I am trying to find the conjugacy classes of the Dihedral group $D_{12}=\{e,a,a^2,a^3,a^4,a^5,b,ba,ba^2,ba^3,ba^4,ba^5\}$   where $a^6=b^2=e$ and $ab=ba^5$. I know that one of the conjugacy classes will be $C_e=\{e\}$ but I don't understand how you find the others. All help appreciated.
I'm trying to understand this proof and I can follow it up the the second last paragraph where it states what happens if n is odd/even. I don't understand why there is just one conjugacy class when n is odd and two when n is even.    Help much appreciated, thanks!
Is there a way that I can configure Visual Studio 2008 to understand CamelCase?  Specifically, I'd like to be able to get Ctrl + right or left cursor to take me to a subsection of a variable or type name. i.e., if my cursor was at the start of this line: LongNamedExampleClass longNamed = new LongNamedExampleClass();  and I hit Ctrl + right, I'd like to jump to &quot;Named&quot; in the class name, rather than the start of the variable name.  If I hit it again, I'd like to go to &quot;Example&quot;.  I hope that makes sense. I used this navigational feature in Eclipse quite a lot, and find it difficult to unlearn something so ingrained. How can I simplify the use of flags with Views Bulk Operations? I have an administrative display (made with views) that lists content that is pending approval.  In the case of my site, approval means flagging with the approval_for_pub flag (which triggers a custom rule).   allows nodes to be flagged/unflagged all at once by clicking a handy button:    However, upon clicking this button, by default it shows all possible flags and the flag/unflag option:    This is a problem on my site because some of the people responsible for approval get confused by the names of the flags and select the wrong option.  So, what I would like to do is create a VBO option that, when the button is pressed, automatically flags all applicable content with the approval_for_pub flag.  How can I approach this?  It seems like a small detail, but this step happens several times a day, so the chance of making a mistake is quite high. I have some files with TikZ graphics that have suddenly changed appearance since I upgraded to PGF-2.10.  Here is a minimal example:  \documentclass{minimal} \usepackage{tikz} \begin{document} \pgfversion\par \begin{tikzpicture} \draw[blue] plot[domain=-1:1] (\x,{\x^2 + \x^3}); \draw[xshift=4cm,blue] plot[domain=-1:1] (\x,{(\x)^2 + (\x)^3}); \end{tikzpicture} \end{document}   In PGF 2.00 I had this:    while in 2.10 I have this:    The one on the left in PGF 2.10 looks like the graph of y=x^3 but could also be y=x^3 + x|x|.   This has been  on sourceforge and  on the pgf-users mailing list.  Although I haven't checked myself yet, the bug remains open and some of the commenters below report that problems still exist in the CVS version.   Geoffrey Jones notes the bug is in the 2.00 CVS version:     Hi Matthew, not an answer, just an FYI   that's too large for the comment box.   My compile of your code through pgf   v2.00 produces the result you see with   v2.10, above:       I.e., the difference you observe seems   to pre-date v2.10.   There is a workaround by inserting parentheses.  But since I have so many old lecture files I'd rather find a solution that doesn't involve rewriting my TikZ code.  And I'm squeamish about using the CVS bleeding edge code even if it were fixed, since I don't want to introduce other problems.  Ideally I'd like a simple patch that I can stick at the preamble, or just fix one file in the PGF distro. Remove sudo privileges from a user (without deleting the user) Date Not Saving in Mysql from Php Registration Form I'm looking to find out if there's any easy way to calculate the number of ways to tile a $3 \times 2n$ rectangle with dominoes. I was able to do it with the two codependent recurrences  f(0) = g(0) = 1 f(n) = f(n-1) + 2g(n-1) g(n) = f(n) + g(n-1)   where $f(n)$ is the actual answer and $g(n)$ is a helper function that represents the number of ways to tile a $3 \times 2n$ rectangle with two extra squares on the end (the same as a $3 \times 2n+1$ rectangle missing one square).  By combining these and doing some algebra, I was able to reduce this to  f(n) = 4f(n-1) - f(n-2)   which shows up as sequence , confirming that this is the correct recurrence.  The number of ways to tile a $2 \times n$ rectangle is the Fibonacci numbers because every rectangle ends with either a verticle domino or two horizontal ones, which gives the exact recurrence that Fibonacci numbers do. My question is, is there a similar simple explanation for this recurrence for tiling a $3 \times 2n$ rectangle? I have a notebook with no Scroll Lock key (or any shortcut using the 'Fn' key).  Somehow Excel 2010 got the Scroll Lock on. I see it left down in the status bar and now I can't navigate through cells using the arrow keys, which is pretty annoying. It wasn't turned on some hours ago, so I think I must have hit some Scroll Lock toggle shortcut in Excel 2010 by mistake.  Anyone knows what hot to turn off Scroll Lock in Excel (shortcut or menu option)? We are currently developing a site that currently has 8 million unique pages that will grow to about 20 million right away, and eventually to about 50 million or more.  Before you criticize... Yes, it provides unique, useful content. We continually process raw data from public records and by doing some data scrubbing, entity rollups, and relationship mapping, we've been able to generate quality content, developing a site that's quite useful and also unique, in part due to the breadth of the data.  It's PR is 0 (new domain, no links), and we're getting spidered at a rate of about 500 pages per day, putting us at about 30,000 pages indexed thus far. At this rate, it would take over 400 years to index all of our data.  I have two questions:   Is the rate of the indexing directly correlated to PR, and by that I mean is it correlated enough that by purchasing an old domain with good PR will get us to a workable indexing rate (in the neighborhood of 100,000 pages per day). Are there any SEO consultants who specialize in aiding the indexing process itself. We're otherwise doing very well with SEO, on-page especially, besides, the competition for our "long-tail" keyword phrases is pretty low, so our success hinges mostly on the number of pages indexed.   Our main competitor has achieved approx 20MM pages indexed in just over one year's time, along with an Alexa 2000-ish ranking.  Noteworthy qualities we have in place:   page download speed is pretty good (250-500 ms) no errors (no 404 or 500 errors when getting spidered) we use Google webmaster tools and login daily friendly URLs in place I'm afraid to submit sitemaps. Some SEO community postings suggest a new site with millions of pages and no PR is suspicious. There is a , too, in order to avoid increased scrutiny (at approx 2:30 in the video).      Clickable site links deliver all pages, no more than four pages deep and typically no more than 250(-ish) internal links on a page. Anchor text for internal links is logical and adds relevance hierarchically to the data on the detail pages. We had previously set the crawl rate to the highest on webmaster tools (only about a page every two seconds, max). I recently turned it back to "let Google decide" which is what is advised. Prove that in any GCD domain every irreducible element is prime Missing processing menu in QGIS? I'm sure there is a really simple explanation to this but haven't found any pointers on this forum or elsewhere. I seem to have lost the 'processing' menu from QGIS and can't figure out how/why!? Any ideas what I might have done to make this happen?  I'm running QGIS 2.8 built against GDAL 1.10 on Ubuntu 14.04. For some reason my Steam just broke today after working perfectly since I installed it a few days ago.  It was originally installed via the deb from the Steam site. So I uninstalled that using the Ubuntu Software Center. Then I installed steam-launcher. 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