I have this group $$G= \left\{ A = (a_1 ,a_2,a_3) \in GL_3(\mathbb{C}) :a_i \in \left\{\pm  \pmatrix{1 \\ 0\\ 0} , \pm  \pmatrix{0 \\ 1\\ 0} , \pm  \pmatrix{0 \\ 0\\ 1} \right\} \right\}.$$ I know that This group is an product of cyclic group (-+1) with S3, so while I'm checking the conjugacy classes to this group, I got stuck of found the number of the elements in the classes that map to elements of order 2 ,their conjugate must be $\left(\begin{array}{ccc}0&amp;*&amp;0\\*&amp;0&amp;0\\0&amp;0&amp;*\\ \end{array}\right)$ the determinant is + or - 1, I know the order of the elements is 2 or 4, and there’s  4 classes from this kind of map, but I get stuck of found the number of the elements of the 4 classes how can I found it, I thought that I must have for least 8 elements but its wrong I’ll be thankful for help.
I want to find the conjugacy classes of this group: $$G=\{(a_1,a_2,a_3):a_i=\pm e_j\}\subset GL_3(\mathbb C)$$ I know that $S_3$ is one of the subgroups of this group, so I can check which elements of $S_3$ continues to be conjugate in our group, I know that in $Z(G)$ there is two elements $I,-I$. I know that I have to check the diagonal matrices on which ones are conjugate, but I don't know how to finish and find the others conjugacy class of this group, and I don't Know how to find the number of the elements of the conjugacy class. I'll be thankful for help
Err http://archive.canonical.com natty InRelease     Err http://security.ubuntu.com oneiric-security InRelease                Err http://extras.ubuntu.com natty InRelease                             Err http://security.ubuntu.com oneiric-security Release.gpg   Temporary failure resolving ‘security.ubuntu.com’ Err http://archive.canonical.com natty Release.gpg   Temporary failure resolving ‘archive.canonical.com’ Err http://extras.ubuntu.com natty Release.gpg   Temporary failure resolving ‘extras.ubuntu.com’ Err http://gb.archive.ubuntu.com oneiric InRelease Err http://gb.archive.ubuntu.com oneiric-updates InRelease Err http://gb.archive.ubuntu.com natty-backports InRelease Err http://gb.archive.ubuntu.com oneiric Release.gpg   Temporary failure resolving ‘gb.archive.ubuntu.com’ Err http://gb.archive.ubuntu.com oneiric-updates Release.gpg   Temporary failure resolving ‘gb.archive.ubuntu.com’ Err http://gb.archive.ubuntu.com natty-backports Release.gpg   Temporary failure resolving ‘gb.archive.ubuntu.com’  Reading package lists... Done W: Failed to fetch http://gb.archive.ubuntu.com/ubuntu/dists/oneiric/InRelease       W: Failed to fetch http://gb.archive.ubuntu.com/ubuntu/dists/oneiric-updates/InRelease   W: Failed to fetch http://gb.archive.ubuntu.com/ubuntu/dists/natty-backports/InRelease   W: Failed to fetch http://security.ubuntu.com/ubuntu/dists/oneiric-security/InRelease   W: Failed to fetch http://archive.canonical.com/ubuntu/dists/natty/InRelease   W: Failed to fetch http://extras.ubuntu.com/ubuntu/dists/natty/InRelease   W: Failed to fetch http://archive.canonical.com/ubuntu/dists/natty/Release.gpg  Temporary failure resolving ‘archive.canonical.com’ W: Failed to fetch http://security.ubuntu.com/ubuntu/dists/oneiric-security/Release.gpg  Temporary failure resolving ‘security.ubuntu.com’ W: Failed to fetch http://extras.ubuntu.com/ubuntu/dists/natty/Release.gpg  Temporary failure resolving ‘extras.ubuntu.com’ W: Failed to fetch http://gb.archive.ubuntu.com/ubuntu/dists/oneiric/Release.gpg  Temporary failure resolving ‘gb.archive.ubuntu.com’ W: Failed to fetch http://gb.archive.ubuntu.com/ubuntu/dists/oneiric-updates/Release.gpg  Temporary failure resolving ‘gb.archive.ubuntu.com’ W: Failed to fetch http://gb.archive.ubuntu.com/ubuntu/dists/natty-backports/Release.gpg  Temporary failure resolving ‘gb.archive.ubuntu.com’ W: Some index files failed to download. They have been ignored, or old ones used instead.   This is what I'm seeing when I try to run sudo apt-get update. I did an update on my instance yesterday and am now experiencing this. ${\rm Aut}(G)$ is cyclic $\implies G$ is abelian I would appreciate if you could please express your opinion about my proof. I'm not yet very good with automorphisms, so I'm trying to make sure my proofs are OK. Proof: Since ${\rm Aut}(G)$ is cyclic, ${\rm Aut}(G)$ is abelian. Thus for any elements $\phi, \psi \in{\rm Aut}(G)$ and some elements $g_i \in G$, $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4=\phi\psi(g_2)\phi\psi(g_1)=\phi\psi(g_2g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian. The proof seems to be quite straightforward, but I'd rather ask for advice. I have problems with my Tilgin Vood Router. As I try to start my router by turning the power on (captain obvious), it generally doesn't work the first 3-4 times. This is getting very annoying.  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I mean rarely drop caches and keep once opened programms in ram longer, preload apps on boot, etc. How do I pack textures for sending off to a render farm Whenever i try to send a project off to a render farm such as sheepit, it says it needs the textures packing with the file in order to render correctly.  How would I do this? Extra blue line appearing in the top bar (Firefox only) Whenever we click on the Inbox or the Achievements in the top bar, an extra blue line appears near by them. This is reproducible on Stack Overflow and Anime &amp; Manga(as I use only those) but not on Meta Stack Overflow. Here is a snapshot of that.    I'm using Firefox 25.0.1 on Win7 Pro.  Edit: Just to be double sure, I restarted the FF and still I'm able to see it. Here is another snapshot, with the 2 distinct sites. Suppose that $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$. How can I show that if $y_1$ and $y_2$ vanish at the same point then they are not linearly independent?   Here is my attempt to prove the problem. Since $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$, it is enough for me to show that the Wronskian of $y_1$ and $y_2$ denoted by $W(y_1,y_2)$ is zero. Now since $y_1$ and $y_2$ vanish at the same point say $p$ we have $y_1(p)=0$ and also $y_2(p)=0$.   Solving for $W(y_1,y_2)(p)$ we have: $y_1(p)y_2'(p)-y_2(p)y_1'(p)=0$ since $y_1(p)=0$ and also $y_2(p)=0$.   Am I correct? Thanks How force Chrome to use google.com search engine without https I'm trying to wrap my head around how SQL Server decides to grant certain rights based on a users login privileges.  If a user is part of a group "READ ONLY" which has access to DB1, but they are also a part of a group "SysAdmin" which has full db owner rights to the same DB1, which rights are they given when they log in?  On a similar note. What if they are part of a group but also have a standalone account with other middle of the road privileges? Server login attempts NOT blocked after configuring fail2ban So, I installed (and configured) fail2ban on three Ubuntu VPS servers. I configured (imo) a maximum of 3 failed login attempts (via Putty as root) within 2 minutes. Then I restarted the fail2ban service (according to ).    But when I look at my auth.log files in /var/log/ I still see 4, 5 or more failed attempts from certain IP's. So obviously I did something wrong as the IP is not blocked after 3 failed attempts.   It might have to do something with iptables, but that's where it gets a bit "foggy" for me..   Not sure how to setup a max-three-failed-login-attempts rule..  Additional: I see that the login attempts mostly have different ports for every attempt. Does that influence the count to max-three?  Because I see a port being "poked" twice and then no more attempts. Then the IP tries a different port (again 2 times). (all bots of course) Why does everyone tell me writing code like this is a bad practice?  if (foo)     Bar();  //or  for(int i = 0 i &lt; count; i++)     Bar(i);   My biggest argument for omitting the curly braces is that it can sometimes be twice as many lines with them. For example, here is some code to paint a glow effect for a label in C#.  using (Brush br = new SolidBrush(Color.FromArgb(15, GlowColor))) {     for (int x = 0; x &lt;= GlowAmount; x++)     {         for (int y = 0; y &lt;= GlowAmount; y++)         {             g.DrawString(Text, this.Font, br, new Point(IconOffset + x, y));         }      }  }  //versus using (Brush br = new SolidBrush(Color.FromArgb(15, GlowColor)))     for (int x = 0; x &lt;= GlowAmount; x++)         for (int y = 0; y &lt;= GlowAmount; y++)             g.DrawString(Text, this.Font, br, new Point(IconOffset + x, y));   You can also get the added benefit of chaining usings together without having to indent a million times.  using (Graphics g = Graphics.FromImage(bmp)) {     using (Brush brush = new SolidBrush(backgroundColor))     {         using (Pen pen = new Pen(Color.FromArgb(penColor)))         {             //do lots of work         }     }  } //versus using (Graphics g = Graphics.FromImage(bmp)) using (Brush brush = new SolidBrush(backgroundColor)) using (Pen pen = new Pen(Color.FromArgb(penColor))) {     //do lots of work }     The most common argument for curly braces revolves around maintance programming, and the problems that would ensue by inserting code between the original if statement and its intended result:  if (foo)     Bar();     Biz();     Questions:   Is it wrong to want to use the more compact syntax which the language offers? The people that design these languages are smart, I can't imagine they would put a feature which is always bad to use. Should we or Shouldn't we write code so the lowest common denominator can understand and have no problems working with it?  Is there another argument that I'm missing? 16.04 Upgrade: Failed to fetch Empty files can't be valid archives Not able to understand the meaning of question? car moving with a speed of 40 km / h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km / h, what is the minimum stopping distance  Does this mean that after 2 m, You will apply brakes and then it stops. What is the retardation value  with which you are brakes.I am not understand the meaning of the values. Had it not been vs If it were not for Probability that no two consecutive heads occur? A fair coin is tossed $10$ times. What is the probability that no two   consecutive tosses are heads?   Possibilities are (dont mind the number of terms):  $H TTTTTTH$, $HTHTHTHTHTHTHT$.  But except for those,  let $y(n)$ be the number of sequences that start with $T$  $T _$, there are two options, $T$ and $H$ so,  $y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)$  Let $x(n)$ be the number of sequences that start with $H$,   $H _$ the next option is only $T$ hence,  $x(n) = y(n - 1)$   The total number of sequences $F(n)$ is:  $$F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)$$  We are after $F(10)$,  $$F(10) = 2y(9) + y(8)$$  $$F(3) = 2y(2) + y(1) = 1 + 4 = 5$$  $$F(4) = 2y(3) + y(2) = 6 + 2 = 8$$  But I'm not quite sure where this will take me though. Decode Base64 data in Java why arcane legacy steps to boot from USB with GPT and UEFI? To run the live AVG rescue "disk" on USB for Windows 10, how is it made to work with a GPT and UEFI system?  (Possible solution listed below.)  I enabled the boot menu so that pressing F12 will display a list to select from.  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How can I make my microSD card writable?