Let $G$ be a finite group. If the order of $G$ is even, prove that there is at least one element $a$ in $G$ such that $a\not= e$ and $a=a^{-1}$.   Here's my idea:  Suppose $\{x_1,\cdots,x_n\}$ is all of the elements in $G$ [edit: I meant to add "such that $x\not =x^{-1}$"]  (but without their inverses or the identity $e$). This set has $n$ elements. Now consider the set of the inverses that correspond to each element: $\{x^{-1}_1,\cdots,x^{-1}_n\}$. Since $G$ is a group, there is a one-to-one correspondence between these two subsets. The total order of these groups is $2n$. Consider the identity $e$, which adds 1 to $\left| G \right|$ to give $G=\{e,x_1,\cdots,x_n,x^{-1}_1,\cdots,x^{-1}_n\}$, which yields: $$|G|=2n+1,$$ which is an odd value. But there may be elements that satisfy $y\in G:y=y^{-1}$. In this case, these elements are redundant and so they each add 1 to the order of $G$. Therefore,  $$|G|=[2n_x+1]+n_y$$  The term in brackets is odd, so $|G|$ is even if and only if the number of elements $y$, or $n_y$, is odd. If $n_y$ is odd, it is clear that $\exists y\in G:y\not = e \wedge y= y^{-1}$. Q.E.D.  My questions are:   Is this proof valid? Any recommendations on how I can make this proof shorter and more elegant? (I struggle with wordiness)   Thank you.
I am working on the following problem from group theory:     If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.   That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$.
Prove that there exists such a set of $n$ positive integers Prove that for any positive integers $m$ and $n$ , there exists a set of $n$ consecutive positive integers each of which is divisible by a number of the form $d^m$, where $d$ is some positive integer not equal to $1$.  I don't know how to approach this question. java.lang.OutOfMemoryError: PermGen space Exception tabular input by columns (i.e. transpose a table) I would like to be able to transpose a tabular(x) table like in the following example. It is not necessary to be completely automated, but I don't want to change the data order, which is a painful and stupid. By "not completely automated" I mean that e.g. the headers can be done seperately etc.  Let's see the MWE:  \documentclass{article}  \usepackage{booktabs} \def\Midrule{\midrule[\heavyrulewidth]}  \begin{document}  \centering  \begin{tabular}{*6c}\toprule \bfseries Name &amp; Alice &amp; Bob &amp; Chuck &amp; Dave &amp; Eve\\\Midrule \bfseries Sex &amp; Female &amp; Male &amp; Male &amp; Male &amp; Female\\\midrule \bfseries Age &amp; 18 &amp; 19 &amp; 20 &amp; 21 &amp; 22\\\bottomrule \end{tabular}  \bigskip  Which I want to look like  \bigskip  \begin{tabular}{*3c}\toprule \bfseries Name &amp; \bfseries Sex &amp; \bfseries Age\\\Midrule Alice &amp; Female &amp; 18\\\midrule Bob &amp; Male &amp; 19\\\midrule Chuck &amp; Male &amp; 20\\\midrule Dave &amp; Male &amp; 21\\\midrule Eve &amp; Female &amp; 22\\\bottomrule \end{tabular}   \end{document}     Note: I know that , but I would like to get it without expl3 (mostly because people would might use my code quite often have an old LaTeX installed and I cannot be convincing them to change this). I masked an image in the UV/Image editor masking section. I then used that mask in the compositor for background replacement. However, the mask size seems to be different. Why is that? Here's a screenshot: i was stopped by my sister from upgrading Ubuntu, for the reason she explained was that i would have to reinstall all the softwares that I've installed in last 6 months. but i think if that's the case, the upgrade itself includes all softwares that i have installed. so please tell me what's the truth?     please note that i just can't afford mentally to reinstall all apps due to slow INTERNET speed.because it may take 3 to 4 days of constant downloading.   if you are an Indian living in an suburban city,you can understand my situation.   i think the question itself explains everything.still I'll explain everything in detail.i want to upgrade my current Ubuntu 12.10 OS to 13.10(now please don't ask what is 13.10). this option is available from "about this computer" and many other places.the upgrade is of about 700mb to 800 MB.    BTW, i don't know why i think it's a rule here to put some question on questions asked here,however detailed or/and simple it may be.at least that's what i feel after looking at many questions. Custom right-click action in Unity Launcher keeps cursor busy for 20 seconds To have a "minimize window" option through right click on an icon in the Unity launcher (Ubuntu 14.04), I followed the detailed instructions  about changing a .desktop file and made a custom Firefox launcher in ~/.local/share/applications/ folder. The relevant part of the .desktop file is:  Actions=NewWindow;NewPrivateWindow;Minimize  [Desktop Action NewWindow] Name=Open a New Window Exec=firefox -new-window OnlyShowIn=Unity;  [Desktop Action NewPrivateWindow] Name=Open a New Private Window Exec=firefox -private-window OnlyShowIn=Unity;  [Desktop Action Minimize] Name=Minimize Windows Exec=sh /home/myusername/Documents/xdotool_sh/minimize.sh firefox OnlyShowIn=Unity;   The desktop action "Minimize" invokes a simple shell script, minimize.sh that has the following content:  #/bin/bash name=$1 for i in $(xdotool search --class "$name"); do     xdotool windowminimize $i done   The script uses xdotool, that can be installed from the official repositories, to find all the firefox windows, iterate over them and minimize them.  The script works and the launcher right menu option "Minimize Windows" works as well, but as soon as the windows are minimized, the mouse pointer gets in "busy" mode and stays like this for about 20 seconds (although the mouse actions are still responsive).   Does anyone know why starting a shell script from a right menu option in Unity could lead to this behaviour?  EDIT: Apparently the wait period is inevitable, as explained in Jacob Vlijm's answer. Since the mouse remains responsive, avoiding the transformation of the pointer in the spinning wheel is a partial esthetic workaround, as explained on .  EDIT2: giving the system a fake window is a better solution, as explained by Jacob below. Most OO languages prefix their interface names with a capital I, why does Java not do this?  What was the rationale for not following this convention?  To demonstrate what I mean, if I wanted to have a User interface and a User implementation I'd have two choices in Java:        Class = User, Interface = UserInterface   Class = UserImpl, Interface = User      Where in most languages:     Class = User, Interface = IUser   Now, you might argue that you could always pick a most descriptive name for the user implementation and the problem goes away, but Java's pushing a POJO approach to things and   most IOC containers use DynamicProxies extensively.  These two things together mean that you'll have lots of interfaces with a single POJO implementation.  So, I guess my question boils down to: "Is it worth following the broader Interface naming convention especially in light of where Java Frameworks seem to be heading?" How do I sort a list of dictionaries by a value of the dictionary? If $A\simeq\prod A_i$ then $\sum e_i=1$ Given $A$ commutative ring with unity. If we have rings $A_1,\dots, A_n$such that $A\simeq A_1\times \dots \times A_n$ there are $e_1,\dots,e_n$ idempotent elements of $A$ such that $e_1+\dots+e_n = 1$ and $e_i e_j =0$ if $i\neq j$.  I'm not sure how general the proof should be written...  The isomorphic relation suggest that if $f$ defines the isomorphism then there is a $(a_1,\dots, a_n)\in \prod A_i$ such that $f(a_1,\dots,a_n)=1\in A$. Could it work thinking of the idemmpotent terms as $e_i=(0,\dots,0, a_i,0, \dots,0)?$ This should satisfy $e_i e_j =0$ if $i\neq j$. But I have two problems with it (1) I don't see why $e_i$ should idempotent as no assumptions in regard to the existence of idempotent elements in $A_i$ has been made. (2) The sum $\sum e_i=1$ has no reason to be true. Uploading video to youtube in right format I am interested in calculating area under the curve (AUC), or the c-statistic, by hand for a binary logistic regression model.  For example, in the validation dataset, I have the true value for the dependent variable, retention (1 = retained; 0 = not retained), as well as a predicted retention status for each observation generated by my regression analysis using a model that was built using the training set (this will range from 0 to 1).  My initial thoughts were to identify the "correct" number of model classifications and simply divide the number of "correct" observations by the number of total observations to calculate the c-statistic. By "correct", if the true retention status of an observation = 1 and the predicted retention status is > 0.5 then that is a "correct" classification. Additionally, if the true retention status of an observation = 0 and the predicted retention status is &lt; 0.5 then that is also a "correct" classification. I assume a "tie" would occur when the predicted value = 0.5, but that phenomenon does not occur in my validation dataset. On the other hand, "incorrect" classifications would be if the true retention status of an observation = 1 and the predicted retention status is &lt; 0.5 or if the true retention status for an outcome = 0 and the predicted retention status is > 0.5. I am aware of TP, FP, FN, TN, but not aware of how to calculate the c-statistic given this information. File changed listener in Java On a "1-D line", the point that minimizes the sum of the distances is the median, WHY? There is a set A which contains some discrete points (1-dimension), for example {1, 3, 37, 59}. And I want to pick one point from A which can minimize the sum of distances between this point and others.  For this "1-D line", many people say the median is the right point, but could anybody give me a concrete proof on it? I'm looking for the title of an anthology of children's science fiction stories from the 1950s. The best of the bunch involved a Martian who visited earth, and realized we'd blown ourselves up in a nuclear war and had fallen into savagery. The Martian is a vegetarian who subsists on biscuits made by a machine he brought with him, and communicates by using one of his tentacles to feed messages direction to/from the brain of the person he's talking to.  Inadvertently left behind by his expedition, he initially befriended some children, and eventually managed to rebuild civilization. Presently he dies, but when another expedition comes a generation later, they find the civilization he rebuilt.  there were other stories, but that was the best one.   If I can't find the name of the book it was in, I'd be happy to settle for the name of the author and the title of the story. What does “” exactly mean in “echo  temp.txt”? This is more of a panic question after reading reviews on Istanbul Atatürk Airport. They said it's messy and hard to navigate, therefore I'm asking questions here in name of my girlfriend who will need to transit through this airport.  My girlfriend (Canadian national) traveling from Toronto (YYZ) to Istanbul (IST) for a couple hours (~4 hrs, so transiting) in order to come to Pristina (PRN). I would like a quick guideline on how to pass through this airport seamlessly without facing any trouble or delay.  Her tickets are both with Turkish Airlines. She purchased it via a third party (Flighthub.ca). It's only one booking and it includes both flights in it.   Once my girlfriend departs, she's obviously required to follow the signs and go to the transit area. Once she's there, will she be required to present a ticket? Where does she get that ticket? Does she get it in YYZ or does she need to visit a counter in IST and get it? After she's got the proper ticket, what does she need to do exactly? Wait in line to have her passport checked? Will they also control her carryon?  After this step, does she simply go to the PRN flight gate and that's it?   We're simply worried about Istanbul Atatürk Airport because of the recent things that happened in Turkey (coup attempt and ISIS attack) and she doesn't wanna somehow end up locked up there. I've read that the airport is a mess lately and a lot of people will be there waiting in line. So, another question: Are 4 hours enough?  I would appreciate a step by step guideline on how she needs to navigate and what she needs to do in that airport to have a smooth experience without facing any delays. Also some misc information on Istanbul Atatürk Airport would be appreciated. Splitting a string into words and punctuation Amongst the reasons for starting a bounty, one can select:     Reward existing answer      One or more of the answers is exemplary and worthy of an additional bounty.   Why not enable one to reward excellent questions in a similar manner? When would you use a WeakHashMap or a WeakReference? The use of weak references is something that I've never seen an implementation of so I'm trying to figure out what the use case for them is and how the implementation would work. When have you needed to use a WeakHashMap or WeakReference and how was it used? Suppose $a_1, a_2,\dots , a_n$ are $n$ positive real numbers with $a_1a_2 \dots a_n = 1$.   Then what is the minimum value of $(1 + a_1)(1 + a_2). . .(1 + a_n)$ ?    I think $(1 + a_1)(1 + a_2). . .(1 + a_n)$ takes its minimum value when $a_1=a_2=\dots=a_n=1$ and thus the minimum value is $2^n$.  I don't know how to prove it. Please help.