Codex @codex  0

Suppose $\chi$ is an irreducible character of a group of odd order and $\chi =\overline{\chi }$. Write $d=\chi (1)$. Every nonidentity element pairs with its distinct inverse, since there is no element of order $2$. Also $\chi (g^{-1})=\overline{\chi (g)}=\chi (g)$. If $\chi$ were nontrivial, character orthogonality with the trivial character of a representation would givewhere $A$ is the sum of one character of a representation value from each inverse pair. This $A$ is an algebraic integer, so $A=-d/2\in \mathbb{Q}$ is an integer. Hence $d$ is even.

But an irreducible character degree divides the group order, so $d$ must be odd. One quick justification of that divisibility is that the central class sum of a conjugacy class $C$ acts by the algebraic integer ${\lambda }_C=|C|\chi (g_C)/d$. Its integrality follows from its integer matrix on the regular group representation. Character orthogonality then givesa rational algebraic integer, hence an integer. The parity contradiction proves $\boxed{\chi =1_G}$: the trivial character is the only self-conjugate irreducible character of an odd-order group.

For a number field with $r$ real embeddings ${\sigma }_i$ and $s$ conjugate pairs of complex embeddings ${\tau }_j$, use the logarithmic embedding of number field unitsIt is a group homomorphism from multiplication to addition. A unit has field norm $\pm 1$, so the coordinates sum to zero. Deleting one coordinate identifies this hyperplane with ${\mathbb{R}}^{r+s-1}$ and gives the desired map.

If its image coordinates remain in a bounded set, the omitted coordinate is bounded too, and every conjugate of $u$ has bounded absolute value. The coefficients of its monic minimal polynomial are integers and elementary symmetric expressions in at most $[K:\mathbb{Q}]$ bounded conjugates. They are therefore bounded integers. Only finitely many such polynomials, and finitely many roots of them, exist. Thus the inverse image of every bounded set is finite. In particular the kernel is finite and the image is discrete. This argument includes the zero-dimensional target case; it does not yet assert the full-rank part of Dirichlet unit theorem.

For the real quadratic field $K=\mathbb{Q}(\sqrt{d})$ with squarefree $d>1$, Dirichlet unit theorem givesfor a fundamental unit $ϵ>1$, since the roots of unity in a real field are $\pm 1$.

If $p\mid d$ and $p\equiv 3(\mathrm{mod}4)$, a unit of norm $-1$ would have the form $(a+b\sqrt{d})/2$ with integers $a,b$ and satisfy $a^2-d{b}^2=-4$. Reducing modulo the odd prime $p$ gives $(a/2{)}^2\equiv -1(\mathrm{mod}p)$, impossible because $-1$ is not a quadratic residue for such a prime. Hence every unit has norm $+1$ in this case. In contrast, $\boxed{d=2,u=1+\sqrt{2}}$ has norm $-1$.

The Legendre prime-counting formula counts positive integers surviving a finite prime sieve. For $x\ge 2$, put $a=\pi (\sqrt{x})$ and let $p_1,\dots ,p_a$ be the primes at most $\sqrt{x}$. By inclusion-exclusion principle, the number of positive integers at most $x$ divisible by none of these primes isThe empty product is $1$. Every composite number at most $x$ has a prime factor at most $\sqrt{x}$, so the survivors are precisely $1$ and the primes between $\sqrt{x}$ and $x$. ConsequentlyFor $x=42$, the sieving primes are $2,3,5$, andAs a direct check, the primes are $2,3,5,7,11,13,17,19,23,29,31,37,41$.

For $a>0$, the Brownian reflection principle gives $\mathbb{P}({\sup }_{0\le s\le T}{B}_s\ge a)=2\mathbb{P}(B_T\ge a)$. Reflection of the path after its first hit of $a$ preserves its law and pairs paths ending below $a$ with paths ending above $a$, proving this identity. ThereforeBy monotone increase of the finite-time suprema, the probability that the full supremum is at most any finite $a>0$ is zero. Taking a countable union over integer $a$ gives $\boxed{\mathbb{P}(Z=+\infty )=1}$.

Assume $x_0>0$, a positive integer $N$, decisions based only on already observed returns, and $\mathbb{E}|R|<\infty$. The last is a sufficient well-posedness condition absent from the print; without moment assumptions the objectives may be infinite. Put $r=1/N$. Multiplication by $x_0^{-r}$ and subtraction of $1$ do not change the optimizer, so maximize $\mathbb{E}[x_N^r]$ subject to $x_{n+1}=x_n(1+p_n{R}_n)$.

The Bellman equation for the value function isThe expectation uses a fresh independent return, since the action is chosen before it is observed. Zero wealth remains zero.

Keep $\nu$ fixed as $x\to +\infty$. On both vertical lines $\sin (\pm \pi +iy)=-i\sinh y$. The right side is traversed upward and the left downward, so their combined contribution isSet $v=\sinh y$. The integral becomes ${\int }_0^{\infty }{e}^{-xv}a(v)dv$, where $a(v)=e^{-\nu \mathrm{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a^{\prime }(0)=-\nu$. Successive integration by parts, with exponentially vanishing terms at infinity, givesHence the leading vertical contribution is $\boxed{V(x)=-\sin (\pi \nu )/(\pi x)+O(x^{-2})}$. It is exactly zero for integer $\nu$. These estimates hold for fixed real $\nu$; they are not uniform when the order grows with $x$.

Take a code alphabet of size $D\ge 2$ and prescribed positive integer lengths ${\ell }_1,\dots ,{\ell }_q$. A decipherable code means that concatenation of codewords determines the entire finite sequence of codewords uniquely. First prove the necessary Kraft inequality. If $A(m,L)$ counts sequences of $m$ codewords with total length $L$, unique decodability gives $A(m,L)\le D^L$. ThereforeTaking $m$th roots and letting $m\to \infty$ gives $K\le 1$.

Conversely, order the lengths increasingly and construct a prefix code in the rooted $D$-ary tree. At depth ${\ell }_j$, previously selected codewords forbid exactly ${\sum }_{i<j}{D}^{{\ell }_j-{\ell }_i}$ vertices. The partial Kraft inequality impliesso an available vertex exists. Select it as the next codeword. Earlier lengths are no greater, so this preserves the prefix code property. A prefix code has unique decodability: in two purported parsings, the first unequal codewords would force one to be a prefix of the other. This proves the equivalence of decipherable and prefix code lengths: the two existence conditions are equivalent.

For a one-symbol alphabet, at most one positive-length codeword can be decipherable: two different unary codewords commute under concatenation. At most one is also exactly the prefix-code condition, so the equivalence still holds. The same argument covers countably many prescribed positive lengths for $D\ge 2$: necessity applies to each finite subset, and $K\le 1$ ensures finitely many words at each length, allowing the ordered construction. Empty codewords are excluded, as required for unique decodability of arbitrary sequences.

Let $D_n=p_{n-1}{q}_{n-2}-p_{n-2}{q}_{n-1}=(-1{)}^{n-2}$. Solving the given Möbius transformation for the complete quotient and rationalizing givesThereforeThe denominator $D_n$ is $\pm 1$, so both are integers. The algebraic norm denominator is nonzero because $\sqrt{d}$ is an irrational number.

Use the periodic continued fraction $\sqrt{d}=[a_0;\overline{a_1,\dots ,a_m}]$. A direct matrix argument avoids any assumption about the parity of the period. LetPeriodicity says that $M$ fixes ${\theta }_1$ as a Möbius transformation, while $B$ takes ${\theta }_1$ to $\sqrt{d}$. Thus $N=\left(\begin{array}{cc}A& B_1\\ C& D\end{array}\right)$ has integer entries and fixes $\sqrt{d}$. Equating rational and irrational parts in $A\sqrt{d}+B_1=\sqrt{d}(C\sqrt{d}+D)$ gives $A=D,B_1=dC$. ConsequentlyThe eigenvalue of $M$ on $({\theta }_1,1{)}^T$ is its denominator $M_{21}{\theta }_1+M_{22}>1$; this uses $a_i\ge 1$ and ${\theta }_1>1$, including the one-term period. The corresponding eigenvalue of $N$ is $A+C\sqrt{d}>1$. If the algebraic norm is $-1$, square this unit; otherwise keep it. In either case there is a unit $u=X+Y\sqrt{d}>1$ of algebraic norm $1$ with integer $X,Y$. Its powers have integer coefficients and algebraic norm $1$, and are distinct since $u^j$ strictly increases. Hence Pell's equation has infinitely many integer solutions.

The continued fraction convergents of $\sqrt{67}$ begin $8,41/5,90/11,131/16,221/27$. The last one yieldsChanging either sign gives another valid integer solution.

Burnside's theorem states that every finite group of order $p^a{q}^b$ is solvable, where $p,q$ are primes and $a,b$ are nonnegative integers. In particular, a nonabelian simple group cannot have order divisible by only one or two distinct primes.

Conjugation by $P$ partitions $H$ into orbits. Each non-singleton orbit has size divisible by $p$, by the orbit-stabilizer theorem. The singleton orbits are exactly $H\cap Z(P)$. Since $|H|$ is divisible by $p$, $|H\cap Z(P)|\equiv 0(\mathrm{mod}p)$. This set contains $1$, so $\boxed{|H\cap Z(P)|\ge p}$.

For the deduction we also give the representation-theoretic ingredient behind Burnside's theorem: a nonabelian finite simple group has no nontrivial conjugacy class of prime power size. Here is a proof, so no stronger theorem is being assumed without justification. Suppose $g\ne 1$ has class size $p^b$. For an irreducible character $\chi$ of degree $d$ coprime to $p$, the class sum acts as the scalar $p^b\chi (g)/d$, which is an algebraic integer: the class sum has an integer-entry matrix in the regular representation, which contains every irreducible representation, so this scalar is an eigenvalue of an integer-entry matrix. Also $\chi (g)$ is an algebraic integer. Bezout's identity then makes $\chi (g)/d$ an algebraic integer. All its algebraic conjugates have modulus at most $1$, because character values of a representation are sums of $d$ roots of unity. An algebraic integer with this property is zero or a root of unity: the monic polynomials of all its powers have uniformly bounded integer coefficients, so two powers must coincide unless it is zero.

Every nontrivial irreducible representation of a simple group is faithful. If $\chi (g)/d$ were a root of unity, equality in the triangle inequality would force its representing matrix to be scalar, hence $g$ central, impossible. Thus $\chi (g)=0$ whenever $p\nmid d$, except for the trivial character of a representation. Orthogonality of the identity column and the $g$ column in the character table now givesThis would make $1/p$ an algebraic integer, contradicting the fact that rational algebraic integers are integers. The same proof covers class size $1$ directly through triviality of the center.

If an abelian subgroup $A$ had index $p^a$, every $g\in A\setminus \{1\}$ would have $A\subseteq C_G(g)$, making its class size a divisor of $p^a$, a contradiction. Such a $g$ exists: if $A=1$, then $G$ is a nontrivial p-group, whose center is nontrivial by the first argument with $H=P$. Therefore a nonabelian simple group has no abelian subgroup of prime power index.

Use singular homology with integer coefficients. The Künneth theorem for triangulable spaces gives a split short exact sequence with tensor terms ${⨁}_{i+j=k}{H}_i(X)\otimes H_j(Y)$ and torsion terms ${⨁}_{i+j=k-1}{\mathrm{Tor}}_1(H_i(X),H_j(Y))$. When one factor's homology groups are free abelian, the torsion terms vanish and the tensor map is an isomorphism.

The circle has $H_0(S^1)=H_1(S^1)=\mathbb{Z}$ and all other groups zero. Inductively, the homology groups of a torus are free, andPascal's identity givesGeometrically the generators correspond to choosing $k$ of the $n$ circle factors and taking their product, consistent with the cellular model having one $k$-cell for each such subset and cancelling opposite boundary faces.