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Proof that $A_n$ the only subgroup of $ S_n$ index $2$. I have what seems to me a very simple proof that $A_{n}$ is the only subgroup of $S_{n}$ of index 2.  Since I've seen other people prove it with what feel like really complicated methods (), I'm wondering if I've overlooked something.  Proof: Let $H$ be any subgroup of index 2 in $S_{n}$.  Then $H\cap A_{n}$ is a normal subgroup of $A_{n}$ and since $A_{n}$ is simple, $H$ is either the trivial or the improper subgroup of $A_{n}$.  If it's the trivial subgroup it doesn't have index 2 in $S_{n}$, and otherwise the theorem is proved. How to find out where an error occurs? Custom kernel: fails to load firmware when module built-in I'd like to have all my modules built-in, but this fails with iwlagn:  iwlagn 0000:03:00.0: request for firmware file 'iwlwifi-6000-4.ucode' failed. iwlagn 0000:03:00.0: no suitable firmware found!   The microcode file exists in /lib/firmware and the whole thing works just fine if I compile iwlagn as module. I have no idea where it's looking for the file or what's wrong - any ideas? Objects disappear when looking through camera Rights if we possess Carta di Soggiorno? Could not find class I just reinstalled my Ubuntu system with 11.10, but ran into a problem when I tried to install Gimp 2.7.  I always get unmet dependencies (it's german, so I can't tell you the exact english translation):     Needs gimp-data (&lt;= 2.6.11-z) but 2.7.3-2011082002~oo should be installed   I already read the follow post, but the answers didn't really help me.  When I type in     sudo apt-get install gimp=2.6   it just says that there is no version 2.6 found. I get the same with 2.7.  I really need Gimp. But installing Gimp 2.6 isn't possible as well.  When I remove the repos for 2.7, I still get the same error. Conditional expectation of $\mathbb{E}(X - Y | (X, Y)\in\mathcal{A})$ Given two independent random variables $X \sim \mathcal{U}[-1,5]$ and $Y \sim \mathcal{U}[-5,5]$, what is $$\mathbb{E}\{Y - X | X \le 1, Y &gt; X, Y \in [-1,1] \}\,?$$  I managed to compute the respective probability   $$ \mathbb{P}\{ X \le 1, Y &gt; X, Y \in [-1,1] \} = \mathbb{P}(X \le 1) \mathbb{P}(Y &gt; X) \mathbb{P}(Y \in [-1,1]) = 1/3 * 1/2 * 1/5 = 1/30$$  How do I continue from there? Every Function in a Finite Field is a Polynomial Function The extension of smooth function What is the progression route in Terraria? Topology: Example of a compact set but its closure not compact Can anyone gives me an example of a compact subset such that its closure is not compact please? Thank you. Understanding Dirac's notation What does "in spades" mean? Prove that $a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$ In Star Wars: "The Force Awakens," how could this lightsaber have turned up where it did? In The Force Awakens:      Maz Kanata tells Rey that the lightsaber she just picked up and had visions from belonged to Anakin and Luke and now it's hers.   The lightsaber that Luke got from Obi-Wan Kenobi in A New Hope was lost/destroyed when Luke fought Darth Vader in Star Wars: The Empire Strikes Back.  In Return of the Jedi Darth Vader compliments Luke on building a new lightsaber and tells Luke that his skills are now complete.   How could that possibly be the lightsaber that Anakin used originally, and that was then passed on to Luke, when that lightsaber no longer exists?  Writers notes, director's notes, interviews, cast interviews, etc. are welcome. Thanks. Zero module of differentials implies finite extension? Prove that $\lim_{n \to \infty} \int_0^1{nx^nf(x)}dx$ is equal to $f(1)$. finding examples for a non negative and continuous function for which the infinite integral is finite but the limit at infinity doesn't exist Question:  a. Find an example for a non-negative and continuous function s.t. $\int _0^\infty f(x)dx$ is finite but the following limit doesn't exist: $\lim_{x\to \infty} f(x)$.  b. Is it possible that $\int _0^\infty f(x)dx$ is bounded but $f(x)$ is not bounded?  What we did with A: We suggested the function $|sin(x^2)|$ which has periods that get smaller and smaller until they no longer imply on the sum. But since it's a trig function it doesn't have a determinate limit. Wolfram didn't have the integral value for that func, and we were wondering if it really converges and if it is really a good example.  What we did with B: We thought about the function: $f(x)=  { x\in \Bbb N },{e^{-x} \notin \Bbb N }$ and we had a disagreement whether f(x) is integrable. I said no because similarly to Dirichlet function, one sum's limit will be 0 while the other one's will be infinite. My partner said that if I find a $\delta&lt;1$ then the definition of Riemann's integral does hold and so this integral is equal to that of $f(x)=e^{-x}$ She was trying to say this functions integral will be 0, but still it won't be bounded. I disagreed. Who's right? How can we disable triggers, Validation Rules and workflow rules in salesforce org. [on off - switch] How can we disable triggers, Validation Rules and workflow rules in salesforce org.  This becomes really needed and important when we are doing data migration(data imports) in salesforce.  The way I handled this is create a custom label and include it in workflow and validation rule logic.   example - Custom label 'Disable workflow' = True then do not fire workflows.            if its False then fire workflow  But wanted to understand if there is any more efficient way to handle these scenarios.