I came across this theorem in many of algebra books. If $G/Z(G)$ is cyclic, then $G$ is abelian. I tried to apply it when $G=p^3$ and $Z(G)=p^2$. What I got is $|G/Z(G)|=p$ so is cyclic. Therefore, $G$ is abelian according to the theorem which yields $G=Z(G)$ contradiction! So I thought either theorem doesn't apply here or we cannot have the centre of degree $p^2$. Am I right? If so, Which one is correct? Thanks.
Let $G$ be a non abelian group of order $p^3$, with $p$ prime.  I'm proving that $Z(G)$ (its center) is of order $p$. I already know how to do it by saying that its order can't be $p^3$, nor 1, and if it's $p^2$ then $G/Z(G)$ is cyclic then G is abelian, Abs!  But I've been reading another way of proving it and the author supposes that $|Z(G)|=p^2$. Then, she says: "$\exists\ x$ outside the center of $G$, of order $p$." Could someone give me a hint? Thanks!
How to toggle or turn off text selection being sent to the clipboard I would like to disable the default behavior that seems to happen with every Linux distribution that I've ever tried that any selected text is immediately sent to the clipboard (Mint, Ubuntu, Fedora, SuSE, etc.) and every window manager (Gnome, KDE, Cinnamon), and instead to behave more like the Windows implementation.    I know that this is a beloved behavior by many in Linux, and I'm sure many will think I'm an idiot.  The reason I want to do this, is that I'm a keyboard junky when navigating a GUI.  (e.g.  when I'm in Linux and I copy a URL and then switch to my browser and type Ctrl+L, it selects the address bar and moves my intended paste down one notch and replaces it with what I'm trying to overwrite.)  I know there are MANY workarounds, but I don't really care about that, what I'd prefer is to be able to toggle the behavior for the clipboard. The difference between sys.stdout.write and print? Story about protagonist finding a world that has died with a museum Prove $\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$ For positive integers Numbers  $a, b, c $ prove that  $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq  2$$ This inequality above take long time to prove it and still couldn't complete it.  How I can prove this inequality?  Any hint will help. Thanks I'm looking to implement a phase lock in an FPGA without using any external components (other than the ADC). For simplicity locking to a simple binary pulse is adequate. The frequency of the signals is ~0.1-1% of the clock. I can't use the onboard clock PLL's because they are typically:   Not configurable (set during synthesis). Jittery. Don't support the frequency I need.   I've been sorting through the literature and found a few binary phase locked loops. Most notable a "pulse steal" design that I can post a link to if desired. I have implemented and synthesized with some success, but its jitter and lock range was not as good as advertised. I've also had success using a external DVCO, but I would prefer if I can implement everything on chip.  A digital circuit design or even a hint in the right direction would be helpful (I've been banging my head against this for a while), a proven FPGA implementation would be wonderful but not expected.   ADDED 10-27-2010   The actual DPLL design that I used has a "Random Walk Filter" as the loop filter (not the "pulse steal" described earlier, going through my notes that didn't work to well), which then drives the clock pulses to the DCO. The lock-in range is set via divider in the DCO. The sensitivity of the loop is established by varying the length of the random walk.   The paper in which this is found is cited at the end of this post. After implementing parts of it myself, I found that it was actually already implemented on OpenCores, however it turns out that in the last few months the project was deleted but I have the Verilog files saved if anyone wants them.  Yamamoto, H.; Mori, S.; , "Performance of Binary Quantized All Digital Phase-Locked Loop with a New Class of Sequential Filter," Communications, IEEE Transactions on , vol.26, no.1, pp. 35- 45, Jan 1978  doi: 10.1109/TCOM.1978.1093972  URL: How can I change media volume when no media is being played? Sometimes I want to play games in (near) silence, and don't like the background music to disturb people. If I use the volume buttons, the ringtone volume is changed instead of media volume. When I'm already in the game, the volume buttons do change the volume of background music, but it's kinda too late.  Is there a way for me to change the media volume to a level I like, from the home screen, before starting apps/games? Word or Phrase for "Easily Swayed" What word or phrase best communicates the idea that a human (or other sentient being) is easily swayed, particularly between moral good and evil? Words I have found that are close are:   Impressionable - Easily influenced because of a lack of critical ability.   Implies that the subject lacks critical skills, and misses the concept of good and evil.  Gullible - Easily persuaded to believe something   Misses the concept of good and evil    I created a typical angels and demons scenario to demonstrate the polar extremes of the the spectrum. In this scenario, I lack the word to describe the humans, who can be influenced towards either end of the spectrum if impressed upon by an external entity.     The angels are devoted to good, and are immovable in their ideals. The demons are wicked, and will feign morality only to manipulate others. Both prey upon the humans, who are ____________.   I think that the key here is the spectrum of good an evil. Here are examples of other spectra and their associated words, for comparison.   Easily pushed from calm to angry: Temperamental Easily moved from determined to hopeless: Weak-willed Easily influenced by propaganda: Why does the following hold:  \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}  Can we generalize the above to      $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?   Are there some values of $x$ for which the above formula is invalid?  What about if we take only a finite number of terms? Is there a simpler formula?     $\displaystyle \sum_{n=0}^{N} x^n$   Is there a name for such a sequence?    This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: . How to remove chromium-browser settings from Ubuntu 14.04? I have already gone through   and I'm really not getting why the solutions given there are not working for me.  What I'm doing is:   sudo apt-get purge chromium-browser   rm -rf ~/.config/chromium     Then reinstalling by running the following command:   sudo apt-get install chromium-browser    This must remove the settings of chromium-browser and do a fresh installation. The browser is uninstalling, but after being installed it is containing the previous settings, moreover it's not signing me out of even Gmail! Same e-mail user, different address? I was wondering why using foo.bar@gmail.com works but not foo.bar@outlook.com.  In Gmail, foo.bar@gmail.com and foobar@gmail.com (and any combination of periods in the username) is equivalent.  Is this a hidden feature in Gmail? What do you call a person who started something that is later followed by everyone? For example, someone started to stay late at night till 8 p.m. and later everyone started to stay till 8 p.m.. Dual booting Windows and Ubuntu My laptop has 500gb hard disk in which for 250gb I installed Windows xp on C drive(primary partition). I want to install Ubuntu 12.04 in the remaing hard drive with dual boot option. When I am going to install Ubuntu it is showing something like  /boot /boot/   Can anybody help me in this regard that where and how to install Ubuntu? Finding a basis for a certain vector space of periodic polynomials I am having a little bit of trouble solving an homework question.  I found that $S={ p(x) \in R_4[x]} \big| p(x)=p(x-1) $ is a vector space.  Now I need to find some set, K that holds ${span(k)=S}$  My idea was to substitute: $ a+bx+cx^{2}+dx^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}$  But here I get something that seems a little bit weird, I'd like to look if I have any mistakes that I did not notice. and if I had mistakes, how to make my answer right. I calculated the right side of the equation and I got to  :  $a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}=a+bx-b+cx^{2}-2cx+c+dx^{3}-3dx^{2}+3dx-d+ex^{4}-4ex^{3}+6ex^{2}-4ex+e= (a-b+c-d+e)+(b-2c+3d-4e)x+(c-3d+6e)x^{2}+(d-4e)x^{3}+ex^{4}  $  From here, $ (a-b+c-d+e)+(b-2c+3d-4e)x+(c-3d+6e)x^{2}+(d-4e)x^{3}+ex^{4}=a+b(x-1)+c(x-1)^{2}+d(x-1)^{3}+e(x-1)^{4}\Rightarrow $  $(-b+c-d+e)+(-2c+3d-4e)x+(-3d+6e)x^{2}+(d-4e)x^{3}=0+0x+0x^{2}+0x^{3}+0x^{4}$  And the solution to this equation is only whan $a=b=c=d=e=0$.  I think that I am maybe not in the right direction. Do you have any Idea how to find a set $K\Rightarrow span(K)=S$ ?  Thanks in advance. Is there a way to show that $\sqrt{p_{n}} &lt; n$?  In , I show that $f_{2}(x)=\frac{x}{ln(x)} - \sqrt{x}$ is ascending, for $\forall x\geq e^{2}$. As a result, $\forall n \geq 3$ $$\frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}}\leq \frac{p_{n+1}}{ln(p_{n+1})} - \sqrt{p_{n+1}}$$ Also (and as a result), $\forall n \geq 3$ $$ \frac{p_{n}}{ln(p_{n})} - \sqrt{p_{n}} &gt; 0$$ Or $$ \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$  According to  $$\displaystyle\smash{\lim_{n \to \infty }}\frac{\pi (p_{n})}{p_{n}/ln(p_{n})}=1$$ Or, $\forall \varepsilon &gt;0$, $\exists N(\varepsilon )$: $\forall n&gt;N(\varepsilon )$ $$1- \varepsilon &lt; \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; 1+ \varepsilon$$ Or $$1- \varepsilon &lt; \frac{\pi (p_{n})}{p_{n}/ln(p_{n})} &lt; \frac{\pi (p_{n})}{\sqrt{p_{n}}}$$ As a result $\forall \varepsilon &gt;0$, $\exists N(\varepsilon )$: $\forall n&gt;N(\varepsilon )$ $$(1 - \varepsilon ) \cdot  \sqrt{p_{n}} &lt; \pi (p_{n}) = n$$  But this is not enough.  Interestingly,  is true iff function $f_{4}(x)=\pi (x) - \sqrt{x}$ is strictly ascending ($x &lt; y \Rightarrow f(x) &lt; f(y)$) for prime arguments.  If $f_{4}(p_{n}) &lt; f_{4}(p_{n+1})$ then $$\pi (p_{n}) - \sqrt{p_{n}} &lt; \pi (p_{n+1}) - \sqrt{p_{n+1}}$$ Or $$\sqrt{p_{n+1}} - \sqrt{p_{n}} &lt; \pi (p_{n+1}) - \pi (p_{n}) =1$$  And vice-versa, if $$\sqrt{p_{n+1}} - \sqrt{p_{n}} &lt; 1$$ Then $$-\sqrt{p_{n}} &lt; -\sqrt{p_{n+1}} + 1$$ Or $$\pi (p_{n})-\sqrt{p_{n}} &lt; \pi (p_{n}) + 1 -\sqrt{p_{n+1}} = \pi (p_{n+1}) -\sqrt{p_{n+1}}$$  So, if Andrica's conjecture is true then $\forall n \geq 3$ $$\pi (p_{n})-\sqrt{p_{n}} &gt; 0$$ Or $$\sqrt{p_{n}} &lt; \pi (p_{n})= n$$ How do I ensure a piece of code runs only once? How do I use variables in a sed command? I tried the following code to replace QQ with ZZ, but it doesn't do what I want:  var1=QQ sed -i 's/$var1/ZZ/g' $file   However, this code does what I want:  sed -i 's/QQ/ZZ/g' $file   How do I use variables in sed? Efficient online linear regression Is there a command like head but for line truncation? Using SQLite3 with Python 2.5, I'm trying to iterate through a list and pull the weight of an item from the database based on the item's name.  I tried using the "?" parameter substitution suggested to prevent SQL injections but it doesn't work. For example, when I use:  for item in self.inventory_names:     self.cursor.execute("SELECT weight FROM Equipment WHERE name = ?", item)     self.cursor.close()   I get the error:     sqlite3.ProgrammingError: Incorrect number of bindings supplied. The current statement uses 1, and there are 8 supplied.   I believe this is somehow caused by the initial creation of the database; the module I made that actually creates the DB does have 8 bindings.  cursor.execute("""CREATE TABLE Equipment      (id INTEGER PRIMARY KEY,      name TEXT,     price INTEGER,      weight REAL,      info TEXT,      ammo_cap INTEGER,      availability_west TEXT,     availability_east TEXT)""")   However, when I use the less-secure "%s" substitution for each item name, it works just fine. Like so:  for item in self.inventory_names:     self.cursor.execute("SELECT weight FROM Equipment WHERE name = '%s'" % item)     self.cursor.close()   I can't figure out why it thinks I have 8 bindins when I'm only calling one. How can I fix it? Gay rights – does Star Trek ever address this issue? I am aware of the picture of the man wearing the 'dress' from TNG, which promotes gender equality, but are gay rights ever addressed in Star Trek at all? (All seasons accepted.)