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First, the orbit-stabilizer theorem applied to a face gives $|G|=20\cdot 3=60$: its stabilizer subgroup consists of the identity and the two rotations through $120^{\circ }$ about the axis through the centres of that face and its opposite. A rotation preserving a face must preserve its outward normal and cyclically permute its three vertices, so there are no further stabilizing rotations.

The conjugacy classes are as follows:

There are $30/2=15$ edge axes, $20/2=10$ face axes, and $12/2=6$ vertex axes. They give respectively $15$, $2\cdot 10=20$, and $4\cdot 6=24$ distinct nonidentity rotations, already accounting for all $59$ nonidentity elements.

To justify the class assertions, conjugation by a rotation carries the rotation axis to its image and preserves the rotation angle about the transported oriented axis. The transitive group action on edges puts all half-turns in one conjugacy class. The transitive group action on faces puts the $120^{\circ }$ rotations defined using outward face normals in one conjugacy class; opposite faces encode opposite signs about the same unoriented axis. The same reasoning with outward vertex directions gives a class of twelve $72^{\circ }$ rotations and a class of twelve $144^{\circ }$ rotations, including their inverses. These last classes cannot merge: the matrix trace of a three-dimensional rotation is $1+2\cos \theta$, which differs for $\theta =2\pi /5$ and $4\pi /5$. Thus the list is complete.

For a $2\times 2$ matrix, the characteristic polynomial is $t^2-\mathrm{tr}(A)t+\det A$. The given matrix trace and determinant therefore make it $t^2+1$. By the Cayley-Hamilton theorem, $A^2=-I$, and hence $A^4=I$. Since $A^2\ne I$, the order of a group element $A$ cannot be $1$ or $2$. If $A^3=I$, then $A=A^4(A^3{)}^{-1}=I$, also impossible. Thus the order is $4$.

The Cayley-Hamilton theorem states that every square complex matrix satisfies its own characteristic polynomial: if ${\chi }_A(t)=\det (tI-A)=t^n+c_{n-1}{t}^{n-1}+\cdots +c_0$, thenFor a proof, expand the adjugate matrix as $\mathrm{adj}(tI-A)={\sum }_{k=0}^{n-1}{B}_k{t}^k$. The adjugate identity givesComparing coefficients yields $-A{B}_0=c_{0}I$, $B_{k-1}-A{B}_k=c_{k}I$ for $1\le k\le n-1$, and $B_{n-1}=I$. Multiply the middle identities on the left by $A^k$, include the first identity, and add $A^n{B}_{n-1}=A^n$. The terms involving the $B_k$ cancel consecutively:Their right sides sum to $c_{0}I+{\sum }_{k=1}^{n-1}{c}_k{A}^k+A^n$. This proves the theorem without any substitution of noncommuting matrices into a scalar polynomial identity.

Every nonempty open set in the given topology contains $1$. Suppose a continuous map $f$ took distinct values $f(m)$ and $f(n)$. Choose disjoint open intervals $U,V\subset \mathbb{R}$ containing these respective values. By continuity, $f^{-1}(U)$ and $f^{-1}(V)$ are nonempty open sets in $\mathbb{N}$, and therefore both contain $1$. This is impossible because the inverse images of disjoint sets are disjoint. Thus every such continuous map is constant. Conversely, constant maps are continuous. This is the continuous real maps from the initial-segment topology argument, using the Hausdorff separation of distinct real points.

Every point has arbitrarily small path-connected open neighbourhoods. Indeed, in a coordinate chart homeomorphic to ${\mathbb{R}}^n$, take the inverse image of a sufficiently small open Euclidean ball. The ball is convex, hence path-connected, and its inverse image is path-connected by the homeomorphism. If the supplied neighbourhood is not itself open, first take an open neighbourhood within it and a ball contained in its chart image. This proves that $X$ is a locally path-connected space.

Fix $x_0\in X$ and let $A$ consist of the points joined to $x_0$ by a path. For $x\in A$, a path-connected open neighbourhood $U_x$ is contained in $A$: concatenate a path from $x_0$ to $x$ with one from $x$ to any point of $U_x$. Hence $A$ is open. For $x\notin A$, the same kind of neighbourhood cannot meet $A$, since a meeting point would provide a path to $x$. Hence $X\setminus A$ is also open. Since $X$ is a connected space and $A$ is nonempty, $A=X$. This proves $X$ is path-connected, including the general principle that connected locally path-connected spaces are path-connected. The empty-space case, if allowed, is vacuous.

The derivative is $f^{\prime }(x)=-4x^3{e}^{-x^4}$. Its absolute value is continuous and tends to zero as $x\to \pm \infty$, so it is bounded. More explicitly,obtained by differentiating $4t^3{e}^{-t^4}$ for $t\ge 0$. The mean value theorem gives $|f(x)-f(y)|\le M|x-y|$ for every $x,y$. Thus $f$ is Lipschitz continuous, and choosing $\delta =\epsilon /M$ proves $e^{-x^4}$ is uniformly continuous on $\mathbb{R}$.

Take $x_n=2\pi n$ and $y_n=2\pi n+1/n$. Then $|y_n-x_n|=1/n\to 0$, whileThe sequential criterion for uniform continuity would force $f(y_n)-f(x_n)\to 0$. These pairs contradict it: for example, for all sufficiently large $n$ their output separation is greater than $\pi$, despite arbitrarily small input separation. Thus $x\sin x$ is not uniformly continuous on $\mathbb{R}$.

The symmetric group $S_5$ has cardinality $5!=120=3\cdot 40$, so its Sylow subgroups for $p=3$ have cardinality $3$. One example isEvery nonidentity permutation of order of a group element $3$ in $S_5$ is a single permutation cycle of length $3$: two disjoint length-three permutation cycles would require six letters. Choose the three moved letters in $\left(\genfrac{}{}{0ex}{}{5}{3}\right)$ ways, then one of their two length-three permutation cycles, giving $20$ such elements. Each subgroup of cardinality $3$ has exactly two nonidentity elements, and distinct such subgroups have no nonidentity element in common. ConsequentlyThis also satisfies the numerical Sylow theorem restrictions $10\mid 40$ and $10\equiv 1(\mathrm{mod}3)$.

Write the cardinality of the finite group as $|G|=p^{a}m$, where $p$ is prime and $p\nmid m$. The Sylow theorems say that a Sylow subgroup of order $p^a$ exists; every p-group is contained in a Sylow subgroup; and all Sylow subgroups are conjugate subgroups. If $n_p$ is their number, thenEquivalently, for a Sylow subgroup $P$, conjugation and the orbit-stabilizer theorem give $n_p=[G:N_G(P)]$, where $N_G(P)$ is the normalizer. The containment statement is often included in the conjugacy form: every p-group lies in a conjugate of any fixed Sylow subgroup.

Regard $B$ as a linear map ${\mathbb{R}}^n\to {\mathbb{R}}^p$ and $A$ as a linear map ${\mathbb{R}}^p\to {\mathbb{R}}^m$ (the same argument holds over any field). ThenIts dimension cannot exceed either $\dim \mathrm{im}A$ or $\dim \mathrm{im}B$, since applying a linear map cannot increase dimension. The rank bound for a matrix product is thereforeThe bound $p$ is attained whenTheir product has an $I_p$ block and zeros elsewhere, so its rank of a matrix is $p$. Thus the bound depending only on the given matrix sizes is sharp.

The row rank of a matrix is the dimension of the linear span of its row vectors. Its column rank is the dimension of the linear span of its column vectors, equivalently the dimension of the image of the corresponding linear map. The equality of row rank and column rank states that these two numbers are equal for every matrix. Their common value is the rank of a matrix.

False. The real line with its usual metric is complete and is homeomorphic to $(0,1)$ with its usual metric, throughIts inverse is $h^{-1}(u)=\tan (\pi (u-1/2))$, so both maps are continuous. However, $(1/n{)}_{n\ge 2}$ is a Cauchy sequence in $(0,1)$ and has no limit there: its only possible real limit is zero. Hence $(0,1)$ is not a complete metric space. Completeness is a property of the metric, not just of its topology.

False. Give $X=\mathbb{R}$ its usual topology, let $Y$ be a one-point space, and take the constant continuous map $f:X\to Y$. The set $K=Y$ is compact, but $f^{-1}(K)=\mathbb{R}$ is not compact. For example, the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. Thus continuity alone does not preserve compactness under inverse images.

True. Let $\{V_{\alpha }\}$ be an open cover of $f(S)$ in its subspace topology. Write $V_{\alpha }=f(S)\cap W_{\alpha }$ with $W_{\alpha }$ open in $Y$. Since $f$ is continuous, the sets $S\cap f^{-1}(W_{\alpha })$ are open in $S$ and cover $S$. By compactness, finitely many cover $S$. Their corresponding $V_{\alpha }$ then cover $f(S)$, proving that the continuous image of a compact space is compact.

Define the Möbius transformationThe three prescribed distinct points go to $0,1,\infty$, respectively, and the specified cross-ratio is exactly $T(z_4)$.

To identify the preimage of the extended real line directly, write $T(z)=(az+b)/(cz+d)$. The condition $T(z)=\overline{T(z)}$ becomesAfter division by $2i$, this has the real form $A|z{|}^2+Bz+\overline{B}\overline{z}+C=0$, with $A,C$ real. If $A=0$ it is a line; otherwise completing the square gives a circle. It contains $z_1,z_2,z_3$, including the pole as a point mapped to infinity, so it is nondegenerate. Three distinct noncollinear points determine one circle; if the three are collinear, their unique generalized circle is their line.

Consequently $T(z_4)$ is real exactly when the fourth point is on the same generalized circle as the first three. This proves the real cross-ratio criterion for a generalized circle:

For the spherical circle, use a rotation to put the center at the north pole. Points at spherical distance $r$ have polar angle $r$ and admit the parametrizationIts speed is $|{\gamma }^{\prime }(\theta )|=\sin r$, so the arc length isFor small $r$, the Taylor series gives $C(r)=2\pi r-(\pi /3)r^3+O(r^5)$. Thus its circumference is smaller than the Euclidean circumference $2\pi r$ for the same intrinsic radius. Their ratio is $\sin r/r=1-r^2/6+O(r^4)$ and tends to one. The leading agreement expresses local Euclidean behavior; the cubic deficit reflects the positive Gaussian curvature of the unit sphere.

Hilbert's basis theorem: if $R$ is a commutative Noetherian ring with identity, then the polynomial ring $R[x]$ is Noetherian. Iteration gives the same assertion for every finite number of variables.

Here is a bounded-degree coefficient proof of Hilbert basis theorem. Let $I$ be any ideal of $R[x]$. For $n\ge 0$, define $J_n\subseteq R$ to be the coefficients of $x^n$ in the polynomials of $I$ of degree at most $n$, including the zero coefficient. Closure under addition and scalar multiplication makes $J_n$ an ideal of $R$. Multiplication of a polynomial by $x$ shows $J_n\subseteq J_{n+1}$. By the ascending chain condition, there is $N$ such that $J_N=J_{N+1}=\cdots$.

Each $J_n$ for $0\le n\le N$ is finitely generated. Choose generators $a_{n,1},\dots ,a_{n,r_n}$ and corresponding polynomials $f_{n,j}\in I$ of degree at most $n$ whose coefficient of $x^n$ is $a_{n,j}$. There are only finitely many chosen polynomials. We prove they generate $I$ by induction on degree.

Take a nonzero $f\in I$ of degree $d$ and leading coefficient $a$. If $d\le N$, express $a={\sum }_j{c}_j{a}_{d,j}$ and subtract ${\sum }_j{c}_j{f}_{d,j}$. If $d>N$, use $a\in J_d=J_N$, express $a={\sum }_j{c}_j{a}_{N,j}$, and subtract ${\sum }_j{c}_j{x}^{d-N}{f}_{N,j}$. In both cases the leading coefficient cancels, leaving a polynomial in $I$ of smaller degree. For degree zero the remainder is zero, and the induction completes the claim.

Thus every ideal of $R[x]$ is finitely generated, proving the Hilbert basis theorem. The proof uses no cancellation of nonzero elements of $R$, so it remains valid when $R$ has zero divisors.

A field, for example $\mathbb{Q}$, is a Noetherian ring: its only ideals are $(0)$ and the whole field, both finitely generated.

For a non-Noetherian example, take $R=\mathbb{Q}[x_1,x_2,\dots ]$, the polynomial ring in countably many variables, each individual polynomial using only finitely many variables. The ascending chain of idealsis strictly increasing. Indeed, if $x_{n+1}$ belonged to $(x_1,\dots ,x_n)$, setting the first $n$ variables to zero would give the impossible equality $x_{n+1}=0$. The chain never stabilizes, contradicting the ascending chain condition. Thus $\mathbb{Q}$ is Noetherian, while $\mathbb{Q}[x_1,x_2,\dots ]$ is not.

For $d$ variables, integrate over the product of $d$ unit circles:For multi-indices $\alpha ,\beta \in {\mathbb{Z}}_{\ge 0}^d$, applying one-variable Fourier orthogonality in each variable givesIf $f={\sum }_{\alpha }{a}_{\alpha }{z}^{\alpha }$, its squared norm is ${\sum }_{\alpha }|a_{\alpha }{|}^2$, a finite sum. The form is therefore a positive definite Hermitian form andThis is the monomial orthonormal basis on the unit torus.

For nonnegative integers $m,n$, Fourier orthogonality givesEvery polynomial is a finite linear combination of these monomials, and distinct monomials are linearly independent. ThereforeThis is an algebraic basis: no infinite linear combinations are needed in the polynomial vector space.