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Normalize the nonvanishing loop to $\gamma (t)/|\gamma (t)|$. A continuous argument lift is a continuous function $\theta :[0,1]\to \mathbb{R}$ such that $\gamma (t)/|\gamma (t)|=e^{i\theta (t)}$. Such a lift exists on an interval; any two differ by a constant integer multiple of $2\pi$. The winding number isIts value is independent of the lift. This definition applies to continuous loops without assuming a differentiable contour.

The homotopy invariance of winding number states: if $H:[0,1{]}^2\to \mathbb{C}\setminus \{0\}$ is continuous and $H(s,0)=H(s,1)$ for every $s$, then $w(H(s,\cdot );0)$ is independent of $s$. Neither the loops nor their moving basepoints may pass through zero. One may justify invariance by uniformly subdividing the homotopy into nearby loops whose normalized values have a continuous argument difference of magnitude less than $\pi$; that argument difference has equal endpoint values, so their winding numbers agree.

If $\mathbb{E}{S}^2<\infty$, then $g^{\prime \prime }(1)={\lambda }^2\mathbb{E}{S}^2$. Expand at $s=1$:Substitution into the previous quotient givesThus the stationary mean isOnly a finite first service moment was stated. If the second moment is infinite and $\lambda >0$, the stationary mean is infinite; the formula has that extended-value interpretation. This follows as well by writing the second-order remainder as an integral of $g^{\prime \prime }(s)$ and taking $s\uparrow 1$, where monotone convergence gives $g^{\prime \prime }(s)\uparrow {\lambda }^2\mathbb{E}{S}^2$. A finite stationary distribution need not have a finite mean.

The service-period arrival count has probability generating functionwith $g(1)=1$ and $g^{\prime }(1)=\rho$. Let $\Pi (s)={\sum }_{j\ge 0}{\pi }_j{s}^j$. In stationarity the recurrence and independence giveinitially for $0<s<1$. Rearrangement gives $\Pi (s)={\pi }_0(s-1)g(s)/(s-g(s))$. Letting $s\uparrow 1$ and using $\Pi (1)=1$ shows ${\pi }_0=1-\rho$. HenceAnalytic continuation gives the identity inside the unit disc; boundary values follow by convergence of the probability series, with removable limiting values wherever the displayed quotient is indeterminate, particularly at $s=1$.

PutThe transition probabilities areThe first and second rows coincide because either an arrival to an empty queue or the single existing customer initiates the next service.

Let $A_{n+1}$ be the number of arrivals during the next customer's service. If $Q_n>0$, that service begins immediately and one of the waiting customers is served; if $Q_n=0$, the next arrival initiates the service, and it is not counted among arrivals during its own service. ThusConditional on a fresh service time $S=t$, $A_{n+1}$ is Poisson with mean $\lambda t$. Independence of service times and the independent-increment/strong Markov property of the Poisson arrival process make these mixed counts identically distributed and independent of the previous departure history. Hence the conditional law of $Q_{n+1}$ depends only on $Q_n$, proving the Markov property of the departure-embedded queue.

Now $f\ge 0$ almost everywhere. The functions $\min (f_n,f)$ converge almost everywhere to $f$ and are dominated by the integrable $f$. The dominated convergence theorem gives $\int \min (f_n,f)\to \int f$. Using $|f_n-f|=f_n+f-2\min (f_n,f)$ and the assumed convergence of the nonnegative integrals yieldsThis is the nonnegative form of Scheffé's lemma: almost-everywhere convergence together with convergence of total mass implies convergence in the L1 norm.

Write $A_n=\int f_n^{+}$, $B_n=\int f_n^{-}$, $A=\int f^{+}$ and $B=\int f^{-}$. Almost-everywhere convergence gives $f_n^{\pm }\to f^{\pm }$, so Fatou's lemma gives $A\le \mathrm{liminf}{A}_n$ and $B\le \mathrm{liminf}{B}_n$. The given norm condition is $A_n+B_n\to A+B$. Consequentlyand the symmetric argument bounds $\mathrm{limsup}{B}_n\le B$. Combining upper and lower limits provesNo common integrable dominating function for $f_n$ has been assumed.

Fatou's lemma states that for nonnegative measurable $h_n$,Indeed $g_n={\inf }_{k\ge n}{h}_k$ is measurable, increases to $\mathrm{liminf}{h}_n$, and satisfies $g_n\le h_k$ for every $k\ge n$. The monotone convergence theorem gives $\int \mathrm{liminf}{h}_n={\lim }_n\int g_n\le {\lim }_n{\inf }_{k\ge n}\int h_k$, proving the claim, including infinite values.

The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ almost everywhere for one integrable nonnegative $g$, then $h$ is integrable and $\int h_n\to \int h$; in fact $\int |h_n-h|\to 0$. Integrability follows from $|h|\le g$. For real functions, Fatou applied to $g+h_n$ gives $\int h\le \mathrm{liminf}\int h_n$, and applied to $g-h_n$ gives the reverse inequality against the upper limit. Thus the integrals converge. To obtain the stronger conclusion, Fatou applied to $2g-|h_n-h|\ge 0$ gives $2\int g\le 2\int g-\mathrm{limsup}\int |h_n-h|$, forcing the latter limit to be zero. Complex functions may be treated through their real and imaginary parts or the same absolute-difference argument.

Since $7\equiv 3(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[\sqrt{7}]$. A unit $a+b\sqrt{7}$ has norm $a^2-7b^2=\pm 1$. Norm minus one is impossible: it would give $a^2\equiv -1(\mathrm{mod}7)$, whereas the squares modulo seven are $0,1,2,4$. The element $\epsilon =8+3\sqrt{7}$ has norm one and exceeds one.

Suppose a positive unit $u$ in the chosen real embedding satisfies $1<u<\epsilon$. Its conjugate is $1/u>0$, soThis expression is strictly increasing in $u$, and at $u=\epsilon$ it equals three. Thus the integer $b$ is one or two. But then $a^2=7b^2+1$ is eight or twenty-nine, neither a square. Therefore there is no positive unit strictly between one and $\epsilon$.

For any positive unit $u$, choose an integer $k$ such that $1\le u{\epsilon }^{-k}<\epsilon$. The remaining unit must be one, so $u={\epsilon }^k$. Multiplication by minus one handles negative units. Hencewhich proves that $8+3\sqrt{7}$ is fundamental, using no continued fractions.

The Dirichlet unit theorem states that for a number field with $r_1$ real embeddings and $r_2$ conjugate pairs of nonreal embeddings,where $\mu (K)$ is its finite cyclic group of roots of unity. Equivalently there exist $r_1+r_2-1$ multiplicatively independent fundamental units such that every unit is a unique product of a root of unity and integer powers of those units.

Embed the ring of integers into ${\mathbb{R}}^2$ by its two real embeddings. Its image is a full Euclidean lattice of some covolume $d>0$. Fix $C>d$ and consider the symmetric rectangle $|{\sigma }_1(\alpha )|\le T$, $|{\sigma }_2(\alpha )|\le C/T$. Its area is $4C>4d$, so the Minkowski convex body theorem gives a nonzero algebraic integer ${\alpha }_T$ in it. Its algebraic norm is a nonzero integer with $|N({\alpha }_T)|\le C$.

As $T\to \infty$, infinitely many distinct ${\alpha }_T$ must occur: a finite collection of nonzero elements has a positive minimum of $|{\sigma }_2(\alpha )|$, whereas here that coordinate tends to zero. The principal ideals $({\alpha }_T)$ have norm $|N({\alpha }_T)|\le C$. There are only finitely many ideals of bounded norm: they are subgroups of bounded index in a rank-two free abelian group, and integer Hermite normal forms give finitely many possibilities at each such index. Thus infinitely many of these elements generate one ideal. Ratios of two generators of the same ideal are units, because inclusion in both directions supplies inverse algebraic integers. Among infinitely many distinct generators choose two whose ratio is not $\pm 1$. This gives a non-trivial unit without assuming Dirichlet's theorem.

For any positive integer $m$, put $N=(m+1)!$. Each of $N+2,N+3,\dots ,N+m+1$ is composite, since $N+j$ is divisible by $j$ and larger than $j$. Let $p$ be the largest prime number at most $N+1$, and let $q$ be the smallest prime number at least $N+m+2$. Such $q$ exists by part (i), and no prime lies strictly between $p$ and $q$. Therefore these are consecutive prime numbers withSince $m$ is arbitrary, gaps between consecutive prime numbers are unbounded.

Suppose the prime numbers were the finite list $p_1,\dots ,p_r$. The integer $N=p_1\cdots p_r+1>1$ has a prime factor, by repeatedly factoring any composite factor until a prime is reached. But division by any $p_j$ leaves remainder one, so that prime factor is absent from the list. This contradiction proves there are infinitely many prime numbers. The argument does not require $N$ itself to be prime.

The invariant positive-definite bilinear form makes every $\rho (g)$ orthogonal in an orthonormal real basis. Every element of $\mathrm{SU}(2)$ has the form $\left(\begin{array}{cc}a& b\\ -\overline{b}& \overline{a}\end{array}\right)$ with $|a{|}^2+|b{|}^2=1$, so this group is homeomorphic to $S^3$ and connected. The continuous function $\det \rho (g)$ takes values in $\{-1,1\}$ and equals one at the identity. It is therefore identically one. Hence $\boxed{\rho (\mathrm{SU}(2))\subseteq \mathrm{SO}(3)}$.

The pairing is real and symmetric: $\overline{\mathrm{tr}(AB)}=\mathrm{tr}(B^{\ast }{A}^{\ast })=\mathrm{tr}(BA)=\mathrm{tr}(AB)$, and cyclicity of the trace exchanges $A,B$. For the coordinates in part (i), $A^2=-(x^2+y^2+z^2)I$, soThus it is a positive-definite real bilinear form, hence nondegenerate. Finallyproving invariance under conjugation.

A traceless skew-Hermitian matrix has the unique formThis real-linear parametrization is bijective, proving $\boxed{V\cong {\mathbb{R}}^3}$ as real vector spaces. It is not a complex vector space: multiplication by $i$ generally turns a skew-Hermitian matrix into a Hermitian one.

Over $K=\mathbb{Q}$, takeIts roots are $\pm \sqrt{2}\pm \sqrt{3}$. For $\alpha =\sqrt{2}+\sqrt{3}$, its inverse is $\sqrt{3}-\sqrt{2}$, so both square roots belong to $\mathbb{Q}(\alpha )$. Since $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree four, $P$ is irreducible and this field is its splitting field. Independently changing the signs of the two square roots gives its Galois group $C_2\times C_2=V_4$, which is abelian and satisfies the hypothesis.

Factorization over $K$ gives the orbits of the Galois group on the three resolvent roots. If $Q$ splits completely, the quotient is trivial and $G=V_4$. If $Q$ is a linear factor times an irreducible quadratic, the quotient is $C_2$ and $G=D_8$.

If $Q$ is irreducible, the quotient acts transitively on three roots, so it is $C_3$ or $S_3$. The product of their pairwise differences is a square root of the nonzero discriminant $\Delta$ and changes by the sign of the permutation. Thus $\Delta$ is a square in $K$ exactly when the quotient contains only even permutations. The criterion isThe assumed containment of $V_4$ is essential to reconstruct $G$ from the quotient by this table.

The three resolvent cubic roots correspond to the partitions $12|34$, $13|24$, $14|23$. Permuting the four roots of the quartic permutes these three quantities and defines a homomorphism $G\to S_3$. They are distinct: because the quartic has zero cubic coefficient, their differences includesince characteristic zero and irreducibility imply separability. Thus the kernel of the action is exactly $V_4$. The splitting field of the resolvent is the subfield generated by its roots; restriction from the quartic splitting field is surjective onto its Galois group, since automorphisms of a normal separable intermediate extension extend to the splitting field. HenceIn the four cases these groups are respectively the trivial group, $C_2$, $C_3$ and $S_3$. Computing the symmetric functions also gives $Q(X)=X^3+2b{X}^2+(b^2-4d)X-c^2$ with the source's root convention.

The Klein four-group $V_4$ is normal in $S_4$, and $S_4/V_4\cong S_3$ through the action on the three partitions into pairs. A subgroup containing $V_4$ is the full preimage of its image in this quotient. Up to conjugacy the subgroups of $S_3$ have orders $1,2,3,6$. Their preimages give the complete listwith respective orders $4,8,12,24$, where $D_8$ denotes the dihedral group of order eight. The order-eight preimage is a Sylow two-subgroup of $S_4$, generated for example by a four-cycle and a reflection reversing it.