The commutator subgroup, or derived subgroup, isA linear character is afforded by a one-dimensional representation $\rho :G\to {\mathbb{C}}^{\times }$. Since ${\mathbb{C}}^{\times }$ is abelian, every commutator lies in $\ker \rho$, so

Let $\pi :G\to G/G^{\prime }$ be the quotient map. Every irreducible representation of a finite abelian group is one-dimensional, so each irreducible character $\psi$ of the abelianization $G/G^{\prime }$ inflates to the linear character $\psi \circ \pi$ of $G$. Conversely, the containment $G^{\prime }\le \ker \chi$ makes every linear character of $G$ factor uniquely through $G/G^{\prime }$. Thus inflation of a group representation gives a bijection

First consider a $K_{t,t}$-free bipartite graph with parts $X,Y$, each of size at most $n$, and $m$ edges. Count pairs $(S,y)$ in which $S$ is a $t$-element subset of the graph neighbourhood of $y\in Y$. Any $t$ vertices of $X$ have at most $t-1$ common neighbours, since $t$ common neighbours would form the complete bipartite graph $K_{t,t}$. ConsequentlyIf the average degree of a vertex is bounded in terms of $t$, then $m=O(n)$ and the required result is immediate. Otherwise, convexity of $x\mapsto \left(\genfrac{}{}{0ex}{}{x}{t}\right)$ and the preceding inequality giveand hence $m=O(n^{2-1/t})$. Every graph has a bipartite spanning subgraph containing at least half its edges, obtained by choosing a random bipartition. Applying the bipartite estimate to that subgraph proves the Kővári–Sós–Turán theorem in the required form:

The stated "nice" subsets of the finite cyclic group ${\mathbb{Z}}_n$ are Sidon sets. Ifthen $a+d=c+b$. Niceness forces $(a,b)=(c,d)$; the alternative pairing would force $a=b$. Thus the $|A|(|A|-1)$ ordered nonzero differences are distinct elements of the $n-1$ nonzero residue classes. Thereforeso

The Erdős-Stone theorem states that if the chromatic number of $H$ is $r+1\ge 2$, thenIf $H$ is a bipartite graph with at least one edge, its chromatic number is two, so $r=1$ and therefore

WriteLet $G$ be an $H$-free graph on $n$ vertices with $\mathrm{ex}(n,H)$ edges, and choose a uniformly random set of $m<n$ vertices. Its induced subgraph is still $H$-free, and each edge survives with probability $\left(\genfrac{}{}{0ex}{}{m}{2}\right)/\left(\genfrac{}{}{0ex}{}{n}{2}\right)$. Henceso $d_m(H)\ge d_n(H)$. Thus the normalized extremal number is nonincreasing and bounded below by zero, proving that

Write $r={\varphi }_0^2/M_{\mathrm{Pl}}^2$. As $t\to 0$,so $\varphi \to -\infty$ andBecause $2/3<r<2$, the power $r/2$ lies in $(1/3,1)$; thus $\ddot{a}<0$ and the early expansion decelerates. As $t\to \infty$, $\varphi \to 0$ and $\sinh (2\lambda t)\sim e^{2\lambda t}/2$, givingso the late universe has accelerating expansion.

The transition satisfies $\ddot{a}/a=H^2+\dot{H}=0$. Substitution givesor $r{\cosh }^2(2\lambda t)=2$. ThereforeThis is the transition described by the exact hyperbolic-sine scalar-field cosmology.

Put $u=\tanh (\lambda t)=e^{\varphi /{\varphi }_0}$. The hyperbolic cotangent double-angle identity givessoAlsoSolving the Friedmann equation for the scalar potential,therefore gives

Differentiating the ansatz givesThe Raychaudhuri equation therefore becomesSince $d[\coth (2\lambda t)]/dt=-2\lambda {\mathrm{csch}}^2(2\lambda t)$, the requested specific solution takes the integration constant to be zero:Now $\dot{a}/a=H$, soand

Differentiate the Friedmann equation:The scalar-field equation gives $\ddot{\varphi }+c^2{V}^{\prime }=-3H\dot{\varphi }$, henceBy continuity this also holds through isolated zeros of $H$, so the Raychaudhuri equation is

For real $z>0$, the exponential decays as $\mathrm{Re}t\to -\infty$, so two allowed contours run from $-\infty$ to $i$ and from $-\infty$ to $-i$. For real $z<0$, it decays as $\mathrm{Re}t\to +\infty$, so reverse the construction: run from $i$ or $-i$ to $+\infty$. These are the infinite contours in the diagram.

For $z>0$ their integrals arefor $z<0$ the corresponding orientation changes only an overall sign. Taking the sum of the two exponentials produces a solution independent of $\sin z/z$:Its nonzero Wronskian with $w_1$ on $z\ne 0$ proves linear independence.

Because $\mathrm{Re}a>0$, the endpoint factor $(1+t^2{)}^a$ vanishes at the branch points $t=\pm i$. A finite contour may therefore run from $-i$ to $i$ on either side of a chosen branch cut; the diagram shows contours bowing through the left and right half-planes.

For $a=1$, $f(t)=1$ is entire, so path independence makes either contour give, up to orientation,Hence a nonzero constant multiple gives the solution

Insertinto the equation. Since $w^{\prime }={\int }_{\gamma }t{e}^{zt}fdt$ and $w^{\prime \prime }={\int }_{\gamma }{t}^2{e}^{zt}fdt$, use $z{e}^{zt}={\partial }_t{e}^{zt}$ and integration by parts:The integral vanishes whenso, up to a constant and on a consistently chosen branch,The remaining contour condition isThis is the Laplace contour-integral method for a linear differential equation.

Fit a Poisson regression to artificial responses $Z_i=1$ with meansWith the logarithmic link, this is the generalized linear modelwhere $\log \Lambda (Y_i)$ is a known offset. Its log likelihood, apart from constants, isThe differencedoes not depend on $\beta$. The Poisson surrogate therefore has exactly the same maximum-likelihood estimate of $\beta$. This is the Poisson surrogate for an uncensored proportional-hazards likelihood.

Letbe the baseline cumulative hazard. Under the proportional hazards model,ThusIndependence makes the log likelihood

Writing $S_i(t)=1-F_i(t)$ for the survival function, the definition of the hazard function givesSince $S_i(0)=1$, integration yieldsThereforeThe exponent is the negative cumulative hazard.