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For a maximal Brauer pair $(D,b_D)$ of a block of a group algebra, the inertial quotient is the displayed quotient of its pair stabilizer by $D{C}_G(D)$. It has order coprime to $p$, and its order is the inertial index of a block. For abelian $D$, conjugation embeds it in $\mathrm{Aut}(D)$.

Let $k$ be an algebraically closed field and let a block of a group algebra $B$ have a central defect group of a block $D$. Its quotient by $J(kD)B$ is a defect-zero block, hence a full matrix algebra over $k$. The algebra $B$ is free over the central local ring $kD$. Lift a primitive idempotent from this matrix quotient; its corner is free of rank one over $kD$, and is therefore $kD$. Lifting matrix units gives $B\cong {\mathrm{Mat}}_r(kD)$.

For a block of a group algebra with identity $e$ and defect group of a block $D$, the trace criterion for defect groups writes $e={\mathrm{Tr}}_D^G(a)$ with $a\in (RG{)}^D$. On any module in the block, multiplication by $a$ is $D$-linear and its relative trace is the identity. The D. Higman criterion therefore makes the module relatively $D$-projective.

Let $P$ be any Sylow subgroup containing a defect group of a block $D$. Regard the block of a group algebra as a permutation module summand for $G\times G$, acting by left and right multiplication. Its vertex is $\Delta D$. Restricting to $P\times P$ retains a summand with this vertex. Transitive permutation modules for a finite p-group are indecomposable, with their point stabilizers as vertices. Thus some orbit in $G$ has stabilizer $\Delta D$ at a suitably chosen point $c$. This means $c$ centralizes $D$ and $P\cap {}^{c}P=D$. Choosing $P$ to contain a Sylow subgroup of $N_G(D)$ shows that $D=O_p(N_G(D))$; every normal p-subgroup of $G$ is contained in both Sylow factors and hence in $D$.

For a block idempotent $b$ with defect $D$, the Brauer quotient of a module of its modular block bimodule at $\Delta D$ is nonzero: on the fixed basis $C_G(D)$, the block projection acts by ${\mathrm{Br}}_D(b)$. Restriction to $D\times D$ splits into transitive permutation modules; their stabilizers have order at most $|D|$. A nonzero quotient at $\Delta D$ therefore forces a summand with stabilizer exactly $\Delta D$, which is the regular $kD$ bimodule. Tensoring on the right with any $kD$-module $U$ gives $U\mid {\mathrm{Res}}_D^G(b{\mathrm{Ind}}_D^{G}U)$. Consequently finite representation type of the modular block forces finite representation type of $kD$.

Rows of the symmetric-group decomposition matrix are all ordinary shapes, columns are regular partitions, and $d_{\mu \lambda }=[S^{\mu }:D^{\lambda }]$. A nonzero column antisymmetrizer of shape $\lambda$ on $M^{\mu }$ forces dominance: the first $r$ rows contain at most ${\sum }_j\min (r,{\lambda }_j^{\prime })$ column entries. For a regular shape its diagonal simple appears once, giving a rectangular triangular matrix and a unitriangular regular-row submatrix.

If a finite group $G$ is generated by $h$ elements, send the $a$-generators of the genus-$h$ surface group to those generators and its $b$-generators to the identity. The surface relation is satisfied, so this is a surjective group homomorphism. Its kernel defines a connected regular covering space, on which the deck transformation group $G$ acts freely. The lifted orientation and multiplicativity of the Euler characteristic give genus $g=1+|G|(h-1)$.

A terminating decimal is a decimal representation with only finitely many nonzero fractional numerical digits. For a reduced fraction $a/q$ with $q>0$, this occurs exactly when every prime factor of $q$ is $2$ or $5$.

If the representation terminates after $k$ fractional digits, write $a/q=M/10^k$ for an integer $M$. Since $a,q$ are coprime integers, $q$ divides $10^k=2^k{5}^k$. This proves necessity. Conversely, if $q=2^u{5}^v$, choose $k=\max \{u,v\}$. Then $q\mid 10^k$, making $M=a10^k/q$ an integer and giving a finite representation.

For example, $3/40=0.075$ because $40=2^3\cdot 5$ divides $10^3$. The reduced fraction $1/6$ has a denominator factor $3$, so its decimal representation cannot terminate. A zero tail is distinct from a nonzero repeating block; the latter is measured by decimal period from multiplicative order.

Let $m=(f(b)-f(a))/(b-a)$ for a function differentiable on a neighborhood of $[a,b]$. If $m=k$, use the whole interval. If $m>k$, the continuous extension of $(f(x)-f(a))/(x-a)$ takes values $f^{\prime }(a)$ and $m$ at the endpoints, so the intermediate value theorem supplies a secant of slope $k$ starting at $a$. If $m<k$, use the continuous extension of $(f(b)-f(x))/(b-x)$ and obtain a secant ending at $b$. The mean value theorem then produces an interior derivative equal to $k$, without requiring a continuous derivative.

For underdamping, let $x_n,x_{n+1}$ be successive same-sign extrema of displacement relative to equilibrium, one full oscillation period apart. The logarithmic decrement isComparing extrema $N$ full cycles apart instead gives $\delta =N^{-1}\ln (|x_n|/|x_{n+N}|)$. Adjacent extrema of opposite sign are only half a cycle apart and give half this decrement. This distinction matters when inferring a damping ratio. Near critical damping, a full second same-sign extremum may be too small to resolve above measurement noise; a full-trace fit with uncertainty is then more informative than a peak ratio.

An ultrasonic motion sensor estimates distance from the return delay of an acoustic wave pulse reflected by a target. The teaching setup places a flat flag on the cart facing the detector above the air track, and specifies a minimum range of about $0.3\mathrm{m}$ for its detector. That apparatus-specific limit should not be assumed for every sensor model. ^[ref]

An air track supports a cart on a cushion of air to reduce contact friction. It provides a nearly one-dimensional platform for a damped harmonic oscillator; residual drag remains part of the baseline damping.

For positive locally constant $A,C$, supplement damped first-order resonant complex eccentricity by $\dot{a}/a=-2(p+1)Ce\sin \varphi -4A/5$. Balancing the two dissipative rates gives $e_f^2=2/[5(p+1)]$, $\sin {\varphi }_f=-A{e}_f/C$ and $B{x}_f=-(C/e_f)\cos {\varphi }_f$. Two fixed points exist when $A<C\sqrt{5(p+1)/2}$. The negative-cosine branch has $x_f>0$ and is stable in this leading model: a phase increase raises the positive resonant torque, raises semi-major axis, lowers mean motion, and reverses the phase drift. The positive-cosine branch has the opposite feedback.

For a coupled verification put $z=u+iv$. The characteristic polynomial at either fixed point is ${\lambda }^3+2A{\lambda }^2+[A^2+B^2{x}_f^2-2B(p+1)C{u}_f]\lambda -4AB(p+1)C{u}_f$. For $u_f<0$ its positive coefficients satisfy the cubic Routh-Hurwitz stability criterion, since the remaining inequality reduces to $2A(A^2+B^2{x}_f^2)>0$. For $u_f>0$ its negative constant term forces a positive real root. At the equality threshold the equilibria merge and an eigenvalue is zero. This is local stability for the frozen smooth-coefficient approximation, not a conclusion for arbitrary rapidly varying dissipation laws.

Half exponents mean the inverse of two in the odd-order root-of-unity group. For $p\nmid ab$, the ratio is a real global unit and is compatible under the norms in the p-power tower. Its Coleman power series is $(1+T{)}^{(b-a)/2}(1-(1+T{)}^a)/(1-(1+T{)}^b)$, with constant term $a/b$. Its higher logarithmic derivatives vanish at odd indices and equal $(a^k-b^k)B_k/k$ at even indices, using $B_1=-1/2$.

For a real cyclotomic field of odd prime-power conductor, the index of its cyclotomic-unit group has the same p-part as its ideal class number. The analytic class number formula compares the regulator of all units with the explicit cyclotomic-unit regulator. The equality here is stated at the p-part level, so harmless prime-to-p normalization factors do not affect it.

For $1\le k<p$, choose $l$ with $kl\equiv 1(\mathrm{mod}p)$. The geometric sum $1+{\zeta }_p^k+\cdots +{\zeta }_p^{k(l-1)}$ is the inverse of $u_k$. Hence these are units, and their residues cover ${\mathbb{F}}_p^{\times }$. In the cyclotomic Milnor square, this makes the $K_1$ map onto the common field surjective and eliminates its connecting obstruction to $K_0$.

Here $a\ge 0$ and $p\nmid m$. The prime-to-$p$ roots of unity generate an unramified extension of degree equal to the multiplicative order of $p$ modulo $m$; for $m=1$ this degree is one. Indeed, their distinct residues lift uniquely by Hensel lemma, and the Frobenius automorphism acts by raising them to $p$. The $p$-power roots of unity generate a totally ramified extension, because the translated cyclotomic polynomial ${\Phi }_{p^a}(1+T)$ is Eisenstein. The two fields have trivial intersection and their compositum has the same residue degree as the prime-to-$p$ field, by unramified base change preserves total ramification. Thus total ramification is equivalent to $p\equiv 1(\mathrm{mod}m)$.

Over a field of characteristic different from two and three, let $D=\{y^{2}z=x^3\}\subset {\mathbb{P}}^2$. Its homogeneous Hessian matrix has determinant $24x{y}^2$. Its common zeros with $D$ are the cusp $[0:0:1]$ and the smooth point $p=[0:1:0]$. The Hessian criterion for a flex therefore leaves exactly one smooth flex.

At $p$, the projective tangent line is $L=\{z=0\}$. Restricting the cubic equation to $L$ gives $x^3=0$, so its intersection multiplicity at $p$ is three. Consequently a triple projective line and a cuspidal cubic meeting only at a smooth point are projectively equivalent to $(3L,D)$ under these field hypotheses.

The characteristic restriction matters. In characteristic three, every smooth point can be written $[a:1:a^3]$ and has projective tangent line $z=a^{3}y$. Substitution gives $(x-ay{)}^3=0$, so every smooth point has triple contact. The unique-flex conclusion then fails.

The new subspace is the orthogonal complement of the old subspace of cusp forms for the Petersson inner product. A form belongs to it precisely when the adjoints of all lower-level oldform by argument dilation maps annihilate it.

The old subspace is spanned by $h(tz)$ for cusp forms $h$ of level $M<N$, where $M\mid N$ and $t\mid N/M$. It is stable under the Hecke operators, but need not have a basis of eigenvectors for every bad-prime operator. Its Petersson inner product orthogonal complement is the new subspace of cusp forms.