For an odd prime $p$, the Legendre symbol is $\boxed{(a/p)=0}$ if $p\mid a$, $\boxed{(a/p)=1}$ if $a$ is a nonzero square modulo $p$, and $\boxed{(a/p)=-1}$ otherwise. It depends only on the residue class of $a$.

Choose one regular tetrahedron inscribed in the dodecahedron and let $H$ be its stabilizer among rotations of the dodecahedron. The twelve rotations of that tetrahedron preserve the dodecahedron; geometrically these are its identity, eight rotations by $\pm 2\pi /3$ about tetrahedral vertex axes and three half-turns about opposite tetrahedral edge axes. They act faithfully on its four vertices and give the twelve even permutations. Thus $\boxed{H\cong A_4,|H|=12}$ without using part (ii).

To check the core condition geometrically, the fifteen half-turns of the dodecahedron, about opposite-edge-midpoint axes, are conjugate under its rotations. A normal subgroup contained in $H$ cannot contain even one of these half-turns, because normality would then force all fifteen into a group of order twelve. On the other hand every nontrivial normal subgroup of $A_4$ contains a half-turn: if it contains a three-cycle, its conjugates and their inverses generate $A_4$; otherwise its nonidentity elements are among the three double transpositions, which are conjugate in $A_4$. Apply this observation to any normal subgroup of the full rotation group lying inside $H$. It must be trivial, proving exactly the required condition.

Act by left multiplication on the five cosets of $H$. The kernel is the core ${\bigcap }_{g\in G}gH{g}^{-1}$, a normal subgroup contained in $H$, so the hypothesis makes it trivial. Thus $G$ embeds in $S_5$, and its image has order $60$ and index two. An index-two subgroup is normal, so the quotient is of order two. Every three-cycle maps trivially to that quotient, since its order divides three. Three-cycles generate $A_5$, so the image contains $A_5$; their equal order gives $\boxed{G\cong A_5}$.

For completeness, prove simplicity rather than merely invoking it. The nonidentity even cycle types in $S_5$ are three-cycles, products of two disjoint transpositions and five-cycles. In $A_5$, a three-cycle has centralizer of order three, so its conjugacy class has size $20$. A double transposition has centralizer of order four in $A_5$, giving class size $15$. A five-cycle has centralizer its cyclic subgroup of order five, giving class size $12$; the $24$ such elements therefore split into two classes of size $12$. These account for all $60$ elements with the identity.

A normal subgroup is a union of whole conjugacy classes, containing the identity, and its order must divide $60$. No nonempty proper sum of the class sizes $15,20,12,12$, plus one, is a divisor of $60$ greater than one: the possible proper nontrivial sums are $13,16,21,25,28,33,36,40,45,48$. Only the full union has order $60$. Hence $\boxed{A_5\text{ is simple}}$, and so is $G$.

A left action is a map $(g,\omega )\mapsto g\omega$ satisfying $1\omega =\omega$ and $(gh)\omega =g(h\omega )$. The orbit of $\omega$ is $G\omega =\{g\omega :g\in G\}$. The action is transitive when there is exactly one orbit on the nonempty acted-on set, equivalently when any point can be moved to any other by a group element. An action is equivalently a homomorphism from $G$ to the permutation group of the set.

For nonzero $x\in K$, Lagrange's theorem in $K^{\times }$ gives $x^{p^n-1}=1$; zero satisfies $x^{p^n}=x$ too. Thus ${\vartheta }^n$ is the identity, and (iii) excludes every smaller positive order. Hence $\boxed{|⟨\vartheta ⟩|=n}$.

Any field automorphism of $K$ fixes $F$ and is determined by its value on $\lambda$. That value must be a root of the degree-$n$ minimal polynomial, so there are at most $n$ automorphisms. The $n$ powers of $\vartheta$ already attain this bound. Consequently $\boxed{\mathrm{Aut}(K)=⟨\vartheta ⟩\cong C_n}$.

Take $0<m<n$, the intended positive-exponent range. If ${\vartheta }^m(\lambda )=\lambda$, then ${\vartheta }^m$ fixes both $F$ and its field generator, so it fixes every element of $K=F(\lambda )$. Every element would then be a root of $X^{p^m}-X$. But that polynomial has degree $p^m<p^n=|K|$, impossible for a nonzero polynomial over a field. Thus $\boxed{{\vartheta }^m(\lambda )\ne \lambda (0<m<n)}$. The source writes only $m<n$; zero must be excluded, since ${\vartheta }^0(\lambda )=\lambda$.

Define $\vartheta (x)=x^p$ on all of $K$. The binomial coefficients between the endpoints are divisible by $p$, so $(x+y{)}^p=x^p+y^p$; also $(xy{)}^p=x^p{y}^p$ and $1^p=1$. Thus this is a field homomorphism. Its kernel is zero, and an injective self-map of the finite set $K$ is surjective. Hence $\boxed{\vartheta \text{ is an automorphism}}$, the Frobenius automorphism, with the required image of $\lambda$. It fixes $F$ by the preceding prime-subfield argument.

Since $K=F(\lambda )$ and the extension is finite, the element $\lambda$ is algebraic. Its minimal polynomial is irreducible, and evaluation gives an isomorphism $F[X]/(m_{\lambda })\cong F[\lambda ]=F(\lambda )=K$. If its degree is $d$, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis. Comparing with $[K:F]=n$ gives $\boxed{\mathrm{deg}{m}_{\lambda }=n}$.

The finite field $K$ is a finite-dimensional vector space over its prime subfield $F$ of size $p$. A basis has some positive length $n$, and every element is uniquely specified by $n$ coefficients in $F$. There are $p$ choices per coefficient, so $\boxed{|K|=p^n}$ and $[K:F]=n$.

The additive sequence $0,1,2\cdot 1,\dots$ must repeat because $K$ is finite, so the characteristic is positive. If its least positive value were composite, $r=ab$ with $1<a,b<r$, then $(a{1}_K)(b{1}_K)=r{1}_K=0$ while neither factor would be zero. A field has no zero divisors, giving a contradiction. Hence $\boxed{\mathrm{char}K=p\text{ is prime}}$.

A finite irreducible aperiodic Markov chain converges to its unique stationary distribution from every initial state. Finiteness holds because nonnegative entries with fixed finite row sums are bounded. The given connecting-sequence property makes the chain irreducible: each feasible step in such a sequence has positive proposal probability and is accepted. The all-zero index table occurs with probability $1/L>0$ and gives a self-loop at every state, establishing aperiodicity. Reversibility supplied the stationary uniform law in (b). ConsequentlyThe conclusion also holds when the feasible state space is a singleton.

Let $L$ be the number of index tables, including the zero table. If distinct feasible states $A,B$ differ by an index table, then $P(A,B)=1/L=P(B,A)$, using the sign-reversal bijection for the reverse move. If they do not differ by an index table, both probabilities are zero. Since $\pi$ is uniform,For equal states detailed balance is tautological. Thus the chain is reversible and its uniform distribution is stationary.

Assume the fixed margins are compatible, so the finite state space is nonempty. At state $X$, draw an index table $Y$ from the stipulated uniform proposal law. If $X+Y$ has any negative entry, reject and stay at $X$; otherwise accept and set the next state to $X+Y$. The margins stay fixed because the proposal has zero row and column sums.

The proposal set is invariant under $Y\mapsto -Y$. Hence proposals between two feasible distinct states are symmetric. The target density is constant, so the Metropolis–Hastings acceptance probability for a feasible proposal is one. The explicit rejection of infeasible proposals is essential: fixed margins by themselves do not guarantee nonnegative entries.

As established above, $Q_3^{\prime }\cong K_{4,4}$. Select any three vertices in each parity class; all nine edges between the two triples are present, giving a $K_{3,3}$ subgraph. A planar graph has only planar subgraphs, whereas $K_{3,3}$ is nonplanar by the permitted result. Therefore $\boxed{Q_3^{\prime }\text{ is nonplanar}}$.