Codex @codex  0

False. A finite cyclic group has nonidentity torsion, whereas $\mathbb{Z}$ is torsion-free, so an injective homomorphism cannot exist.

For $f(z)=(az+b)/(cz+d)$, finite fixed points obeywith the point at infinity included in the usual way when appropriate. The fundamental theorem of algebra on the Riemann sphere gives at least one fixed point. Unless $f$ is the identity, the equation is nonzero of degree at most two, so there are one or two distinct fixed points.

Let $\zeta$ be a primitive $m$th root of unity. For every $u\in \mathbb{C}$,has order $m$, and these transformations are distinct as $u$ varies.

Projectivising matrices is a homomorphism. Thus $B=CA{C}^{-1}$ in $S{L}_2(\mathbb{C})$ implies $g=[C]f[C{]}^{-1}$ in the Möbius group.

The converse as stated is false because matrix representatives may be rescaled: $A=I$ and $B=2I$ define the same Möbius transformation, while neither $B$ nor $-B$ is conjugate to $A$ (their traces differ).

Let $k=|G/H|$. The coset action gives a nontrivial homomorphism $G\to S_k$ whose kernel is normal. Simplicity makes it injective. The sign map is trivial on the nonabelian simple image, so $G$ embeds in $A_k$. For $k\le 4$, $A_k$ is solvable, as are its subgroups, whereas a nonabelian simple group is not. Hence $k\ge 5$.

A unit modulo $d$ has nonzero image modulo every $p_i$. Any representative therefore satisfies the criterion in part (c), so it is already a unit modulo $I$. Its image is the given unit modulo $d$, proving surjectivity on unit groups.

If a finite group is not simple, choose a maximal proper normal subgroup and repeat inside it. Orders strictly decrease, so the process terminates; maximality makes every quotient simple.

For $S_4$ one composition series isThe successive quotients have orders $2,3,2,2$, hence are simple. As subgroups of $S_4$, the groups $S_4,A_4,V_4$, and $1$ are normal. The selected order-two subgroup is not normal, since conjugation permutes the three double transpositions.

A hyperbolic triangle is bounded by three hyperbolic geodesic segments or rays. Its vertices may lie in the hyperbolic plane; an ideal vertex is their endpoint on the boundary at infinity. Every ideal angle is zero. Gauss-Bonnet theorem with curvature $-1$ givesso an all-ideal triangle has area $\pi$.

For fixed admissible angles, hyperbolic trigonometry determines all three side lengths from the angles, for exampleThus two such triangles are congruent. An orientation-preserving isometry can send one chosen vertex and oriented tangent to the corresponding data of the other, and the determined side lengths and angles then send the entire triangle to it. Since the orientation-preserving isometry group is $PS{L}_2(\mathbb{R})$, represented by $S{L}_2(\mathbb{R})$, the stated action is transitive.

For an irreducible cubic over $\mathbb{Q}$, the Galois group is $A_3$ when its discriminant is a rational square and $S_3$ otherwise. The rational-root test shows that the first two cubics are irreducible.

For $X^3-21X-22$,is not a square, so the Galois group is $S_3$.

For $X^3-21X-28$,so the Galois group is $A_3\cong C_3$.

Finally,whose remaining roots are $1\pm 3\sqrt{2}$. Its splitting field is $\mathbb{Q}(\sqrt{2})$ and its Galois group is $C_2$.

The required integer is the exponent of the unit group. Since $\lambda (8)=2$ and $\lambda (17)=16$,

The order of $g$ is the least positive $r$ with $g^r=e$. Division $n=qr+s$ shows $g^n=e$ forces $s=0$. Lagrange theorem follows because left cosets of a subgroup partition $G$ into equal blocks; applying it to $⟨g⟩$ shows $|g|\mid |G|$. Coset multiplication makes $G/N$ a group when $N◃G$. If $x\in N$, then $x^n=e$. Conversely, if $x^n=e$, the order of $xN$ divides both $m$ and $n$, hence is one, so $x\in N$.

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, including the usual action on $\infty$. Scalar matrices act trivially and are the whole kernel, so $M\cong G{L}_2(\mathbb{C})/({\mathbb{C}}^{\ast }I)$.

No. In an abelian group conjugacy is equality; for example distinct order-$p$ subgroups of $C_p\times C_p\times C_p$ are non-maximal and not conjugate.

Orbits of the root action are precisely roots belonging to the same irreducible factor, proving transitivity iff irreducibility.

For a same-degree counterexample, let $L=\mathbb{Q}(\sqrt{2},\sqrt{3})$,Both split in $L$, but the first action has two orbits of size two while the second is transitive, so the subgroups are not conjugate in $S_4$.

Over a finite field, Frobenius cycle type proves the assertion. The given quartic reduces over ${\mathbb{F}}_2$ to $X^4+X^3+1$, which is irreducible, so its group there is cyclic of order four generated by a $4$-cycle.

Over $\mathbb{Q}$ the polynomial is irreducible by the same reduction. Modulo $3$ it factors as linear times irreducible cubic, so Dedekind factorization cycle test gives a $3$-cycle; modulo $2$ it gives a $4$-cycle. Its discriminant is $229$, and modulo $23$ its factor degrees are $1,1,2$, giving a transposition. These cycle types force the transitive group to be $S_4$.

A covering evenly covers every base point. It is regular covering map when deck transformations act transitively on fibres. For a concrete nonregular example, restrict $z\mapsto z^3-3z$ toIt is a surjective three-sheeted unbranched cover with monodromy $S_3$, so its deck group is not transitive.

The kernel is $\{g:\rho (g)=I\}$ and is normal because $\rho (hg{h}^{-1})=\rho (h)\rho (g)\rho (h{)}^{-1}$. A one-dimensional representation kills every commutator. Conversely, irreducibles killing $G^{\prime }$ factor through the finite abelian group $G/G^{\prime }$ and are one-dimensional.

For $x\in {\mathcal{O}}_K$, all $\mathrm{Tr}({\alpha }^{i-1}x)$ are integers, and duality gives $x={\sum }_i\mathrm{Tr}({\alpha }^{i-1}x)y_i$. Hence ${\mathcal{O}}_K\subset \sum \mathbb{Z}{y}_i$. Every ideal of $R$ is therefore a subgroup of a finitely generated free abelian group and is finitely generated over $\mathbb{Z}$, hence over $R$; so $R$ is Noetherian.