Finite group of even order has an element $g \neq e$ such that $g^ 2 = e$ I am trying to prove the following result.  Let $G$ be a finite group of even order. Prove that there exists $g \in G$ where $g^2 = e$ and $g \neq e$.  Here is my attempt.  Since $G$ has even order, $|G| \geq 2$. Hence, there exists some $g \neq e$. Since $G$ is of a finite order, there must exist some power, possibly not minimal, such that $g^m = e$. (Otherwise, the order is infinite.) Let $n$ be the order of $G$. Then $n \mid m$, so $m = nk$ for some $k \in \mathbb{N}$. But $G$ is of even order, so $n = 2j$ for some natural number $j$, so $m=nk=(2j)k = 2(jk)$. We have $$g^m = e \iff g^{2jk} = (g^{jk})^2.$$  The one remaining thing to show is that $g^{jk} \neq e$, but I'm having trouble accomplishing this. (I worry, actually, that we may have $jk = n$, in which case this wouldn't work.)
Group of even order contains an element of order 2 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 $G$.
Does the proof of Bolzano-Weierstrass theorem require axiom of choice? When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of explanation. I want to prove this statement: Let $a_1, a_2, \ldots \in \mathbf{R}$, and $(a_n)_{n\in\mathbf{N}}$ is bounded, then $(a_n)$ has some convergent subsequence.  The proof is as follows. Since $(a_n)$ is bounded, for all $n\in\mathbf{N}$,   $a_n \in I = [b, c]$.  Now, let $I_0 = I$ and if $I_n = [b_n, c_n]$,  we define $d_n = (b_n+c_n)/2$ and  if infinite terms of $(a_n)$ is included in $[b_n, d_n]$(resp. $[d_n, c_n]$), we will define  $I_{n+1} = [b_n, d_n]$(resp. $[d_n, c_n]$).If both intervals contain infinite terms, let $I_{n+1}$ be $[d_n, c_n]$. For all $n\in \mathbf{N}$, infinite numbers of $m \in \mathbf{N}$ exist such that $a_m \in I_n$ suffices. We take the sequence of natural numbers $(n(k))_{k\in\mathbf{N}}$which suffices $n(0) &lt; n(1) &lt; \cdots &lt; n(k) &lt; \cdots$ following this procedure: Now we have already selected $a_{n(1)}, \ldots, a_{n(k)}$, there are infinite numbers of $m\in \mathbf{N}$ which suffices $n(k)&lt;m, a_m \in I_{k+1}$, so let's take the minimum m out of it. Applying this process recursively, we obtain a infinite convergent subsequence(?). I think intuitively, by only repeating this process we can't obtain countable infinite terms of subsequence because we have to repeat infinite times. How can I bypass Activation Lock? Asymptotics of uniformly continuous functions over $\mathbb R$ Steward badge can now be awarded multiple times -- backfill details As we announced in our post on  to the Review Queues, the Steward badge can now be earned more than once per queue. The purpose of this post is not to debate the merits of this decision ( already exists), but to explain the nuts and bolts of how it will be backfilled. This is how it will appear on your profile if you have earned it multiple times:  Once we push the visual design changes live across the network, this is how badge attainment will look on /review/close/stats:  For those of you who have already completed enough reviews to earn multiple Steward badges, we will be retroactively awarding them to you, starting today. The backfill script will run about twice an hour, and your achievements trophy will light up as each one is awarded. For those of you who have done thousands and thousands of reviews, things may be a bit noisy, so we thank you in advance for your patience. It may take up to two days for all of your badges to show up. Is it possible to identify what the parent program/app is from a running process? Find an equation in spherical coordinates for the surface represented by the rectangular equation The rectangular equation is $$x^2+y^2-8z^2=0$$ $$x^2+y^2=8z^2$$ Know in the relationship between rectangular and spherical coords. we can manipulate our given to fit the form: $$x^2+y^2+z^2=9z^2$$ $$\rho=x^2+y^2+z^2, \space z=\rho\cos(\phi)$$ $$\rho^2=9\rho^2\cos^2(\phi)$$ $$1=9\cos(\phi)$$ $$\frac{1}{3}=\cos(\phi)$$ $$\arccos(\frac{1}{3})=1.23 \space rads$$ And so the equation in spherical coords. is $\phi=1.23$ I know my math is correct but I have the wrong answer so I'm not sure where I went wrong. Preposition 'to' with gerund (-ing)? Check this sentence from  (US Department of Education)     Technology ushers in fundamental structural changes that can be integral to achieving significant improvements in productivity.   Cannot it be better as...     Technology ushers in fundamental structural changes that can be integral to achieve significant improvements in productivity.   The preposition to takes gerund (-ing) here. Any rule? How do I change the default file manager back to Nautilus? Does $\lim_{n\to \infty} \int^{\infty}_{-\infty} f_n(x) dx\, = \int^{\infty}_{-\infty} f(x)\,dx\,$? I'm wondering whether it's possible to downgrade sfdx-cli to use an earlier version?  I'm trying to debug an issue with using it in pipeline scripts.  When I run sfdx version --core I get sfdx-cli 7.28.7 (core) .    If I run  sfdx version I get  sfdx-cli/7.28.7-7bceba6f24 darwin-x64 node-v10.15.3 .  If I do cat package.json inside the actual package dir for sfdx-cli after doing npm install sfdx-cli@7.26.0 I get  "sfdx-cli": "^7.26.0"  Why does it then show that the version is 7.28.7?  How can I force the lower version to have preference?  I do understand that this question is not really salesforce specific but hopefully someone has run into the same problem. Annoying FrontEnd bugs in Mathematica 10 What is the maximum value of the Process ID?  Also, is it possible to change a Process ID? How can I set default value in ActiveRecord?  I see a post from Pratik that describes an ugly, complicated chunk of code:   class Item &lt; ActiveRecord::Base     def initialize_with_defaults(attrs = nil, &amp;block)     initialize_without_defaults(attrs) do       setter = lambda { |key, value| self.send("#{key.to_s}=", value) unless         !attrs.nil? &amp;&amp; attrs.keys.map(&amp;:to_s).include?(key.to_s) }       setter.call('scheduler_type', 'hotseat')       yield self if block_given?     end   end   alias_method_chain :initialize, :defaults end   I have seen the following examples googling around:    def initialize      super     self.status = ACTIVE unless self.status   end   and    def after_initialize      return unless new_record?     self.status = ACTIVE   end   I've also seen people put it in their migration, but I'd rather see it defined in the model code.  Is there a canonical way to set default value for fields in ActiveRecord model? In  (the original movie), Marty comes back from 1955 10 mins early to prevent Doc from being shot (which he fails at). He then rolls up to the Lone Pine Mall and sees himself hit 88 mph and shoot back to 1955 (again). Marty then carries on with this life as usual. What happens to the Marty that goes back to 1955 at the end of the film?  This obviously is a recursive problem where it would create another timeline and Marty would keep coming back 10 mins early to view another Marty zip back to 1955? ensurepip is disabled in Debian/Ubuntu for the system python I am trying to make a virtual environment for the development of my Django application. The commando's which I am using:   vagrant@vagrant:/var/www/djangogirls$ python3 -m venv myvenv The virtual environment was not created successfully because ensurepip is not available.  On Debian/Ubuntu systems, you need to install the python3-venv package using the following command.      apt-get install python3-venv  You may need to use sudo with that command.  After installing the python3-venv package, recreate your virtual environment.  Failing command: ['/var/www/djangogirls/myvenv/bin/python3', '-Im', 'ensurepip', '--upgrade', '--default-pip']   vagrant@vagrant:/var/www/djangogirls$ sudo apt-get install python3-venv Reading package lists... Done Building dependency tree        Reading state information... Done python3-venv is already the newest version (3.5.1-3). The following packages were automatically installed and are no longer required:   javascript-common libjs-jquery libjs-sphinxdoc libjs-underscore python-pbr python-pkg-resources   python-six python-stevedore python3-virtualenv virtualenv virtualenv-clone Use 'sudo apt autoremove' to remove them. 0 upgraded, 0 newly installed, 0 to remove and 108 not upgraded.  vagrant@vagrant:/var/www/djangogirls$ python3 -m ensurepip ensurepip is disabled in Debian/Ubuntu for the system python.  Python modules for the system python are usually handled by dpkg and apt-get.      apt-get install python-&lt;module name&gt;  Install the python-pip package to use pip itself.  Using pip together with the system python might have unexpected results for any system installed module, so use it on your own risk, or make sure to only use it in virtual environments.   vagrant@vagrant:/var/www/djangogirls$ rm -r myvenv/   vagrant@vagrant:/var/www/djangogirls$ python3 -m venv myvenv The virtual environment was not created successfully because ensurepip is not available.  On Debian/Ubuntu systems, you need to install the python3-venv package using the following command.      apt-get install python3-venv  You may need to use sudo with that command.  After installing the python3-venv package, recreate your virtual environment.  Failing command: ['/var/www/djangogirls/myvenv/bin/python3', '-Im', 'ensurepip', '--upgrade', '--default-pip']   As you can see, I am trying to make a myvenv, which cannot be created due to the python3-venv missing. I have already installed this, but the ensure pip is missing. After searching it seems that the system (Ubuntu 16.04) discourages the usage of the package. Could someone help me work around this problem? I have a Student object and have a pick list field  "Student Status" in this object. Status can be Full-Time OR Part Time. Now my requirement is i want this student status list, Part-Time or Full-Time in apex controller. how can i do this?  If i get a student record it will only give me that particular student status Full-Time or Part-Time. But i need all options that are available in student status field.  Please help me Topology: Proof that a finitely generated cone is closed In Java, are there clear rules on when to use each of access modifiers, namely the default (package private), public, protected and private, while making class and interface and dealing with inheritance? Clicking to reload an updated answer removes the code snippet buttons and removes syntax highlighting How can I retrieve mp3 metadata in Python?