Use a normalization proof of Young's inequality. For arbitrary $x,y\ge 0$, putIf $A=0$, both coordinates are zero and the desired inequality is equality. If $A>0$, defineThen $X^p/p+Y^q/q=1$. Part (c) gives $XY\le 1$, and the conjugate exponent identity gives $XY=xy/A$. Therefore Young's inequality for products isThe unique equality case in part (c) says $X=Y=1$. Translated back, and including the zero case, this gives the sharp conditionConversely, if both powers equal $r\ge 0$, then $xy=r^{1/p+1/q}=r$ and the right side is $r(1/p+1/q)=r$, directly verifying equality.

The feasible set is closed and contained in the rectangleIt is therefore compact, and the continuous function $xy$ attains a global maximum there by the extreme value theorem. At the two boundary endpoints where a coordinate is zero, the product is zero. Meanwhile $(1,1)$ is feasible because the conjugate exponents satisfy $1/p+1/q=1$, and its product is one. Hence every maximizer has $x,y>0$.

On that positive region the constraint is $C^1$ with nonzero gradient. The Lagrange multiplier equations areThe first two identities imply $xy=\lambda x^p=\lambda y^q$. Since $x,y>0$, the multiplier is nonzero, so $x^p=y^q=r$. The constraint now givesThus the sole interior candidate is $(1,1)$. Existence of a maximum and the boundary comparison prove the unique global resultThis conjugate-power constrained product maximum is valid for all real $p,q>1$ with $1/p+1/q=1$; no integrality of the exponents is needed.

At a constrained critical point, the directional derivative of $f$ vanishes along every admissible tangent velocity:For a regular zero locus, $T_{a}M=\ker D{g}_a$. A linear functional vanishing on this kernel lies in the row space of $D{g}_a$. Equivalently there are Lagrange multipliers ${\lambda }_1,\dots ,{\lambda }_k$ such thatGeometrically, $\nabla f$ is orthogonal to the tangent space, so it lies in its orthogonal complement, spanned by the independent constraint gradients. Thus the level set of $f$ has no first-order change in a tangent direction. The equations provide candidates for constrained extrema; they must still be classified and compared, and boundary or singular cases must be treated separately. The condition is necessary for a regular constrained extremum, not a guarantee of a maximum.

A map $g$ is differentiable at $a$ when a linear map $D{g}_a:{\mathbb{R}}^n\to {\mathbb{R}}^k$ approximates its increment to first order:This Fréchet derivative is unique, and its matrix in the standard bases is the Jacobian matrix. Differentiability on a domain means differentiability at each point there.

A standard sufficient condition for a zero locus to be a submanifold is the regular level set theorem: if $g$ is $C^1$ on an open domain and $D{g}_a$ is surjective at every $a\in g^{-1}(0)$, thenFor smooth $g$, the submanifold is smooth. The full-rank condition is the relevant constraint qualification.

Under this regularity hypothesis, its tangent space isIndeed, if a differentiable curve $\gamma$ lies in $M$ and passes through $a$, the chain rule gives $D{g}_a({\gamma }^{\prime }(0))=0$. Thus all tangent velocities lie in the kernel of a linear map. The rank-nullity theorem makes this kernel $(n-k)$-dimensional, the same dimension as the tangent space, proving equality. In coordinates, solve the homogeneous linear equations given by the Jacobian matrix.

The regularity qualification matters: $g(x)=x^2$ has zero set $\{0\}$, a zero-dimensional submanifold, but $D{g}_0=0$. Its kernel is all of $\mathbb{R}$, whereas its tangent space is zero. This is the phenomenon that singular defining maps can overestimate tangent spaces.

The cross-ratio permutation orbit is generated by the two operations $\lambda \mapsto 1/\lambda$ and $\lambda \mapsto 1-\lambda$ from part (d). The full list isThese are all the possibilities: the four permutations generated by interchanging both pairs and by exchanging the two pairs leave the cross-ratio unchanged. They form the Klein four-group inside the symmetric group on four symbols, leaving at most $24/4=6$ different values; the displayed operations attain all six rational expressions.

There are fewer than six values exactly when a nonidentity operation in this six-element action fixes $\lambda$: if two operations give the same value, composing one with the inverse of the other gives such a fixed point. For real $\lambda \ne 0,1$, the fixed-point equations areThe other two equations reduce to ${\lambda }^2-\lambda +1=0$, which has no real root. HenceEach exceptional value has the same three-element orbit $\{-1,\frac{1}{2},2\}$. Every other permitted real value gives six different values.

For the geometric interpretation, put $P_0=[1:0]=\infty$ and $P_i=[x_i:1]$ for $i>0$. ThenThe three exceptional cases becomeThus the three finite points are collinear and equally spaced, with one their midpoint, even when the $x_i$ are viewed as points of the complex plane. Equivalently they form an arithmetic progression in some order. These are the harmonic conjugate points configurations with the fourth point equal to the point at infinity of the complex projective line.

By invariance under projective transformations and part (b), it is enough to calculate on the normalized quadruple $(\infty ,0,1,\lambda )$. Direct evaluation of the cross-ratio givesFor instance, $(0\infty :1\lambda )=(0-1)/(0-\lambda )=1/\lambda$, and $(\infty 1:0\lambda )=(1-\lambda )/(1-0)$. ThusAll denominators are legitimate because $\lambda \notin \{0,1\}$. The first identity explains why some distinct permutations always give the same cross-ratio.

Choose three distinct projective points $A,B,C$. Their images under the injective map $\tau$ are distinct. Sharp triple transitivity gives a projective transformation $S$ agreeing with $\tau$ on these three points. The map $\sigma =S^{-1}\circ \tau$ fixes $A,B,C$ and preserves their cross-ratios.

For any other projective point $P$, invariance givesLet $T$ normalize $(A,B,C)$ to $(\infty ,0,1)$. By part (b), the left side is $T(\sigma (P))$ and the right side is $T(P)$. Since $T$ is a bijection, $\sigma (P)=P$. It fixes the original three points too, so $\sigma$ is the identity map everywhere. ThereforeNo continuity or prior surjectivity assumption was needed. This is the rigidity assertion that cross-ratio preserving maps are projective.

The needed general-position theorem is sharp triple transitivity of the projective linear group on the projective line: for any two ordered triples of distinct projective points, exactly one projective transformation takes the first triple to the second.

In particular, let $T_P$ take $(P_0,P_1,P_2)$ to $(\infty ,0,1)$. In these coordinates,so cross-ratio invariance gives $T_P(P_3)=\lambda$, where $\lambda =(P_0{P}_1:P_2{P}_3)$. Distinctness implies $\lambda \ne 0,1$ and $\lambda$ is finite. The entire ordered quadruple is thus normalized toFor a second quadruple $(Q_0,Q_1,Q_2,Q_3)$ with the same cross-ratio, its normalizing map $T_Q$ yields the same quadruple. ConsequentlyThis proves the equivalence. Uniqueness on the first three points also shows that the projectivity carrying one ordered quadruple to the other is unique. The cross-ratio normalization of four projective points therefore gives a complete invariant for this ordered classification.

Use the cross-ratio convention compatible with the permutation identities in this paper. Represent a projective point $P_i\in {\mathbb{CP}}^1$ by a nonzero vector $p_i\in {\mathbb{C}}^2$, and write $[p_i,p_j]=\det (p_i,p_j)$. DefineThe denominator is nonzero because the four projective points are distinct. Rescaling any one representative multiplies numerator and denominator by the same nonzero scalar, so the definition is independent of the chosen homogeneous coordinates. In finite affine coordinates it readswith the point at infinity of the complex projective line handled by the homogeneous formula.

A projective transformation is represented by an invertible matrix $A\in {\mathrm{GL}}_2(\mathbb{C})$. Since $[A{p}_i,A{p}_j]=\det (A)[p_i,p_j]$, the two factors of $\det (A)$ in the numerator cancel the two in the denominator. Therefore every projective transformation preserves this cross-ratio. This also shows that the formula descends from matrices to the projective linear group, whose representatives differ by a nonzero common scalar.

The given unitriangular group of degree three over F3 has $27$ elements, just as $G$ does, so counting alone cannot distinguish them. Instead compare element orders.

Write an element as $I+U$, with $U$ strictly upper triangular of size $3$. Then $U^3=0$. Over the finite field ${\mathbb{F}}_3$, the binomial theorem for commuting matrices givesEvery nonidentity element of $H$ consequently has element order $3$. In contrast, $x\in G$ has element order $9$, by part (b). An isomorphism preserves element order, soEquivalently, the exponent of a finite group is $3$ for $H$ and $9$ for $G$.

The cyclic subgroup $⟨x^3⟩$ is normal: conjugation by $y$ sends $x^3$ to $x^{12}=x^3$, and conjugation by $x$ fixes it. Modulo this subgroup, the relation $yx{y}^{-1}=x^4$ becomes $yx{y}^{-1}=x$, so the quotient group is Abelian. Thus the commutator subgroup $G^{\prime }$ is contained in $⟨x^3⟩$.

Conversely the displayed relation gives the commutator $yx{y}^{-1}{x}^{-1}=x^3$, so $⟨x^3⟩\subseteq G^{\prime }$. HenceSince $x$ has element order $9$, $x^3\ne e$, so the first commutator subgroup is nontrivial. ThereforeHere derived length counts the number of steps needed for the derived series to reach the trivial group.

For the given group homomorphism $\phi :H\to \mathrm{Aut}(N)$, the external semidirect product has underlying Cartesian product $N\times H$ and operationThe notation $\mathrm{Aut}(N)$ denotes the automorphism group of $N$. Closure follows because $\phi (h)$ is an automorphism of $N$. To check associativity, calculate both bracketings:They agree because $\phi (hk)=\phi (h)\phi (k)$. The identity is $(e_N,e_H)$, using $\phi (e_H)={\mathrm{id}}_N$, and the two-sided inverse isIndeed multiplication in either order gives the identity, using $\phi (h)\phi (h^{-1})={\mathrm{id}}_N$. This verifies the group axioms.

For the specified group presentation, take $N=C_9=⟨X⟩$ and $H=C_3=⟨Y⟩$. The rule $\alpha (X)=X^4$ defines an automorphism because $4$ is coprime to $9$. Furthermore $4^3\equiv 1(\mathrm{mod}9)$, so $Y\mapsto \alpha$ defines the required group homomorphism $C_3\to \mathrm{Aut}(C_9)$. In the resulting semidirect product $S$, the canonical generators satisfyConsequently the group presentation defines a surjective group homomorphism $G\to S$, sending $x$ to $X$ and $y$ to $Y$. The underlying Cartesian product gives $|S|=27$, whereas part (a) gives $|G|\le 27$. Surjectivity forces equality, so the map is an isomorphism:This also proves that the $27$ normal forms in part (a) are distinct and that $x$ has element order exactly $9$.

The defining relation $yx=x^{4}y$ lets us move every occurrence of $y$ to the right of every occurrence of $x$. More generally, repeated conjugation givesNegative powers cause no problem because $x^{-1}=x^8$ and $y^{-1}=y^2$. Thus every word in the group presentation can be put in the required form, reducing powers using $x^9=y^3=e$:At this stage existence is proved; uniqueness will follow from the concrete semidirect product in part (b). Moving $x^r$ past $y^j$ gives the multiplication lawThe action is consistent with reduction of $j$ modulo $3$ because $4^3\equiv 1(\mathrm{mod}9)$.

First simplify the specified residue by modular exponentiation. Modulo the prime number $113$,Therefore $102^{70}\equiv 28\cdot 64\cdot 8\equiv 98$, so the required equation becomes $x^{37}\equiv 97(\mathrm{mod}113)$. Its right side is nonzero, so any solution is a unit modulo $113$.

By Fermat's little theorem, exponents of such units can be reduced modulo $112$. Sinceraising to the $109$th power is the inverse of raising to the $37$th power. There is therefore exactly one residue class of solutions:Now $97\equiv -16$ and $97^3\equiv -4096\equiv 85$. Since $4\cdot 85\equiv 1$, we obtainAs a direct check, $4^4\equiv 30$, $4^8\equiv 109$, $4^{16}\equiv 16$, and $4^{32}\equiv 30$, whence $4^{37}\equiv 30\cdot 30\cdot 4\equiv 97$.

In Pollard's p-1 algorithm, a large exponent containing many small factors makes $a^M-1$ divisible by a prime factor whose predecessor divides $M$, by Fermat's little theorem. A greatest common divisor with the number being factored can then separate that prime factor from the others.

Choose base $a=2$ and build factorial exponents using modular exponentiation. The successive exponents $2!$, $3!$, and $4!$ giveThe first two greatest common divisors are $\gcd (3,1003)=\gcd (63,1003)=1$. At the next step the Euclidean algorithm givesHenceHere the element order of $2$ modulo $17$ is $8$, which divides $24$, even though $17-1=16$ does not divide $24$. The algorithm only needs the element order to divide the chosen exponent; divisibility by the full predecessor is a sufficient condition, not a necessary one.