For representatives $p=(a,b,c)$ and $p^{\prime }=(a^{\prime },b^{\prime },c^{\prime })$, a point $[x:y:z]$ belongs to their projective line exactly when these three vectors are linearly dependent. The determinant criterion givesExpanding along the first row givesThe coefficient vector is the cross product $p\times p^{\prime }$. It is nonzero because the projective points are distinct. Rescaling either representative merely rescales all three coefficients, leaving the projective line unchanged. This is the join formula in projective joins and meets by cross products.

The real projective plane is the set of one-dimensional linear subspaces of ${\mathbb{R}}^3$. Equivalently,A projective point is written in homogeneous coordinates $[x:y:z]$. A projective line is the set of projective points whose representatives belong to one fixed two-dimensional linear subspace of ${\mathbb{R}}^3$. Such a linear subspace is the kernel of a linear map of a nonzero linear functional, so its equation has the form $\alpha x+\beta y+\gamma z=0$ with $(\alpha ,\beta ,\gamma )\ne 0$.

Two distinct projective points have linearly independent representatives $p,q$. Their linear span has dimension two and hence defines a projective line containing both. Conversely, any two-dimensional linear subspace containing both representatives contains their entire linear span and therefore equals it. Thus exactly one projective line passes through two distinct projective points.

Put $N=HK=⟨r⟩$, the normal subgroup of order $35$. By Cauchy's theorem for finite groups, $G$ contains an element $t$ of order two. Such an element cannot belong to the odd-order subgroup $N$. Thus $N\cap ⟨t⟩=\{1\}$ and $G=N⟨t⟩$, giving the semidirect productEvery automorphism of a finite cyclic group sends $r$ to $r^a$ for a unit $a$ modulo $35$. The relation $t^2=1$ forces the conjugation action of $t$ to square to the identity:The Chinese remainder theorem identifies a congruence class modulo $35$ with its pair of classes modulo $5$ and $7$. In each of these prime fields, the factorization $(a-1)(a+1)=0$ forces $a=1$ or $a=-1$. The choices are independent, giving exactly four homomorphisms from a finite cyclic group $C_2\to \mathrm{Aut}(C_{35})$. Explicitly,The factors $C_5$ and $C_7$ of $N$ are preserved by every automorphism, since each is its unique subgroup of that order. The four conjugation actions therefore give:

Here $D_n$ has group order $n$, as in the paper. The center of a group distinguishes these four groups of order seventy: its orders are $70,1,5,7$, respectively, so the four groups are pairwise nonisomorphic. Each construction is a semidirect product with $70$ elements, so every listed possibility occurs.

The Sylow theorems also give $n_5\mid 14$ and $n_5\equiv 1(\mathrm{mod}5)$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a normal subgroup of $G$ as well.

By Lagrange's theorem, $H\cap K=\{1\}$, since its group order divides both $5$ and $7$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $H$ by the normal subgroup property of $H$, and in $K$ by the normal subgroup property of $K$. It is therefore the identity element, so all elements of $H$ commute with all elements of $K$. This is the fact that normal subgroups of coprime order commute.

The subgroup $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have prime orders, they are cyclic groups. Write $H=⟨h⟩$ and $K=⟨k⟩$. The product of commuting elements of coprime order has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a normal subgroup because both its factors are normal subgroups; equivalently, it has subgroup index two. Therefore

The Sylow theorems give $n_7\mid 10$ and $n_7\equiv 1(\mathrm{mod}7)$. Among the divisors $1,2,5,10$, only $1$ satisfies the modular congruence. HenceUniqueness and the conjugacy assertion of the Sylow theorems make $K$ a normal subgroup.

Write the group order as $|G|=p^{a}m$, where $p$ is a prime number and $p\nmid m$. A Sylow p-subgroup is a subgroup of group order $p^a$, the largest power of $p$ dividing $|G|$.

The three Sylow theorems assert:

In particular, a unique Sylow p-subgroup is a normal subgroup, since all its conjugates must equal it.

Use the fact that $K$ is a normal subgroup to move a factor of $K$ past a factor of $H$. For $h_1,h_2\in H$ and $k_1,k_2\in K$,whileThe identity element belongs to $HK$. Hence $HK$ is a subgroup, by the subgroup test. This proves the product of a subgroup with a normal subgroup construction; $H$ need not be a normal subgroup.

For each $h\in H$, conjugation maps $K$ to itself and respects its multiplication. The inverse of $k\mapsto h^{-1}kh$ is $k\mapsto hk{h}^{-1}$, so both are group automorphisms of $K$.

The printed formula requires a composition convention. For the printed ${\varphi }_h(k)=h^{-1}kh$, ordinary function composition givesIndeed, $(h_1{h}_2{)}^{-1}k(h_1{h}_2)=h_2^{-1}(h_1^{-1}k{h}_1)h_2$. Thus the displayed assignment is a group antihomomorphism with the usual multiplication in the automorphism group. It is a group homomorphism if the automorphisms are written as right operators and multiplied in the corresponding reverse order; this is precisely why right conjugation reverses composition.

With the usual left function composition, the corresponding group homomorphism is insteadDirect expansion gives ${\psi }_{h_1{h}_2}(k)=h_1(h_{2}k{h}_2^{-1})h_1^{-1}=({\psi }_{h_1}\circ {\psi }_{h_2})(k)$. This states the intended conjugation action without silently reversing the order in the printed formula.

The signed rapidity corresponding to velocity $u$ isEquivalently $u/c=\tanh \varphi$, $\gamma =\cosh \varphi$, and $\gamma u/c=\sinh \varphi$. Thus the Lorentz boost from part (i) becomesMultiplying two such matrices gives diagonal entries $\cosh \varphi \cosh \psi +\sinh \varphi \sinh \psi =\cosh (\varphi +\psi )$ and off-diagonal entries $\cosh \varphi \sinh \psi +\sinh \varphi \cosh \psi =\sinh (\varphi +\psi )$, by the addition identities for hyperbolic functions. HenceThe resulting rapidity is $\varphi +\psi$. In velocity variables this is the relativistic velocity-addition formula $u\oplus v=(u+v)/(1+uv/c^2)$.

Now use $(ct,x{)}^T$ and ${\eta }_2=\mathrm{diag}(1,-1)$. WritePreservation of the Minkowski metric says that the first column has squared norm $1$, the second has squared norm $-1$, and the columns are orthogonal for the Minkowski inner product:Orthochronousness gives $a\ge 1$. From the orthogonality equation, $b=de/a$; substituting in the second norm identity gives $e^2=a^2$. Hence $(b,e)=\epsilon (d,a)$ with $\epsilon =\pm 1$. The determinant is then $\epsilon (a^2-d^2)=\epsilon$, so properness forces $\epsilon =1$. Thus $b=d$ and $e=a$.

Set $\beta =d/a$. The first norm identity implies $|\beta |<1$ and $a=(1-{\beta }^2{)}^{-1/2}$. With $u=c\beta$, this givesConversely, the displayed matrix has ${\gamma }^2-(\gamma u/c{)}^2=1$, equal off-diagonal entries, determinant one, and positive time-time entry. Direct multiplication therefore gives $L^T{\eta }_{2}L={\eta }_2$, proving both directions. The plus signs are the Lorentz boost convention used in the paper; reversing the boost direction replaces $u$ by $-u$.

For every matching in a graph $M^{\prime }$ and every vertex cover $C^{\prime }$, the disjoint edges of $M^{\prime }$ need distinct covering vertices. ThereforeNow assume the assertion in part (ii). It supplies a matching in a graph $M$ and a vertex cover $C$ formed by choosing one endpoint of each edge of $M$. Because these edges are disjoint, the chosen endpoints are distinct, so $|C|=|M|$. By the definitions of maximum and minimum,Every inequality must be equality, proving Kőnig's theorem for bipartite matching:

Choose a maximum matching $M$ and a minimum vertex cover $C$. By Kőnig's theorem for bipartite matching, $|M|=|C|$. Every edge of $M$ must have at least one endpoint in $C$, because $C$ is a vertex cover. Distinct edges of the matching in a graph have disjoint endpoints, so these requirements already use at least $|M|$ distinct vertices of $C$.

Since $C$ has exactly that many vertices, each edge of $M$ has exactly one endpoint in $C$, and no vertex of $C$ lies outside the endpoints of $M$. Thus selecting the unique endpoint in $C$ from each edge of $M$ produces precisely the vertex cover $C$:The argument also covers an edgeless graph: take both sets empty.

A matching in a graph is a set of edges no two of which share a vertex. A cover here means a vertex cover: a set $C$ of vertices such that every edge has at least one endpoint in $C$. It is not an edge cover.

Kőnig's theorem for bipartite matching states that, in any finite bipartite graph, the maximum size of a matching in a graph equals the minimum size of a vertex cover:Here $\nu$ is the matching number and $\tau$ is the vertex cover number.

Here each $T_i$ is a nonempty connected subgraph of a tree, rather than a descendant-closed data-structure subtree. We prove the Helly property for connected subgraphs of a tree by mathematical induction on the number of vertices of $T$.

For a one-vertex tree, every nonempty $T_i$ contains that vertex. For a larger tree, choose a leaf of a tree $v$ using part (i), and denote its unique neighbour by $w$. If some $T_i$ consists only of $v$, the assumed pairwise intersections force every $T_j$ to contain $v$, and we are done.

Otherwise, remove $v$ from every $T_i$ that contains it, leaving all other $T_i$ unchanged. Any affected $T_i$ has at least two vertices; connectedness forces it to contain $w$ and the edge $vw$. Thus $v$ is a leaf of a tree of $T_i$, and part (ii) shows that its deletion leaves a nonempty tree. All reduced $T_i$ are therefore nonempty connected subgraphs of a tree $T\setminus v$.

They still intersect pairwise. An original common vertex other than $v$ survives. If a pair had only $v$ in common, both members would also contain $w$, by the previous paragraph, contradicting that it was their only common vertex. The induction hypothesis now supplies a common vertex of the reduced family, which is also common to the original family. Consequently

Deleting a vertex and its incident edge cannot create a graph cycle, so $T\setminus v$ is acyclic. It remains to establish connectedness and nonemptiness.

Since the leaf of a tree $v$ has one neighbour, $T$ has at least two vertices, so $T\setminus v$ is nonempty. For any two remaining vertices $a,b$, take their unique simple graph path in $T$. It cannot pass through $v$: an internal vertex on such a graph path has two incident edges, whereas $v$ has vertex degree one. Neither endpoint is $v$, so the entire graph path remains after deletion. The remaining graph is connected and acyclic, givingThis includes the two-vertex case, where deletion leaves the one-vertex tree.

A tree is a nonempty connected graph with no graph cycle. A leaf of a tree is a vertex of vertex degree one. As usual in this finite graph theory problem, trees are finite; the assertion would fail for a doubly infinite graph path.

Take a longest path $v_0,v_1,\dots ,v_m$ in a tree with at least two vertices. Connectedness gives $m\ge 1$. If $v_0$ had a neighbour outside this graph path, the path could be extended, contradicting maximal length. If it had a neighbour $v_j$ with $j\ge 2$, that edge together with the segment $v_0,v_1,\dots ,v_j$ would make a graph cycle. Thus its only neighbour is $v_1$, and it is a leaf of a tree. The same argument at the other end makes $v_m$ a second, distinct leaf of a tree. Hence

An impulse at $a$ shifts the causal sine response. Substituting the Dirac delta in the convolution from part (i), or using the Laplace transform time-shift rule, giveswhere $H$ is the Heaviside step function. The answer is continuous at $a$, while its first derivative jumps from $0$ to $1$. Consequently its second distributional derivative contains exactly one unit Dirac delta at $a$, giving $f^{\prime \prime }+f=\delta (x-a)$. Since $a>0$, both initial conditions vanish.

Let $K(p)=\mathcal{L}[k](p)$ and $F(p)=\mathcal{L}[f](p)$. Apply the Laplace transform of a derivative twice and use the two zero initial conditions. The differential equation becomesThe factor $1/(p^2+1)$ is the Laplace transform of $\sin x$. The convolution theorem for Laplace transforms therefore givesThis is a causal convolution: the value at $x$ uses only the forcing up to $x$. For ordinary continuous $k$, direct differentiation gives $f^{\prime }(x)={\int }_0^x\cos (x-t)k(t)dt$ and $f^{\prime \prime }(x)=k(x)-f(x)$, verifying both the differential equation and the initial conditions. For an impulsive $k$, the same formula is interpreted using distributions.

The defining evaluation rule of the Dirac delta givesThe impulse lies strictly inside the integration domain, so there is no endpoint convention to settle. This Laplace transform exists for every complex $p$.

With the same unilateral Laplace transform convention, the standard sine formula is

Use the unilateral Laplace transform convention $\overline{f}(p)={\int }_0^{\infty }{e}^{-px}f(x)dx$. The Laplace transform of a derivative isThe formula holds in the common Laplace convergence region under the usual endpoint and growth hypotheses. The initial value is the right-hand value at zero.