Oliver Kayende
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If $\int\limits_0^1\frac{x^n}{1+x}\,\mathrm{d}x=\xi_n^n\,\ln 2,$ prove that $\xi_n^n \to 0$
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5 votes

This is equivalent to the problem of showing $$\lim_{n\rightarrow\infty}\int_0^1\frac {x^n}{1+x}\;dx = 0$$ Use the compound inequality $$0\leq\int_0^1\frac {x^n}{1+x}\;dx\leq\int_0^1 x^n\;dx =\frac{1}{...

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Does the Frobenius endomorphism apply to polynomials?
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4 votes

Yes. Note that $(x+a)^p=\sum_{r=0}^p\binom{p}{r}x^ra^{p-r}=x^p+a^p$ because $p$ divides $\binom{p}{r}$ except when $r=0,p$.

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Show that $\cos x=x^3+x^2+4x$ has exactly one root in $[0,\frac{\pi}{2}]$
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4 votes

Let $f(x)=-\cos(x)+x^3+x^2+4x$. Note $f(0)=-1<0$ and $f(\frac{\pi}{2})>0$. Therefore, by the Intermediate Value Theorem $f$ must admit a zero in $[0,\frac{\pi}{2}]$. If $f$ had another zero in $[...

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Showing that union over a collection of connected sets $\{F_n\}$ with $F_i\cap F_j \neq \varnothing, i\neq j$ is connected.
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3 votes

The OP claim is false. The singleton subsets $\{1\},\{2\},\{3\},\dots$ are connected in $\Bbb R$ and pairwise disjoint but $\Bbb N:=\bigcup_{n=1}^\infty\;\{n\}$ is not a connected subspace of $\Bbb R$ ...

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A prime ideal of $\mathbb{Z} \times \mathbb{Z}$ that is not maximal.
3 votes

$\;\;\;\;\;$ From the OP then $I=\Bbb Z\times\{0\}$. By the first homomorphism theorem then $$\frac{\Bbb Z\times\Bbb Z}{\Bbb Z\times\{0\}}\approx\Bbb Z$$ as $\Bbb Z\times\{0\}$ is the kernel of the ...

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Why is $A \rtimes B \simeq \mathbb Z \rtimes \mathbb Z/2\mathbb Z$
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3 votes

$B$ and $\frac{\Bbb Z}{2\Bbb Z}$ act faithfully on $A$ and $\Bbb Z$, respectively, via $A$ automorphisms and $\Bbb Z$ automorphisms. The map $$\Psi:\langle f_2\circ f_1\rangle\rtimes \langle f_2\...

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Let $G$ be a finite abelian group, and let $n$ divide $|G|$. Let $m$ be the number of solutions of $x^n=1$. Prove that $n\mid m$.
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3 votes

By the fundamental theorem of finite abelian groups we may choose a $G$ subgroup $G_n$ of size $n$. Lagrange's theorem gaurantees $G_n\leq\ker(\varphi_n)$ where $\varphi_n$ denotes the $G$ ...

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Group of units of a non-finitely generated ring
3 votes

Consider maybe the the polynomial ring $\Bbb Z[X_1,X_2,\dots]$ with infinitely many indeterminates. This ring is not finitely generated and the only units are $\pm 1$.

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Finding a generator for a principal ideal
3 votes

$X,Y$ are coprime in $\Bbb Q[X,Y]$. Therefore any generator of $I:=\;${$Xa+Yb | a,b\in\Bbb Q[X,Y]$} would be a common divisor of $X,Y$ and thus constant. Since $I$ is a proper ideal, i.e. $f(0,0) = 0$ ...

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Isomorphic subgroups such that the quotient is isomorphic.
3 votes

Under pointwise addition the family of functions $\Bbb F_2^{X}$ from any set $X$ to the finite field $\Bbb F_2$ of two elements is an abelian group all of whose non-trivial elements have finite order ...

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In search for easy counterexample of $L^2$ ineq. - if it exists
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2 votes

Define $u:[0,3]\to[0,1]$ by $u(x)=\sin(\frac{x}{2})$. Note $u''(x)^2=\frac{1}{16}\sin^2(\frac{x}{2})$ and thus $u(0)=u''(0)=0$. Finally, $$\Vert u\Vert^2=\int_0^3\sin^2(x)\;dx=16\cdot\int_0^3\frac{1}{...

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Prove that $F[x,y] / (y^2-x)$ is an integral domain
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2 votes

$$f(x,y)=y^2-x$$ is irreducible in $\Bbb F[x][y]$ because it is a monic quadratic polynomial in $y$ with no roots in $\Bbb F(x)$. Equivalently, $f(x,y)$ is irreducible over $\Bbb F[x,y]$ because $$g(x,...

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Is the order of a quotient of the multiplicative group of integers a prime power if the group is cyclic?
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2 votes

No. The size of the quotient need not be a prime power. Consider the field $\Bbb F_{31}:=\frac{\Bbb Z}{31\Bbb Z}$ and denote by $\mathbf x$ the residue class $x+31\Bbb Z$. As presecribed in the OP the ...

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When two elements of a group generate the same subgroup
2 votes

Suppose $3<n<p$ and $\Bbb F_p^+$ is the additive group of the finite field $\Bbb F_p:=\frac{\Bbb Z}{p\Bbb Z}=\{\mathbf 0,\mathbf 1\,\dots,\mathbf{p-1}\}$ of size $p$. Let $$x=(u,\mathbf 1)\;\;\;\...

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To Prove $ \bigcap_{i \in I} A_i \in \bigcap_{i \in I} P(A_i) $
2 votes

Let $A=\bigcap_{i\in I}A_i$. By definition, $A\subseteq A_j$ for each $j\in I$ and thus $j\in I\implies A\in\mathcal P(A_j)$. $$\therefore A\in\bigcap_{i\in I}\mathcal P(A_i)$$

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Let $G$ be a nonabelian group of order $p^{3},$ where $p$ is a prime. Show that $G$ has exactly $p^{2}+p-1$ distinct conjugacy classes.
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2 votes

Because $\mathcal Z(G)\subset\mathcal C(a_i)\neq G$ we must have $p$ $|$ $|\mathcal C(a_i)|$ $|$ $p^3$. Therefore $|\mathcal C(a_i)|=p^2$. $$\therefore p^3=p+np$$ $$\therefore n=p^2-1$$

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Finding points of discontinuity of $f(x)=\lim_{t\to\infty}\frac{|a+\sin(\pi x)|^t-1}{|a+\sin(\pi x)|^t+1}$
2 votes

$f(x)=1-2\cdot\lim_{t\to\infty}\frac{1}{|a+\sin(\pi x)|^t+1}= \begin{cases} -1 & \text{$|a+\sin(\pi x)|<1$} \\ 0 & \text{$|a+\sin(\pi x)|=1$}\\ 1 & \text{$|a+\sin(\pi x)|>1$} \end{...

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First isomorphism theorem $G=\mathbb{Z}/60\mathbb{Z}=<\overline{1}>$
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2 votes

From the OP we have $G=\Bbb Z_{60}^+$ ; $G$ is the additive group of the ring $\frac{\Bbb Z}{60\Bbb Z}$ ; $\phi:\bar a\mapsto\overline {3a}$. Since $G$ is cyclic, i.e. $G=\langle\bar 1\rangle$, then $\...

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Why is it impossible for the mapping $x\mapsto 1/x$ to be a polynomial in an infinite field?
2 votes

From Jose, $xp(x)$ would have positive degree and thus $x(p(x))-1$ would be a non-constant polynomial with infinitely many roots which is impossible.

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Preimage of subgroup $H \subseteq G/N$ has order $|H| \cdot |N|$
2 votes

Let $H'=\varphi^{-1}[H]$ and note $\ker(\varphi)=\varphi^{-1}[\{N\}]\subseteq H'$. Now, $\varphi|_{H'}$ is a group epimorphism from $H'$ to $H$ with $\ker(\varphi|_{H'})=H'\cap\ker(\varphi)=\ker(\...

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The number of subgroups of $Z_{p^2}$ $\oplus$ $Z_{p^2}$ which are isomorphic to $Z_{p^2}$.
2 votes

$\pi:g\mapsto \langle g\rangle$ defines a map from $G-\text{ker}(\psi)$ onto the family $\mathcal F$ of all cyclic $G$ subgroups of size $p^2$ where $G$ is an abelian group of exponent $p^2$ and $\psi:...

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Epimorphism from $\mathbb{Z}[i]$ to $\mathbb{F}_{p}$ where $p\equiv1 \pmod 4$
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2 votes

$\;\;\;\;$$\Bbb F_p^*$ is cyclic and of size $p-1$ which is divisible by $4$ and therefore we may fix $\alpha\in\Bbb F_p^*$ of order $4$ so that $\alpha^2$ has order $2$ ; i.e. $\alpha^2=\mathbf{-1}$....

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Show that $\langle x,y,z \rangle$ has order $42$ for $x,y,z$ with given properties
2 votes

Except for the common identity element between them the cyclic subgroups $\langle x\rangle,\langle y\rangle,\langle z\rangle$ are pairwise disjoint because their sizes are coprime. $\langle x\rangle$ ...

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Evaluating $\int_{0}^{1}\int_{0}^{1}\sqrt{x^{2}+y^{2}}dxdy$ using polar coordinates.
2 votes

The integrand $\sqrt{x^2+y^2}$ is symmetric about the line $y=x$ yielding $$\iint_{[0,1]\times [0,1]}\sqrt{x^2+y^2}\;dydx=2\int_0^1\int_0^x\sqrt{x^2+y^2}\;dydx=2\int_0^\frac{\pi}{4}\int_0^{\sec(\theta)...

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Understanding the biconditional logical connective in the set: $K = \{x \in G: x \circ a \circ x^{-1} \in H \iff a \in H \}$
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2 votes

Firstly, the biconditional statement $xax^{-1}\in H\iff a\in H$ is apparently meant to be read as $$\mathbf B(x,a):(\forall a\in G \;\; xax^{-1}\in H\iff a\in H) $$ with $\forall$ unnecessary as $G$ ...

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In $\mathbb Z[x]$, find an ideal $P$ such that $R/P$ consists of $4$ elements.
2 votes

Let $P := <2 > + <x^2+x+1>$ be the $\Bbb Z[x]$ ideal generated by the polynomial $x^2+x+1$ and the constant $2$. $$\frac {\Bbb Z[x]}{P}\approx\frac{\frac {\Bbb Z[x]}{<2>}}{<x^2+x+...

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Determine the accumulation points of a set of complex numbers
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1 votes

Given an $S$ accumulation point $z$ choose a convergent one-to-one $S\setminus\{z\}$ sequence $(\frac{i^{A_n(z)}}{A_n(z)})_{n=1}^\infty$ and a monotonic subsequence $(a_n(z))_{n=1}^\infty$ of $(A_n(z))...

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Proving Group of Odd Order cannot be Doubly Transitive
1 votes

Proceed by contradiction and assume $(G,\cdot)$ is a doubly transitive permutation group on $\Omega$ but $2\nmid|G|$ where $|\Omega|>1$. By the double transitivity of $G$ we may fix distinct $a,b\...

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Question about abelian groups and count of elements
1 votes

Given a finite abelian group $(G,\cdot)$ with identity $\mathbf 1$ then $$\psi(|G|)=\prod_{j=1}^m{\frak p}(e_j)$$ using the prime factorization $|G|=\prod_{j=1}^mp_j^{e_j}$ where $\psi(n)$ is the ...

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Artin Proposition 2.4.2.
1 votes

Denote by $e$ the $G$ identity and let $n=|\langle x\rangle|=\text{ord}(x)$. (a) becomes clear because $$\ker(\varphi)=\mathcal S:=\{k\in\Bbb Z:x^k=e\}$$ where $\varphi:k\mapsto x^k$ defines the group ...

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