Suppose a group $G$ has no proper subgroups (that is, the only subgroup of $G$ is $G$ itself and the trivial subgroup $\{e\}$. Show that $G$ is cyclic.
There is a lemma that says if a group $G$ has no proper nontrivial subgroups, then $G$ is cyclic. And here is the proof of the lemma:     Suppose $G$ has no proper nontrivial subgroups. Take an element $a$ in $G$ for which $a$ is not  equal to $e$. Consider the cyclic subgroup $\langle a \rangle$. This subgroup contains at least $e$ and $a$, so it is not trivial. But $G$ has no proper subgroups, so it must be that $\langle a \rangle = G$. Thus $G$ is cyclic, by definition of a cyclic group.   But here i do not understand the following: Why must $\langle a \rangle$ be a subgroup of $G$? For every single element $a$ in $G$, if $\langle a \rangle$ is a subgroup of $G$, then every group should have at least as many subgroups as the number of its elements. I would appreciate any help. Thanks
I started by Time Series Analysis by Hamilton, but I am lost hopelessly. This book is really too theoretical for me to learn by myself.   Does anybody have a recommendation for a textbook on time series analysis that's suitable for self-study? Everywhere I get to read statements like EM radiations and EM fields carry "energy as well as momentum " . I wonder if energy and momentum are inalienable . Is it possible for a field to carry only energy without having any momentum term ? I mean isn't it sufficient to say that fields or EM waves carry energy or Doesn't the ability to transfer energy, of the fields, automatically imply that they can change the momentum of another particle while transferring energy to them? Show that $u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$ is in $W^{1,n}(U)$, where $U=B(0,1)\subset\mathbb{R}^n$. The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback! I am looking for a package to create bullet gauge graphs, but I couldn't find any.  A bullet graph looks like this:    and an ample description/specification can be found   It is available in Mathematica 9 ()  Is it possible create this kind of graphs in a easy way with TeX (like Matehmatica 9 or similar)? Particular nodes are missing from node editor Covariance between sum of iid random variables and sum of indicator functions The question is-     Let $X_1,X_2,..,X_n$ be iid random variables from a continuous distribution whose density is symmetric about $0$. Suppose $\mathbb{E}(|X_1|)=2$ and define $Y=\sum_{i=1}^{n}X_i$ and $Z=\sum_{i=1}^{n}I(X_i&gt;0)$. Then calculate covariance between $Y$ and $Z$.   My attempt:  $E(X_i)=0$ for all $i=1(1)n$ because $X$ is symmetric about $0$ and $E(|X|) $ exists.   Now,  $Cov (Y,Z)=E(YZ)-E(Y)E(Z)$ $=E(YZ)-0$ $=E[(\sum_{i=1}^{n}X_i)(\sum_{i=1}^{n}I(X_i&gt;0)]$  $=(\sum_{i=1}^{n}E[(X_i.I(X_i&gt;0))]$ $+\sum\sum_{i \neq j}E(X_i)E(I(X_j&gt;0)$ as $X_i,X_j$ are independent.   $=\sum_{i=1}^{n}E[(X_i.I(X_i&gt;0)] +0 $ as $E(X_i)=0$  $ =\sum_{i=1}^{n}\{E[X_i.I(X_i&gt;0)|I(X_i&gt;0)=1]×1/2] + E[X_i.I(X_i&gt;0)|I(X_i&gt;0)=0]×1/2]\}$  $=\sum_{i=1}^{n}E[X_i.I(X_i&gt;0)|I(X_i&gt;0)=1]×1/2] +0$  $=\sum_{i=1}^{n}E[X_i|X_i&gt;0]×1/2]$  $=2n×(1/2)$ $=n$  Is my reasoning correct ? Thanks in advance! Calculate intersection of a line with a plane I was reading something about: “Intersection of a line with a plane”. But I don’t fully understand how to calculate it on their way.  The goal is to calculate where the plane and the line will “cross”. These items are given:   The eye is in position: O(0,0,-10)   A point of a 3D-object is P(10,20,30) The screen has the equation z=0   I have a questions about how they calculate this:    This part is clear. That is the location form the eye (vector representation).   They specify the location of the plane, there is nothing strange here (vector representation).   Now they calculate what part of the line will cross, how did they calculate this? How did they know a = 1/4? And b = 10/4?  This is everything that was given. Short story about a woman that travels to other planets. Read online recently. She just manages to live long enough to embrace new technology Can all real polynomials be factored into quadratic and linear factors? Do I need a firewall for my desktop? I don't have a  router, but I mostly use my desktop for online shopping and torrenting. So, do I have to enable the firewall? I hope not, because I'm really dumb and I don't think I will be able to configure the firewall. What is the difference between sector and cluster? I have gone through wikipedia and other sites and here's the information I have obtained.  Sector: Smallest Addressable Unit (mostly 512 bytes) Cluster: Smallest Allocation Unit (equals to 1 or more sectors and the number of clusters depends on the file system).  Let's say in the below example:  sector =512 bytes and a cluster =8 sectors=4096 bytes. Now, I am having a hard time figuring out, when I am trying to create a file what does the file system do? Does it consider 4096 bytes as the smallest write/read unit or 512 bytes?  If the answer is 4096 bytes: When the manufacturer of the disk drives has provided 512 bytes as the smallest addressable unit, why does the file system consider cluster as the smallest unit? I mean, what are the advantages in doing so? I'm using a custom class file.  My main file looks like this:  \documentclass[12pt,a4paper]{thesis} \begin{document} \frontmatter     \pagenumbering{alph}      \pagenumbering{roman}     \clearpage      \cleardoublepage     \tableofcontents     \clearpage     \listoffigures     \clearpage     \listoftables   \mainmatter        \part{test}       \input{test} \end{document}   If I put this in my test.tex it does work:  \chapter{test} \label{chapter:test} test    When it becomes more text (lets say 40x the word "test") I get the error stated in my question title.  To test it I brought my class file down to this:  \ProvidesClass{thesis} \NeedsTeXFormat{LaTeX2e}  % Based on the memoir class \DeclareOption*{\PassOptionsToClass{\CurrentOption}{memoir}} \ProcessOptions \LoadClass{memoir}   I'm totally confused what is causing my error now.  This is the complete error:  ! pdfTeX error (font expansion): auto expansion is only possible with scalable  fonts. \AtBegShi@Output ...ipout \box \AtBeginShipoutBox                                                    \fi \fi Python gives us the ability to create 'private' methods and variables within a class by prepending double underscores to the name, like this: __myPrivateMethod(). How, then, can one explain this  >>> class MyClass: ...     def myPublicMethod(self): ...             print 'public method' ...     def __myPrivateMethod(self): ...             print 'this is private!!' ...  >>> obj = MyClass() >>> obj.myPublicMethod() public method >>> obj.__myPrivateMethod() Traceback (most recent call last):   File "", line 1, in  AttributeError: MyClass instance has no attribute '__myPrivateMethod' >>> dir(obj) ['_MyClass__myPrivateMethod', '__doc__', '__module__', 'myPublicMethod'] >>> obj._MyClass__myPrivateMethod() this is private!!   What's the deal?!  I'll explain this a little for those who didn't quite get that.  >>> class MyClass: ...     def myPublicMethod(self): ...             print 'public method' ...     def __myPrivateMethod(self): ...             print 'this is private!!' ...  >>> obj = MyClass()   What I did there is create a class with a public method and a private method and instantiate it.  Next, I call its public method.  >>> obj.myPublicMethod() public method   Next, I try and call its private method.  >>> obj.__myPrivateMethod() Traceback (most recent call last):   File "", line 1, in  AttributeError: MyClass instance has no attribute '__myPrivateMethod'   Everything looks good here; we're unable to call it. It is, in fact, 'private'. Well, actually it isn't. Running dir() on the object reveals a new magical method that python creates magically for all of your 'private' methods.  >>> dir(obj) ['_MyClass__myPrivateMethod', '__doc__', '__module__', 'myPublicMethod']   This new method's name is always an underscore, followed by the class name, followed by the method name.  >>> obj._MyClass__myPrivateMethod() this is private!!   So much for encapsulation, eh?  In any case, I'd always heard Python doesn't support encapsulation, so why even try? What gives? When is the order of convergence of Newton-Raphson greater than 2? If some function $f$ has a simple root at $x_*$ then I know that the order of convergence of the Newton-Raphson iteration is at least 2. But when is this order strictly greater than 2? If object A is moving at velocity $v$ (normalized so that $c=1$) relative to a ground observer emits object B at velocity $w$ relative to A, the velocity of B relative to the ground observer is  $$ v \oplus w = \frac{v+w}{1+vw} $$  As expected, $v \oplus 1 = 1$,  as "nothing can go faster than light".  Similarly, $v \oplus -1 = -1$. (same thing in the other direction)  But what if object A is moving at the speed of light and emits object B at the speed of light in the exact opposite direction? In other words, what is the value of $$1 \oplus -1?$$ Putting the values in the formula yields the indeterminate form $\frac{0}{0}$. This invites making sense of things by taking a limit, but $$ \lim_{(v,w)\to (1,-1)} \frac{v+w}{1+vw}$$ is not well-defined, because the limit depends on the path taken.  So what would the ground observer see? Is this even a meaningful question?  Edit: I understand $1 \oplus -1$ doesn't make sense mathematically (thought I made it clear above!), I'm asking what would happen physically. I'm getting the sense that my fears were correct, it's physically a nonsensical situation. Have there been any recent attempts to falsify the  hypothesis alternative to the Doppler shift explanation for the red shift, or is it simply ignored like the pilot wave quantum mechanical hypothesis alternative to the Copenhagen hypothesis of quantum mechanics has been? What new evidence would it take to get the tired light hypothesis for the red shift reexamined as the recent fluid dynamics oil drop experiments have done for the pilot wave hypothesis of quantum mechanics? Why does your reflection suddenly turn upside down while slowly walking away from a curved mirror? Many people should have realised, when looking into a concave curved mirror (or even a rather reflective spoon in that fact) at a close distance, you will see a slightly distorted reflection. But as you move further away, the image will suddenly become upside-down.  why is this? What does “would not have been possible had I remained” mean? What does this mean?      I am sure this would not have been possible had I remained a typical Anglophone North American.   I have never heard that before and I really find it strange.  I deduce that it means      I am sure this would not have been possible if I kept remaining Anglophone North American.   But I might be wrong in this. Proving that $n$-cycle has order $n$ Solve a congruence $1978^{20}\equiv x\pmod{125}$