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True. The group has order six. A Sylow $3$-subgroup $P$ has order three, while the given normal subgroup $N$ has order two. Their intersection is trivial and $NP=G$. Conjugation by $P$ on $N$ is trivial because $\mathrm{Aut}(C_2)$ is trivial, so $G=N\times P\cong C_2\times C_3\cong C_6$.

The Lagrange theorem states that $|H|$ divides $|G|$ for every subgroup $H$ of a finite group $G$, with $|G|=|G:H||H|$. The left cosets partition $G$, and multiplication by a representative bijects $H$ with each coset, proving the formula.

The intersection $H\cap K$ contains the identity and is closed under $x{y}^{-1}$, so it is a subgroup. Its order divides both $|H|$ and $|K|$; if these are coprime, $H\cap K=\{e\}$.

The order $o(x)$ is the least positive $m$ with $x^m=e$. Division $k=qm+r$, $0\le r<m$, shows that $x^k=e$ exactly when $r=0$, so exactly when $o(x)\mid k$.

If $x,y$ commute and have coprime orders $m,n$, then $(xy{)}^{mn}=e$. Conversely $(xy{)}^r=e$ implies $x^r=y^{-r}$ lies in $⟨x⟩\cap ⟨y⟩=\{e\}$, so $m,n\mid r$ and $mn\mid r$. Hence $o(xy)=mn$.

Cauchy theorem for groups says that every prime divisor $p$ of $|G|$ occurs as the order of an element of $G$. For $|G|=26$, the Sylow $13$-subgroup $P\cong C_{13}$ is unique and normal, and Cauchy's theorem supplies a complement $C_2$. Thus $G=C_{13}⋊C_2$. The homomorphism $C_2\to \mathrm{Aut}(C_{13})\cong C_{12}$ is either trivial or has the unique image of order two, inversion. These give exactly $C_{26}$ and the dihedral group of order $26$.

The subgroup $A_5^{\prime }$ is normal in $A_5$. It is nontrivial because $A_5$ is nonabelian; simplicity of $A_5$ therefore gives

Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.

The relation preserves the equality pattern among positions. The identity permutation gives reflexivity, inverses give symmetry, and compositions give transitivity.

For $W_3$, representatives are the restricted-growth wordswhen $n=1$, $n=2$, and $n\ge 3$, respectively. Here digits denote distinct symbols, and each displayed word represents one class.

For $W_4$ with $n=3$, the fourteen classes have representativesThe cyclic subgroup $F$ gives finer classes than all of $P_4$. For example, $x_1{x}_1{x}_2$ and $x_1{x}_1{x}_3$ have the same equality pattern and are equivalent under $P_4$, but no power of the four-cycle fixes $x_1$ while sending $x_2$ to $x_3$. Thus the two equivalence-class decompositions of $W_3$ differ.

Suppose the additive group of the integral domain $R$ is isomorphic to ${\mathbb{Z}}^n$. Every ideal $I$ is then an additive subgroup of a finitely generated free abelian group, so it has finitely many additive generators $x_1,\dots ,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb{Z}$ in $R$. Thus every ideal is finitely generated, proving that $R$ is Noetherian by noetherianity from finite additive rank.

For an example that fails (i)--(iv), takeIt is an integral domain and has additive group ${\mathbb{Z}}^2$. The idealhas index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is ${\mathbb{F}}_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have determinant and absolute indexwhich has no integer solution. Hence $I$ is not principal, as detailed in nonprincipal ideal in the integers adjoined a square root of minus five. Condition (i), and therefore all four equivalent conditions, fails.

Count pairs $(x,P)$ where $P$ is a Sylow $2$-subgroup fixing $x$. Each of the seven point stabilizers contains three such subgroups, giving 21 pairs. Conversely, a $2$-group acting on seven points has a fixed point, and a Sylow $2$-subgroup cannot fix two points because a two-point stabilizer has order four. Hence every Sylow $2$-subgroup occurs in exactly one pair, so

Similarly, each point stabilizer contains four Sylow $3$-subgroups, giving 28 pairs. A group of order three acting on seven points has a fixed point, and it cannot fix two because a two-point stabilizer is a $2$-group. Thus

The Sylow congruence and divisibility conditions give $n_7\in \{1,8\}$. Regard the faithful action as an embedding $G\le S_7$. If $n_7=1$, its Sylow $7$-subgroup $P$ is normal, so $G\le N_{S_7}(P)$. The stated fact gives $|N_{S_7}(P)|=7\cdot 6=42$, impossible for a subgroup $G$ of order 168. HenceThese are the Sylow counts in a faithful degree-seven action with S4 point stabilizers.

By transitivity and the orbit-stabilizer theorem,Inside $G_x\cong S_4$, there are three Sylow $2$-subgroups, each of order eight, and four Sylow $3$-subgroups, each of order three. These are also Sylow subgroups of $G$.

Let $N⊴G$ be proper. If $7$ divides $|N|$, then $N$ contains a Sylow $7$-subgroup of $G$, and normality makes it contain all eight of them. Thus $n_7(N)=8$, so the Sylow divisibility theorem givesSince $|N|$ divides 168 and $N$ is proper, this forces $|N|=56$.

Every nontrivial normal subgroup is transitive by normal subgroup orbits in a faithful prime-degree action. HenceBut $N\cap G_x$ is normal in $G_x\cong S_4$, whereas $S_4$ has three, rather than one, Sylow $2$-subgroups. This contradiction proves that no proper normal subgroup has order divisible by seven.

If instead $3$ divides $|N|$, normality makes $N$ contain all 28 Sylow $3$-subgroups of $G$. Hence $n_3(N)=28$, so $28$ divides $|N|/3$ and therefore $84$ divides $|N|$. A proper such subgroup would have order 84, which is divisible by seven and has just been ruled out.

Finally, if $N$ is any nontrivial normal subgroup, the prime-degree orbit argument makes $N$ transitive, so seven divides $|N|$. This is impossible for a proper normal subgroup. Therefore $G$ is simple.

Fordefine its action on the projective line $F\cup \{\infty \}$ by the Möbius transformationwith the usual conventions when the denominator vanishes or $x=\infty$. Matrix multiplication agrees with composition, so this is a group action. If $A$ fixes every projective point, then it fixes $0$ and $\infty$, forcing $b=c=0$, and fixing $1$ then gives $a=d$. Thus the kernel consists exactly of the nonzero scalar matrices $Z$. The induced group homomorphismis therefore injective. This is the projective general linear group action on the projective line.

For $F={\mathbb{F}}_4$,The scalar subgroup has order $|Z|=3$, soThe four classes represented byform a subgroup $P$ of order four, hence a Sylow subgroup for the prime two. Its action is translation $x\mapsto x+b$ on ${\mathbb{F}}_4$ and fixes $\infty$. For $b\ne 0$, characteristic two makes this permutation a product of two disjoint transpositions on the four finite points. It is therefore even, andThis realizes the Sylow 2-subgroup of PGL2 over F4.

The Lagrange root bound over a field says that a nonzero polynomial of degree $r$ over a field has at most $r$ roots. In particular, a polynomial modular congruence of degree $r$ modulo a prime number has at most $r$ incongruent solutions unless all its coefficients vanish modulo that prime.

Suppose that $d$ is good and that the positive integer divisor $e$ divides $d$. The $d$ roots of $X^d-1$ form a finite multiplicative subgroup $H$ of ${\mathbb{F}}_p^{\times }$. By the fact that every finite multiplicative subgroup of a field is cyclic, $H=⟨g⟩$ is a cyclic group of order $d$. The solutions in $H$ of $x^e=1$ areso there are at least $e$ of them. The Lagrange root bound over a field gives at most $e$ roots in all of ${\mathbb{F}}_p$, hence exactly $e$. Thus every divisor of a good number is good.

Now put $n=pq$. By the Chinese remainder theorem for unit groups, a base $b$ is a Fermat-pseudoprime base precisely when its two components satisfyThe power roots in a finite field andshow that each component has ten choices. Therefore there areFermat-pseudoprime bases.

To impose the strong pseudoprime condition, write $n-1=2^{s}m$ with $m$ odd. Since $10\mid n-1$, we have $5\mid m$. In each cyclic group of ten Fermat components, raising to the $m$th power sends five elements to $1$ and five elements to $-1$. A pair of components passes the strong test exactly when their signs agree: the pair $(1,1)$ satisfies the first alternative, and $(-1,-1)$ satisfies the second at $j=0$. Components of opposite sign become $(1,1)$ after squaring and can never jointly equal $-1$. Hence the number of strong-pseudoprime bases is

A Möbius transformation is generated by translations $z\mapsto z+b$, nonzero scalings $z\mapsto az$, and inversion $z\mapsto 1/z$. Translations and scalings plainly take circles and lines to circles and lines. More uniformly, every circle or line has an equationwhere $\alpha ,\gamma \in \mathbb{R}$; the case $\alpha =0$ is a line. Under inversion, put $z=1/w$ and multiply by $|w{|}^2$ to obtainwhich is again such an equation. The generators, and hence every Möbius transformation, therefore preserve Generalized circle under a Möbius transformation.

Now letIf $c\ne 0$, its finite pole $-d/c$ lies on some Euclidean circle, and that circle maps to a line because its image contains infinity. Thus $f$ takes every circle to a circle exactly when $c=0$, so the subgroup in question is the Affine subgroup of the Möbius group, consisting of $z\mapsto az+b$ with $a\ne 0$.

It is not normal. If $I(z)=1/z$ and $T_b(z)=z+b$ with $b\ne 0$, then $T_b$ is affine buthas a finite pole and is not affine. Hence conjugation by $I$ does not preserve the subgroup.