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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 $\boxed{\mathrm{Aut}(K)=⟨\vartheta ⟩\cong C_n}$.

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 $G$.

In SI units Maxwell's equations areThe given vector identity yieldsDivide by ${\mu }_0$ and collect the two energy derivatives to obtain the local identity ${\partial }_t\mathcal{U}+\nabla \cdot P+E\cdot J=0$. Integration over the fixed region and the divergence theorem give the required integral conservation law. Its terms are, respectively, the rate of change of stored electromagnetic energy, the outward electromagnetic energy flux through the boundary, and work per unit time done by the field on charges. The last term transfers energy from the field to matter when positive.

In vacuum put ${\rho }_e=0$, $J=0$ and write $\varphi =kz-\omega t$. Faraday's law gives the plane-wave magnetic fieldBoth divergence equations are satisfied, and the remaining curl equation gives $k^2={\mu }_0{ϵ}_0{\omega }^2$. Hence $\boxed{\omega =ck,c=({\mu }_0{ϵ}_0{)}^{-1/2}}$, with $B_y=E_x/c$. Here the corresponding field is the pure wave, with no added static background.

Its electric and magnetic energy densities are equal. Explicitly,The energy flux points in the propagation direction, toward increasing $z$.

For $i\notin A$, condition on the first step. The Markov property makes the chance of eventual entry after reaching $j$ equal to $h_j$, including $h_j=1$ when $j\in A$. Total probability givesThe countable sum is justified by its nonnegative terms. This first-step identity alone need not uniquely determine a hitting probability on an infinite state space; one must choose the minimal nonnegative solution or supply the appropriate boundary behavior.

Unbiasedness as an estimator of the parameter means $DX\theta =\theta$ for every $\theta \in {\mathbb{R}}^p$, rather than just one particular value. Thus $(DX-I_p)\theta =0$ for every vector, forcing $\boxed{DX=I_p}$. Conversely that identity plainly suffices for unbiasedness.

Proceed column by column. At stage $j=1,\dots ,n-2$, consider the entries below the diagonal in column $j$, indexed $j+1,\dots ,n$. If this tail is nonzero, part (c) provides a Householder matrix sending it to a multiple of its first coordinate vector. Embed this matrix as an orthogonal matrix acting identically on coordinates $1,\dots ,j$, and apply the similarity transformation $A\mapsto Q_j^{T}A{Q}_j$. The tail's entries below row $j+1$ become zero. Symmetry is preserved, so the corresponding entries in row $j$ vanish too. If the tail is already zero, use the identity transformation.

Earlier reduced columns remain unchanged in their required zero positions: all their entries in the newly mixed tail coordinates were already zero, and the first $j$ coordinates are fixed. Thus each stage preserves the previous zeros while creating the next ones. After $n-2$ stages, all entries at distance greater than one from the diagonal vanish. With the product $Q=Q_1\cdots Q_{n-2}$ one has $Q^{T}Q=I$ andThis constructive reduction uses only reflections on successive trailing subspaces.

For $\alpha \ne 0$, Lagrange's theorem in the multiplicative group of order $q^m-1$ gives ${\alpha }^{q^m-1}=1$; at zero the desired identity is immediate. Thus all elements satisfy ${\alpha }^{q^m}-\alpha =0$.

For an irreducible degree-$m$ polynomial $f\in F[X]$, the quotient $L=F[X]/(f)$ is a field of dimension $m$, hence has $q^m$ elements by part 1. Its element $\alpha =[X]$ therefore satisfies ${\alpha }^{q^m}-\alpha =0$. The evaluation homomorphism $F[X]\to L$ has kernel $(f)$, so

Compose the second map first. Direct substitution givesThe result is affine because the product matrix is invertible. The identity is $(I,0)$ and the inverse is $(A^{-1},-A^{-1}b)$. Associativity follows from composition of maps, or by expanding this product three times. Distinct pairs give distinct maps, since evaluating at zero gives $b$ and evaluating the difference on basis vectors gives all columns of $A$. Thus these pairs faithfully describe a group.

Every pair factors as $(A,b)=(I,b)(A,0)$, and a linear map which is also a translation fixes zero and has linear part $I$, hence is the identity. Thus $G=NH$ and $N\cap H=\{1\}$. The conjugation formula in (c) says that the action needed for reconstruction is simply the usual matrix action $A:t\mapsto At$ on the additive group $F^n$. Therefore

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 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

A group presentation of $G$ consists of a set $S$ of symbols and a subset $R\subseteq F(S)$ of relators, together with a group isomorphismHere $F(S)$ is the free group on $S$, and $⟨⟨R⟩⟩$ is the normal closure of $R$. The images of $S$ form a generating set of a group; the relators prescribe which words become the identity element.

For the canonical presentation of a group, use one symbol $s_g$ for every group element $g\in G$. The universal property of a free group gives a surjective group homomorphismLet $R(G)=\ker \epsilon$, namely all words whose evaluation in $G$ is the identity element. ThenIndeed $R(G)$ is already a normal subgroup, so its normal closure is itself, and the first isomorphism theorem identifies the resulting quotient group with $G$. This is the course's canonical presentation of a group.

An equivalent, smaller collection of relators is $s_g{s}_h{s}_{gh}^{-1}$ for all $g,h\in G$. These encode the multiplication table. They imply $s_1=1$ and $s_g^{-1}=s_{g^{-1}}$, so every word reduces to the single symbol recording its product. This proves directly that they normally generate $R(G)$, although the canonical collection above consists of every word in that group homomorphism kernel.

The nonnegative cone determines the partial order by $a\le b⟺a^{-1}b\in G_{\ge 1}$. It contains the identity element, is closed under products and conjugation, and intersects its inverse only in the identity. Conversely these conditions define a two-sided invariant partial order. Its strictly positive part omits the identity; conventions for $G^{+}$ differ, so specify which is intended.

Suppose a binary operation on the underlying set $\mathbb{Z}$ makes it a group, and $\rho (n,x)=x^n$ is a group action on another group $G$. Every acting element must induce a bijection, with inverse function given by its acting inverse element. The integer $0$ instead induces the constant function $x\mapsto e$, where $e$ is the identity element of $G$, so $G$ must be the trivial group. Conversely, on the trivial group, ordinary addition on $\mathbb{Z}$ supplies such a group action.

A prefix scan computes successive accumulated results for an associative operation $\oplus$ while keeping input order fixed. A scan based on addition produces prefix sums; one based on matrix multiplication produces successive ordered matrix products. Associativity allows regrouping, while commutativity is unnecessary because the order stays fixed. If an initial empty prefix is included, its value is the identity element for the chosen operation.

The ability to compute accumulated results does not by itself allow subtraction of two prefixes. For example, two accumulated minima cannot reveal a block minimum: the lists $(1,7,8)$ and $(1,2,3)$ both have successive minima $(1,1,1)$, but the minimum of positions two through three is seven for one and two for the other. Prefix sums support block recovery because addition has additive inverses; an arbitrary associative operation need not have that property.

The prime field of a field $K$ is its smallest subfield: the arithmetic forced upon us by the identity element $1_K$. Every subfield must contain $1_K$, its repeated sums, their additive inverses, and the multiplicative inverses of nonzero results. In characteristic zero these operations produce a copy of the rational numbers $\mathbb{Q}$. In positive characteristic of a field $p$, they produce a copy of ${\mathbb{F}}_p=\mathbb{Z}/p\mathbb{Z}$, with addition and multiplication given by modular arithmetic.

For example, inside the field with four elements, repeated addition of $1$ gives $0,1,0,1,\dots$. Thus its prime field is $\{0,1\}$, and the other two elements cannot be reached by repeatedly adding $1$. This is why a finite field need not consist entirely of ordinary integer residues: a larger finite field extends the arithmetic of its prime field with additional elements.

The positive characteristic of a field must be a prime number. If its least vanishing multiple of $1_K$ were $rs$ with $1<r,s<rs$, then $(r{1}_K)(s{1}_K)=0$ would exhibit zero divisors. The absence of zero divisors rules this out. The prime field supplies the scalars when viewing $K$ as a vector space; its basis then explains the possible sizes of a finite field.

If $H$ is a subgroup and $K$ a normal subgroup of a group $G$, their product set $HK=\{hk:h\in H,k\in K\}$ is a subgroup. Normality gives $(h_1{k}_1)(h_2{k}_2)=h_1{h}_2(h_2^{-1}{k}_1{h}_2)k_2\in HK$ and $(hk{)}^{-1}=h^{-1}(h{k}^{-1}{h}^{-1})\in HK$. Together with the identity element this proves the subgroup test. For finite subgroups, $|HK|=|H||K|/|H\cap K|$: the pairs representing one product differ exactly by multiplying the first factor by an element of $H\cap K$ and its second factor by its inverse.

For a group $G$ and a normal subgroup $N$, the quotient group $G/N$ is the set of cosets $gN$, with operationIts identity element is the coset $N$, and the inverse element of $gN$ is $g^{-1}N$. A quotient group treats all elements within one coset as the same input while retaining a consistent operation on the classes. The normal subgroup condition is what makes this operation independent of which representatives are chosen. ^[ref].

Take the additive group $G=\mathbb{Z}/6\mathbb{Z}$ and $N=\{0,3\}$. Since $G$ is an abelian group, $N$ is a normal subgroup. There are three elements in the quotient group:Adding two copies of $2+N$ gives $4+N=1+N$; adding $1+N$ and $2+N$ gives $3+N=N$. Hence the quotient group has the addition table of $\mathbb{Z}/3\mathbb{Z}$. It does not select one representative and discard the others: its elements are the entire cosets.

This is useful whenever distinctions within a normal subgroup should be ignored. For example, the sign of a permutation retains only even or odd parity of a permutation in $S_3$. Its group kernel is the alternating group $A_3$, and the quotient group $S_3/A_3$ has two elements. The first isomorphism theorem explains why a group homomorphism's image is a quotient group of its domain.

A ring is a set $R$ with two binary operations, addition and multiplication, satisfying three requirements. Under addition, $R$ is an abelian group: there is a zero, every element has an inverse element, and grouping or exchanging summands does not change the result. Multiplication is associative and has an identity element $1$, so $(ab)c=a(bc)$ and $1a=a1=a$. Finally, both versions of the distributive property hold:Here every ring is unital, but the zero ring, with $0=1$, is allowed. Commutativity of multiplication is an additional condition defining a commutative ring. These conventions agree with ^[ref].

The integers form a ring with their usual operations. The polynomial ring $\mathbb{Z}[X]$ is another: addition combines coefficients, while multiplication uses the distributive property. For example,The same rules justify symbolic calculations in both rings; this is why one definition can organize apparently different kinds of arithmetic. In every ring, $a0=0a=0$, and $a(-b)=-(ab)$. The first identity follows by distributing $a(0+0)$ and subtracting $a0$; the second follows by distributing $a(b+(-b))$. Thus familiar sign rules are consequences of the structure.

Order still matters in some rings. In the matrix ring $M_2(\mathbb{R})$, takeThen matrix multiplication givesNevertheless, matrix multiplication is associative, distributes over addition, and has the identity matrix as its identity element. This is a valid ring. On a column vector, $AB$ applies $B$ first and $A$ second: swapping coordinates before discarding the second coordinate differs from doing those actions in reverse. The failure of commutativity records a meaningful difference in function composition.