I'm running Ubuntu 18.04.  I first downloaded (and installed) a Debian image of Skype from skype.com. But because that installation wasn't working, I downloaded an alternative version through the "Ubuntu Software" app, which ended up working just fine.   Now I want to remove the first, non-working version, but it seems to be caught in a weird state. The removal command seems to run to successful completion:  &gt; sudo dpkg --remove skype:i386 (Reading database ... 170025 files and directories currently installed.) Removing skype:i386 (4.3.0.37-1) ... Processing triggers for gnome-menus (3.13.3-11ubuntu1.1) ... Processing triggers for desktop-file-utils (0.23-1ubuntu3.18.04.1) ... Processing triggers for mime-support (3.60ubuntu1) ... Processing triggers for hicolor-icon-theme (0.17-2) ...   ...but:   the app still shows up in the list of installed applications when I click "Show Applications" the listing in "Show Applications" appears without an icon selecting the listing from "Show Applications" doesn't launch anything (nor present any sort of error prompt)   How can I remove this zombie entry? Where is the registry that drives "Show Applications"?  UPDATE 1  Per @GabrielaGarcia's suggestion, the results of running sudo dpkg -P &lt;filename&gt; are below. For some reason the I couldn't use auto-complete to fill in the filename. When I type "s" and then tab for the autocomplete, it actually just writes out "skype" (as opposed to the filename), but there is no Skype installed on my system at the moment.  homerman@~/Downloads&gt; ls -l total 1050536 drwxr-xr-x  2 homerman homerman       4096 Nov 11 19:56 ./ drwxr-xr-x 23 homerman homerman       4096 Nov 11 21:04 ../ -rw-rw-r--  1 homerman homerman   20112698 Nov 11 16:11 skype-ubuntu-precise_4.3.0.37-1_i386.deb  homerman@~/Downloads&gt; sudo dpkg -P skype-ubuntu-precise_4.3.0.37-1_i386.deb dpkg: error: you must specify packages by their own names, not by quoting the names of the files they come in  Type dpkg --help for help about installing and deinstalling packages [*]; Use 'apt' or 'aptitude' for user-friendly package management; Type dpkg -Dhelp for a list of dpkg debug flag values; Type dpkg --force-help for a list of forcing options; Type dpkg-deb --help for help about manipulating *.deb files;  Options marked [*] produce a lot of output - pipe it through 'less' or 'more' !   UPDATE 2  I've determined the package name for this archive is skype:i386. Below are the results of trying to run the dpkg commands to remove and purge - the system doesn't recognize this as an installed application.  homerman@~/Downloads$ sudo dpkg -r skype:i386 dpkg: warning: ignoring request to remove skype:i386 which isn't installed  homerman@~/Downloads$ sudo dpkg --purge skype:i386 dpkg: warning: ignoring request to remove skype:i386 which isn't installed
I am using Ubuntu 16.04, with Gnome 3.18. Using extensions I have gotten rid of the dash panel and only use the Applications Menu in the upper left-hand corner of the screen. (I don't have Unity.) What I can't figure out is how do you add a new program, move software around to different categories, and create new categories?  Also, any keyboard shortcut to access the Applications Menu?
Why does Lang describe a field as a union and compositum of its subfields in this manner? I am reading Serge Lang's Algebra, revised 3rd edition, and on page 226, the author makes the following definitions and observations:   If $E,F \subset L$, the compositum of $E$ and $F$ in $L$ is defined to be the smallest subfield of $L$ containing both $E$ and $F$. It is denoted $EF$. If $k \subset E$ and $\alpha_1,\dots,\alpha_n \in E$, then $k(\alpha_1,\dots,\alpha_n)$ is defined to be the smallest subfield of $E$ containing $k$ and $\alpha_1,\dots,\alpha_n$. Observe that $$ E = \bigcup k(\alpha_1,\dots,\alpha_n), $$ where the union is taken over all finite families $\{ \alpha_1,\dots,\alpha_n \}$ of elements of $E$. The compositum of an arbitrary family of subfields of a field $L$ is defined as the smallest subfield of $L$ containing all fields in the family. $E$ is finitely generated over $k$ if there is a finite family $\{ \alpha_1,\dots,\alpha_n \}$ of elements of $E$ such that $E = k(\alpha_1,\dots,\alpha_n)$. Observe that $E$ is the compositum of all its finitely generated subfields over $k$.   My question is regarding the two observations (point nos. 3 and 6). It appears to me that it is enough to take union (and compositum) over all subfields generated by a single element. For example for point no. 3, I can write $$ E = \bigcup_{\alpha \in E} k(\alpha) $$ because $E$ is clearly contained in the RHS, and each $k(\alpha)$ is contained in $E$ and hence so is the union, implying that $E$ contains the RHS. Similarly for point no. 6.  So, why does Lang emphasise to take the union and compositum over all finitely generated subfields? Is there some perspective that he wishes to emphasise that I am missing? Any help in understanding this will be appreciated. $\mathbb Q$ is not the direct product of two non-trivial groups. I know this has been asked before, in a bunch of places, so I expect it to be closed as a duplicate at some point, but I'd like to know if my own proof works. If the proof is incorrect, please limit answers to pointing out the errors. If the proof is correct, then thoughts on improving it would be appreciated.  Prove that $\Bbb Q$ is not the direct product of two non-trivial groups.  —Aluffi exercise II.3.5. My proof Suppose that $\Bbb Q = G \times H$, neither $G$ nor $H$ trivial. First note that $0_G$ is the only element of $G$ with finite order: If $m\ne 0$ and $m\cdot g=0_G$, then $m\cdot(g,0_H)=0_{G\times H}$, contradicting the fact that $\Bbb Q$ has no non-zero elements of finite order. Suppose that for some $m\ne 0$ and some $n\ne 0$, $\pi_G(m/n)=0_G$. Then $0_G=n\cdot\pi_G\left(\frac m n\right)=\pi_G\left(n\cdot\frac m n\right)=\pi_G(m)=m\cdot\pi_G(1)$. Since $\pi_G(1)$ has finite order, we conclude that $\pi_G(1)=0$. Since $\Bbb Z=\langle 1\rangle$, $\pi_G$ maps every integer to $0_G$. Now let $a$ and $b$ be non-zero integers. Then $b\cdot\pi_G\left(\frac a b\right)=\pi_G\left(b\cdot \frac a b\right)=\pi_G(a)=0_G$. As above, this shows that $\pi_G\left(\frac a b\right)=0_G$. Thus if $\pi_G$ is not an injective function, then it must map everything to $0_G$, so $G$ will be a trivial group. The same result holds for $\pi_H$. Since by assumption, $G$ and $H$ are non-trivial, $\pi_G$ and $\pi_H$ are both injective functions, which is of course impossible. On a series involving harmonic numbers How do you pull first 100 characters of a string in PHP I am looking for a way to pull the first 100 characters from a string variable to put in another variable for printing.    Is there a function that can do this easily?  For example:  $string1 = "I am looking for a way to pull the first 100 characters from a string variable to put in another variable for printing."; $string2 = 100charfunction($string1); print $string2   To get:  I am looking for a way to pull the first 100 characters from a string vari $H$ is normal subgroup of $G$ iff $gHg^{-1}=H$ for every $g$ of $G$ Given the following definition of normal subgroup     A subgroup $H$ of a group $G$ is said to be normal if, for every $g\in G$: $$gH=Hg$$   I've tried to show that $H\mathrel{\unlhd} G$ if and only if we have $gHg^{-1}=H$ for every $g\in G$:  (if) Let $H\mathrel{\unlhd} G$, $g\in G$; if $l\in gHg^{-1}$, there exists $h\in H$ such that $l=ghg^{-1}$ and since $gh$ = $h'g$ for some $h'\in H$, then $l=ghg^{-1}=h'(gg^{-1}) = h'\in H$. Let $h\in H$; we can write $h$ as $hgg^{-1}$, and since $hg = gh'$ for some $h'\in H$, we obtain $h = hgg^{-1} = gh'g^{-1}\in ghg^{-1}$. So $gHg^{-1}=H$.  (only if) Let $gHg^{-1}=H$, $g\in G$; if we take an $l\in Hg$, $l = hg$ for some $h\in H$, we note immediately that $l=hg=(gh'g^{-1})g=gh'\in gH$. What should I do to show that $gH\subset Hg$?  I'm also wondering if this mess is correct, and if it is possible a more elegant proof of this fact. Elsarticle - Abstract two columns wide I am currently trying to create an article in elsarticle style, twocolumn. The title and text end up in two columns, the abstract however is made two columns wide. Any ideas what I am doing wrong? Obviously I want my abstract to also be only in the first column.  \documentclass[a4paper,10pt,twocolumn,preprint]{elsarticle}  \usepackage{lipsum}  \journal{TexExchange}   \begin{document}     \begin{frontmatter}       \title{Two-column abstract}       \begin{abstract}         \lipsum[1]      \end{abstract}       \begin{keyword}        abstract      \end{keyword}    \end{frontmatter}   \section{Outline}    \lipsum[2-3]  \end{document} "Belongs on" comments Remote Desktop Session Black after Minimize PROBLEM: When I minimize a remote desktop session and restore it, the remote desktop screen shows up black. This only happens when connecting to a particular computer.   DETAILS: If I start clicking around in the black area, portions of the screen will start redrawing and showing up correctly. For example, if I leave a window open in the remote session and click where that window is located on the remote computer, then that window--and only that window--will redraw, and sometimes a portion of that window won't redraw (usually the toolbar).  And to clarify--the window only has to be minimized momentarily, so it doesn't seem to be a timeout issue. Clicking or typing in the remote session still causes the remote computer to respond appropriately. Disconnecting from the session and reconnecting restores the whole screen image, as does clicking all over the place in the black image (causing each section to redraw).  CONFIGURATION: This problem only happens for me when connecting to a particular computer (a W2K Server box configured to allow remote administration) and only with certain client computers. I've tried 7 different client computers with various versions of Remote Desktop (the OSes were: Win2K, Server 2003, Server 2008, Windows 7 RC, 3 XP) and two of them exhibit the problem (one is one of the XP boxes and the other is Windows 7). Those same computers can RDP to other computers without problem.  RESOLUTION ATTEMPTS: I have tried the following:   Disable the LOCAL screen saver as mentioned on  Turned off bitmap caching in the client, as mentioned on many forums. Updated to version 6.1 of the remote desktop client Using mRemote (I doubted this would work since it uses MS's code for connecting to RDP servers) Turning off all video acceleration.   QUESTION: Any ideas on what is causing this? If GCD $(a_1,\ldots, a_n)=1$ then there's a matrix in $SL_n(\mathbb{Z})$ with first row $(a_1,\ldots, a_n)$ Let $n \geq 2$. Let $a_1, a_2, \ldots, a_n$ be $n$ integers such that $\gcd\left(a_1, a_2, \ldots, a_n\right) = 1$. Prove that there exists a matrix in $\operatorname{SL}_n\left(\mathbb{Z}\right)$ whose first row is $\left(a_1, a_2, \ldots, a_n\right)$.   Since the gcd of the integers $a_1,\ldots, a_n$ is $1$, there exists weights $x_i \in \mathbb{Z}$ such that $a_1x_1+\cdots+ a_nx_n=1$. My two ideas are (a) to brute force construct an $n\times n$ matrix with first row $a_1,\ldots ,a_n$ and to construct the remaining rows such that the determinant is $\sum a_ix_i=1$ or (b) to use induction.  (a) (Constructive) This is tedious since once I find a way to construct the remaining $n-1$ rows to ensure that $a_1x_1$ appears in the determinant, I am not sure how to modify these $n-1$ rows to ensure  that only the terms $a_ix_i$ appear in the cofactor expansion. If such a matrix exists, I'd like to see it.  (ii) (Non-constructive) If I proceed by induction then the base case $n=2$ is settled since I can choose the 2nd row to be $-x_2, x_1$ so that the determinant is $a_1x_1-a_2(-x_2)=1$. However, I'm not sure how to use the inductive hypothesis to show that if I can construct such an $n\times n$-matrix then I can construct an $\left(n+1\right) \times \left(n+1\right)$-matrix with the desired property. In particular, if the gcd $(a_1,\ldots ,a_{n+1})$ is $1$, it is not necesarry that the gcd of any $n$ of these terms is $1$, so induction may not even apply here.  How can I construct such a matrix or prove that one exists (without necessarily constructing it)? I'm looking for a phrase that describes trying to fix problems caused by underlying issues, with the result of compounding the issue. I usually call this "targeting surface problems rather than root problems", but my terminology is verbose.  Example 1: A government bans buying and selling bananas in the state. Later, it passes a clause allowing citizens to purchase imported bananas.  Example 2: A divided highway is erected with two lanes going in each direction but, since it gets lots of traffic, the shoulders are widened into new lanes. This leaves no shoulder room.  What would be a more concise phrase? Should there always be a comma after "therefore","However" etc.? Do SSD's really last 285 years? Some Sold State Disks (SSD) have a  (MTBF) of 2,500,000 hours. If you divide that by the rough number of hours in a year you get 2500000÷(365×24) ~= 285 years before there is a problem.  I know that the first generation SSD's used to have a short life - but now the estimated time before they fail seems absurdly high. Most warranties only cover them for 1-3 years it seems.  How long are the modern, SATA III drives supposed to last? \overbrace and \underbrace with square bracket How align 2 adjacent divs horizontally WITHOUT float? QGIS apply style to shapefile Duplicate notifications I receive every push notification from the iOS app twice on iPhone. The second comes immediately after the first. This happens every single time.   I also have the app installed on an iPad and signed in to the same account - could this be a possible cause?   App Version: 1.4.0 Device: iPhone7,2 OS Version: Version 9.1 (Build 13B143) I am creating an ZIP file with ZipFile in Python 2.5, it works ok so far:  import zipfile, os  locfile = "test.txt" loczip = os.path.splitext (locfile)[0] + ".zip" zip = zipfile.ZipFile (loczip, "w") zip.write (locfile) zip.close()   but I couldn't find how to encrypt the files in the ZIP file. I could use system and call PKZIP -s, but I suppose there must be a more "Pythonic" way.  I'm looking for an open source solution. Hide files from root in ubuntu I got 2 ghast tears after I thought I deflected a fireball back at the ghast (with my looting 3 sword).  But I was pretty close, so I might have just hit it. I'm not sure.  If I deflect a fireball with a looting sword and it kills the ghast, do I still get the looting bonus? Python: Finding differences between elements of a list