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For $1<N◃G$ proper, the displayed condition is equivalent to $N$ being a Frobenius kernel. For a normal subgroup the induction norm is $|I_G(\phi ):N|$. Trivial inertia for each nonprincipal character and Brauer's permutation lemma imply that every element outside $N$ has trivial centralizer in $N$. This forces $N$ to be a normal Hall subgroup; a complement is then fixed-point-free on $N$.

For a connected smooth algebraic group and a finite-field Frobenius $F$, the Lang map $g\mapsto g^{-1}F(g)$ is surjective. One proof uses the twisted action $x\cdot y=xyF(x{)}^{-1}$. Since $dF=0$, every orbit map has surjective differential, so each orbit is open; connectedness leaves only one orbit. The same differential calculation makes the Lang map étale, with finite fibres equal to left cosets of $G^F$.

Let a monic integer polynomial have distinct roots in its splitting field. Outside the finitely many primes dividing its polynomial discriminant or ramifying in that field, its roots reduce to distinct residue roots. The Frobenius automorphism permutes them by $a\mapsto a^p$; a reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed. Distinctness of reduction identifies this with a fixed root of the Galois permutation itself. Combining this observation with a derangement in a transitive group action and the Chebotarev density theorem proves that an irreducible polynomial of degree at least two has rootless reduction at a positive Dirichlet density of primes. The degree bound is essential: every monic linear integer polynomial has a root modulo every prime.

If a Freiman 2-isomorphism into ${\mathbb{F}}_2^m$ uses minimal $m$, every nonzero vector lies in the fourfold sumset. Otherwise quotienting by the line it spans would preserve all additive quadruple relations and would still be injective, contradicting minimality. Since zero also belongs to the fourfold sumset, that sumset is the full model group. Plünnecke-Ruzsa inequality then bounds its size by $K^4|A|$. A finite subset of the direct sum of countably many binary fields has a finite-dimensional model to start with.

For planar strong anchoring and a quarter-turn base profile, ${\varphi }^{\prime }=\pi /(2L)$ and $\theta =0$. Expanding the one-dimensional twisted nematic energy through quadratic order gives $\Delta F=\frac{1}{2}\int [K_1{\theta }^{\prime 2}+\{(K_3-2K_2)(\pi /(2L){)}^2-\Delta ϵE^2/(4\pi )\}{\theta }^2]dz$. The lowest zero-endpoint mode is $\sin (\pi z/L)$. Its coefficient vanishes at the displayed threshold, with $V=EL$, provided the bracket and anisotropy are positive. Halving the electric coupling doubles $V_c^2$.

For a Neumann screen on the negative horizontal axis, the Helmholtz equation has an exact solution in Fresnel integrals. With time dependence $e^{i\omega t}$, a wave incident from angle $0<{\theta }_0<\pi$ has total field $e^{ikr\cos (\theta -{\theta }_0)}\mathcal{F}(s_{-})+e^{ikr\cos (\theta +{\theta }_0)}\mathcal{F}(-s_{+})$, where $s_{\pm }=\sqrt{2kr}\cos [(\theta \pm {\theta }_0)/2]$ and $\mathcal{F}(s)=\frac{1}{2}\mathrm{erfc}(-e^{i\pi /4}s)$. The angular derivatives cancel on both faces of the screen. Away from the two shadow boundaries, the Fresnel tail expansion splits this into incident/reflected geometrical optics fields and an outgoing edge wave proportional to $r^{-1/2}{e}^{-ikr}$.

A normal form for free groups and free product of groups represents elements by reduced words. In a free group, adjacent inverse letters cancel; in a free product, alternating nonidentity syllables from different factors are unique. In an amalgamated free product, chosen coset representatives produce the corresponding reduced form relative to the common subgroup.

A subgroup $A$ is a free factor of $G$ if $G$ is isomorphic to a free product $A\ast B$ by an isomorphism restricting to the inclusion of $A$, for some subgroup $B$. The normal form theorem for a free product ensures that both factors embed. A free factor need not be a normal subgroup.

Suppose $P$ has a soluble word problem for a group and $z$ generates an infinite cyclic group. In $P\ast ⟨z⟩$, for $g\ne 1$ the words $gz$ and $[g,z]$ are cyclically reduced with two and four syllables respectively. Their nonzero powers have syllable lengths $2|k|$ and $4|k|$. Given an input's reduced free product form, its length bounds the possible exponents; compare the finitely many candidates using the base word algorithm. Membership and the exponent in each of these infinite cyclic subgroups are therefore computable, with no algorithm for the order of $g$ required.

A finite generating set and finitely many relators in a free pro-p group, divided out by their closed normal closure. Relators are elements of the free pro-p group; they need not be finite ordinary words. A pro-p group with such a presentation is finitely presented in the pro-$p$ category.

Embed a free pro-p group on $x_1,\dots ,x_d$ into the units of the noncommutative power-series algebra ${\mathbb{F}}_p⟨⟨X_1,\dots ,X_d⟩⟩$ by $x_i\mapsto 1+X_i$. Its Magnus degree is the lowest total degree occurring in $\iota (w)-1$, with $D(1)=\infty$. Since $(1+v{)}^{p^e}=1+v^{p^e}$ in characteristic $p$, one has $D(w^{p^e})\ge p^{e}D(w)$. High powers therefore impose relations with small weight at any fixed parameter $0<\tau <1$.

Writing the characteristic equation with reciprocal hyperbolic cosine avoids exponentially large products. For the positive roots indexed from one, $q_n=(n+1/2)\pi +2(-1{)}^{n+1}{e}^{-(n+1/2)\pi }+O(e^{-2(n+1/2)\pi })$. The apparent root at zero represents the separate affine kernel, not another positive bending mode. Evaluating hyperbolic pieces of high modes with decaying exponential forms also avoids cancellation.

For bounded operators with trace-class commutator, the multiplicative commutator differs from identity by a trace-class operator. Along $F(t)=e^{tA}{e}^B{e}^{-tA}{e}^{-B}$, the trace of its logarithmic derivative is the constant $\mathrm{Tr}[A,B]$. This follows by writing $e^{B}A{e}^{-B}-A$ as the integral of conjugates of $[B,A]$ and using trace invariance under bounded similarities. Integrating the Fredholm determinant derivative from $F(0)=I$ proves the formula. Individual traces of $A$ and $B$ need not exist.

Suppose $f(a+h)=f(a)+Lh+r(h)$ and $|r(h)|/\Vert h{\Vert }_1\to 0$. If $c\Vert h{\Vert }_1\le \Vert h{\Vert }_2\le C\Vert h{\Vert }_1$, then $|r(h)|/\Vert h{\Vert }_2\le c^{-1}|r(h)|/\Vert h{\Vert }_1\to 0$. The boundedness of $L$ is also preserved, and the same argument in reverse proves equivalence of differentiability. The derivative is the same linear map, not a norm-dependent replacement.

A subset generates a finite p-group precisely when its images span the Frattini quotient. Indeed $⟨S⟩\Phi (G)=G$ implies $⟨S⟩=G$, since otherwise a maximal subgroup contains both factors. Minimal generating sets correspond to vector-space bases in the quotient.

For a vector bundle carrying a smooth nondegenerate symmetric bilinear form of constant metric signature, a pseudo-orthonormal frame is a smooth local frame of a vector bundle whose Gram matrix is the constant diagonal signature matrix $\eta$. It identifies the metric bundle locally with a fixed model ${\mathbb{R}}^{p,q}$. Two such frames differ by a smooth map into the corresponding orthogonal group. Positive-definite signature is included as the ordinary orthonormal case. A global pseudo-orthonormal frame is a section of a fiber bundle of the orthonormal frame bundle and trivializes that principal bundle.

Factor subsets multiply by symmetric difference; their sign columns multiply pointwise. The subgroup generated by independent defining words has $2^k$ elements. On a selected fractional factorial design, every word in this subgroup is constant, with sign determined by the chosen generator signs. The cosets classify identical or opposite contrast columns.

The resolution of a regular fraction is the shortest nonidentity word in its defining contrast subgroup. Two effects alias only when their word product belongs to that subgroup. Thus resolution VI separates all main effects and two-factor interactions from one another and from three-factor interactions, though some three-factor interactions alias in complementary pairs.

On the unit square with exact Dirichlet boundary data, let $D_{9}U=B_{h}f$ and suppose the normalized residual on the exact smooth solution has supremum norm at most $\epsilon$. The error satisfies $D_{9}e=-\tau$ and vanishes on the boundary. The barrier $q=[x(1-x)+y(1-y)]/4$ has $D_{9}q=-1$, $q\ge 0$ on the boundary, and maximum $1/8$. Apply the discrete maximum principle to $e-\epsilon q$ and $-e-\epsilon q$, obtaining $|e|\le \epsilon q$. Thus $\Vert e{\Vert }_{\infty }\le \epsilon /8$, and fourth-order residual gives fourth-order supremum norm nodal convergence. The proof needs a uniform smoothness bound and a compatible boundary discretization.

The nine-point finite-difference stencil for the Laplacian has expansion $D_{9}u=\Delta u+h^2{\Delta }^{2}u/12+h^4(u_{xxxxxx}+u_{yyyyyy}+5u_{xxxxyy}+5u_{xxyyyy})/360+O(h^6)$. If $\Delta u=f$, differentiating this equation gives $u_{xxxxyy}+u_{xxyyyy}=f_{xxyy}$ and $u_{xxxxxx}+u_{yyyyyy}={\Delta }^{2}f-3f_{xxyy}$. Hence the corrected compact equationhas sixth-order normalized consistency of a numerical method. Source derivatives can be supplied analytically; numerical approximation of $\Delta f$ must be fourth order and of the fourth derivatives second order. With exact square-grid Dirichlet boundary data and uniformly smooth solution, the discrete maximum principle inverse bound converts this residual into sixth-order nodal convergence. The unknown stencil stays compact even if more distant known source values are used.