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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.

Let $H={\mathrm{Stab}}_F(x)$. The identity belongs to $H$ because $F(e,x)=x$. If $g,h\in H$, the group action law givesso $gh\in H$. Finally, if $g\in H$, thenso $g^{-1}\in H$. The subgroup criterion therefore proves that ${\mathrm{Stab}}_F(x)$ is the stabilizer subgroup of $x$.

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 $x\in N$.

An action is a map $G\times X\to X$ satisfying $ex=x$ and $(gh)x=g(hx)$; it induces $G\to S_X$. Acting on the $n$ left cosets of $H$ gives a homomorphism with kernel $K◃G$ contained in $H$. If $K$ were trivial, $|G|$ would divide $n!$, so under the stated hypothesis $K$ is nontrivial. For $[G:L]=p$ with $p$ the least prime divisor of $|G|$, the transitive image in $S_p$ has order divisible by $p$; its other possible prime divisors are smaller, hence absent, so its order is $p$. Thus the kernel has index $p$ and, being contained in $L$, equals $L$. Prime index alone is insufficient: a subgroup generated by a transposition in $S_3$ has index $3$ and is not normal.

Sylow I gives a subgroup of order equal to the full $p$-part of $|G|$. Sylow II says every $p$-subgroup lies in a conjugate of any Sylow $p$-subgroup; act by the subgroup on the Sylow cosets and use a fixed orbit because the number of cosets is prime to $p$.

Let an order-$p^{d-1}$ subgroup $H$ act on $G/H$. Since the number of cosets is divisible by $p$, orbit counting and Cauchy produce a fixed coset with stabiliser extending $H$ by index $p$. Induction yields subgroups of every $p$-power order. Applying the same orbit count to the action of $H$ on $G/H$ leaves fixed cosets indexed by $N(H)/H$, giving $|G:H|\equiv |N(H):H|(\mathrm{mod}p)$.

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 primitive root modulo $N$ is a unit of order $\phi (N)$; a nonzero quadratic residue modulo an odd prime $p$ is a square.

If a primitive root $g$ modulo $p$ were $x^2$, then its order would divide $(p-1)/2$, the order of the subgroup of squares, contradicting ${\mathrm{ord}}_p(g)=p-1$.

Modulo $11$, $2$ has order $10$, so the primitive roots are $2^k$ for $\gcd (k,10)=1$:Finally $2^{10}=1024\equiv 56\not\equiv 1(\mathrm{mod}121)$, so the lifting criterion makes $2$ primitive modulo $121$. It is plainly the smallest possible positive primitive root.

Identify $G\simeq (\mathbb{Z}/15\mathbb{Z}{)}^{\times }\simeq C_4\times C_2$, with $\sigma (\zeta )={\zeta }^2$ and $\tau (\zeta )={\zeta }^{-1}$. Its subgroups areTheir fixed fields, in the same layers, areThe real fields are $\mathbb{Q}$, $\mathbb{Q}(\sqrt{5})$, and $L^{+}$. This is the Galois correspondence.

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.

Homotopy equivalence means maps $f:X\to Y$, $g:Y\to X$ with $gf\simeq {\mathrm{id}}_X$ and $fg\simeq {\mathrm{id}}_Y$. Radial retraction of ${\mathbb{C}}^{\ast }$ and contraction of $\mathbb{C}$ give ${\mathbb{C}}^{\ast }\times \mathbb{C}\simeq S^1$. Coordinates $(a,b)\mapsto (a-b,b)$ identify $X\simeq {\mathbb{C}}^{\ast }\times \mathbb{C}$, so ${\pi }_1(X)=\mathbb{Z}$. The map to unordered roots is a two-sheeted covering. Its induced subgroup is $2\mathbb{Z}$ inside ${\pi }_1(Y)$, while exchanging roots is a generator; hence ${\pi }_1(Y,p_0)\simeq \mathbb{Z}$.

The square gives one vertex, edges $a,b$, and one 2-cell attached by $ab{a}^{-1}b$, hence ${\pi }_1(K)=⟨a,b\mid ab{a}^{-1}=b^{-1}⟩$. Based connected covers correspond contravariantly to subgroups of ${\pi }_1$. The subgroup $⟨a^2,b⟩\simeq {\mathbb{Z}}^2$ gives the two-sheeted torus cover; $⟨b⟩\simeq \mathbb{Z}$ gives an infinite cylindrical cover. If the total space is compact, fibres over a point are compact discrete and hence finite, so the subgroup has finite index. Conversely a finite-index cover has finitely many compact lifted cells and is compact.

The degree is the sum of divisor coefficients; principal divisors are $(f)$ for nonzero rational functions. $\mathrm{Cl}(C)=\mathrm{Div}(C)/\mathrm{Prin}(C)$ and ${\mathrm{Cl}}^0$ is its degree-zero subgroup. On ${\mathbb{P}}^1$, $K\sim -2P$; on a hyperelliptic genus-$g$ curve with degree-two fibre $H$, $K\sim (g-1)H$. If ${\mathrm{Cl}}^0=0$, any point divisor has $\ell (P)=2$ by Riemann-Roch, giving a degree-one map to ${\mathbb{P}}^1$; the converse follows because every degree-zero divisor on ${\mathbb{P}}^1$ is principal. For $g\ge 1$, $[p-p_0]=[q-p_0]$ would give a function with divisor $p-q$, hence a degree-one map unless $p=q$, proving injectivity. If $p+q\sim r+s$ and the pairs differ, the resulting degree-two pencil makes $C$ hyperelliptic; therefore on a nonhyperelliptic curve the unordered pairs coincide.

The restricted direct sum of a family of groups $G_i$ is the subgroup of their direct product of groups consisting of tuples with only finitely many nonidentity coordinates. Multiplication is coordinatewise. If every factor is a finite group, each subgroup that is a finitely generated group lies in a finite subproduct, so the restricted direct sum is a locally finite group. This gives concrete infinite torsion groups without a uniform bound on element orders.