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A double coset is generally not a subgroup, so Lagrange's theorem does not force its size to divide the ambient group order. In $S_3$, take $H=K=⟨(12)⟩$. Then $HeH$ has size two, while $H(13)H$ has size four, which does not divide six. The general size formula $|HgK|=|H||K|/|H\cap gK{g}^{-1}|$ explains why different double cosets can have different sizes.

The doubling constant of a nonempty finite set $A$ is $|A+A|/|A|$. A small doubling constant indicates that $A$ behaves approximately like a coset of a subgroup.

If $G$ has $d$ generators, there are at most $(n!{)}^d$ group homomorphisms $G\to S_n$. Each subgroup of index $n$ is a point stabilizer in a transitive coset group action, so there are at most $n(n!{)}^d$ such subgroups. Consequently finitely many subgroups have index at most any fixed bound. Intersecting them gives a finite-index characteristic subgroup, useful in proving residual finiteness of semidirect products.

A nonzero finitely generated subgroup of the additive rational numbers is an infinite cyclic group. Put the generators over a common positive denominator. Their integer numerators generate the ideal given by their greatest common divisor, and Bezout identity expresses that divisor as an integer combination of them. Dividing back by the denominator gives a single generator.

Every finite subgroup $H$ of the multiplicative group of a field is cyclic. If $m$ is the exponent of $H$, commutativity lets one combine elements whose orders realize the prime-power factors of $m$, producing an element of order $m$. Every element of $H$ is a root of $X^m-1$, so the Lagrange root bound over a field gives $|H|\le m$. Since $m\le |H|$, equality holds and that element generates $H$.

A finitely generated group has only finitely many subgroups of any prescribed index of a subgroup $n$. Every such subgroup is a point stabilizer for a transitive action on $n$ points, while a homomorphism from a finitely generated group to the finite symmetric group $S_n$ has only finitely many possible images of the generators.

For a finite-index subgroup of the modular group, take the union of translates of the standard fundamental domain of the modular group over left coset representatives. Boundary identifications are inherited from the subgroup action. The truncated region is compact, and its finitely many cusp ends have exponential parameters; this proves compactness of the compactified modular curve and boundedness of invariant cusp-form norms.

A subset $S$ of a group $G$ is a generating set when every element of $G$ is a finite product of elements of $S$ and their inverses, equivalently when the subgroup $⟨S⟩$ generated by $S$ equals $G$.

The connected component of the identity in a Lie group is an open normal subgroup. Connected coordinate neighbourhoods prove openness, and conjugation preserves the component of the identity. Every open identity neighbourhood generates it: the generated subgroup is open, its other cosets are open, and connectedness forces it to be the whole component.

A group is locally finite when every subgroup that is a finitely generated group is finite. It is a torsion group, since the cyclic subgroup generated by one element must be finite. A restricted direct sum of groups with finite factors is locally finite. A locally finite group can be infinite, while a locally finite finitely generated group is finite by definition.

The least size of a finite set on which the quaternion group acts faithfully is eight. Every nontrivial subgroup of $Q_8$ contains its central involution $-1$. In an action on fewer than eight points, the orbit-stabilizer theorem makes every stabilizer subgroup nontrivial, so $-1$ fixes every point. This rules out a faithful group action, even with several group orbits. The regular action on eight points is faithful by Cayley theorem.

A monomial matrix has exactly one nonzero entry in each row and column. It factors as an invertible diagonal matrix times a permutation matrix. Such matrices permute the coordinate axes and normalize the group of invertible diagonal matrices; over very small fields that normalizer can be larger if the diagonal subgroup fails to distinguish the axes.

Restriction of an irreducible representation to a subgroup is multiplicity-free when its semisimple representation decomposition has no repeated irreducible summands. Over the complex numbers, for finite groups, this is equivalent to the restricted module's endomorphism algebra being commutative. This follows by writing that algebra as a product of matrix algebras of sizes equal to the multiplicities.

Every subgroup of a free group is itself a free group. The theorem follows by representing the subgroup by a connected covering graph of a rose and choosing a maximal subtree.

For an odd prime $p$, embed the elementary abelian group $(C_p{)}^3$ and the Heisenberg group over a prime field regularly in $S_{p^3}$. Every nonidentity element has order $p$ and hence regular permutation cycle type $p^{p^2}$. The two subgroups have equal intersection counts with every symmetric-group conjugacy class, so they are Gassmann equivalent, but they are not isomorphic because one is abelian and the other is not. Using a compact simply connected isometric cover with ambient deck group $S_{p^3}$ gives quotients with these nonisomorphic fundamental groups; Sunada theorem makes them isospectral while their topology differs.

Every proper subgroup $H$ of a finite p-group is properly contained in its normalizer. Let $H$ act on the left cosets $G/H$ by left multiplication. The fixed cosets are those represented by elements of $N_G(H)$, so there are $[N_G(H):H]$ of them. All other orbits of a group action have sizes divisible by $p$. Since $[G:H]$ is divisible by $p$ and at least one coset is fixed, the fixed-coset count is a positive multiple of $p$. This also proves $[N_G(H):H]\ge p$.

Over the reals or a finite field with more than two elements, the normalizer of all invertible diagonal matrices in $G{L}_2$ consists of monomial matrices and has quotient $C_2$ over that diagonal subgroup. Over ${\mathbb{F}}_p$ its size is $2(p-1{)}^2$ for $p>2$. For $p=2$ the diagonal subgroup is trivial, so its normalizer is all of $G{L}_2({\mathbb{F}}_2)$, of order six.

For covers branched over three real points with a real base point, choose the two generating loops so complex conjugation reverses each. A sheet permutation satisfying the displayed identities lifts that conjugation to an anticonformal equivalence of the covers, including the filled branch points. For closed hyperbolic Riemann surfaces this is a Riemannian isometry. Such an isometry need not come from conjugating the subgroups inside the original deck group, so a Gassmann equivalent pair can be isometric despite nonconjugacy there.