonabelian finite group of order $27$ whose nonidentity elements all have order three.