A cyclic normal subgroup commutes with the commutator subgroup
Commutator subgroup and cyclic normal subgroup
I think it would be useful if we could join across databases in the data-explorer.  For example, I'd like to query up a list of questions that I had a hand in migrating away from SO, and see what percentage of them were subsequently closed or re-migrated.  I realise I could download the various data dumps and do this on my own machine, but it would be nice if it were allowed on the online version.  Is this possible/practical? When I have equations, as in  \begin{equation} ... \end{equation}   Can I arbitrarily assign the numbers to the equations? Instead of the usual order (1), (2), (3), ..., I'd like something like (1), (2.1), (2.2), (2.3), (3.1), (3.2), (4), ... How to change font-size, number of rows / columns in the console I'm running a Linux server without X11 or any kind of GUI, with a modern 22" LCD monitor.  Given the huge size of my monitor, I would like to increase the number of rows and columns in my console, how can I go about achieving it? Words that mean ''to say something unintentional'' Can anybody please suggest how I can compute the moment generating function of the inner product of two Gaussian random vectors, each distributed as $\mathcal N(0,\sigma^2)$, independent of each other? Is there some standard result available for this? Any pointer is highly appreciated. Is there a way to detect if a browser window is not currently active? Number Theory: Prove that $m | \phi(a^m - 1)$ Dynamic loading of python modules algebraic element over a field Let $F$ be a subfield of a field $K$, $a$ an element of $K$.  Prove that $a$ is algebraic over $F$ iff $F[a]=F(a)$.  For the first direction I've tried this: $a$ is algebraic over $F$ $\Rightarrow$ there is a polynomail $f$ over $F$ such that $f(a)=0$ $\Rightarrow$ $1,a,a^2,...,a^m$ are linearly dependent $\Rightarrow$ $\deg(F[a])&lt;m$.    but it contains $f(a)$ so they are equal..  is that correct? about the other direction.. i simply have no clue Bounded sequence which is not convergent, but differences of consecutive terms converge to zero I have a question that says "Show that there is a bounded sequence $x_n$ which is not convergent but has the property that $x_n - x_{n+1} \to 0$ as $n \to 0$.  What does this mean? Do I need to come up with an example or does the problem actually want me to prove such proposition?  By the way, I see that this sequence looks like Cauchy because of $x_n - x_{n+1} \to 0$ as $n \to 0$, but it is obviously not. If the dual spaces are isometrically isomorphic are the spaces isomorphic? I am currently using a MacBook Pro 13&quot; Mid 2018 with Touch Bar. Today macOS Big Sur was released and I tried to download it but it's getting stuck at 484.4 MB. Initially, it shows 16-17 minutes left, but as soon as it reaches 484.4 MB it's stuck and the timer goes till 10 hours! My Wi-Fi is totally stable with a speed of 100 Mbps. I tried to restart my Mac but it still gets stuck. Earlier there were similar issues in the App Store (even now they are). The issue was I was not able to download an app unless I connect to the US VPN. I even contacted support but they were not able to help out. It wasn't a big deal as a VPN could be used. But now even with VPN, it gets stuck. Will this schematic work as a four-bit shift register? &ndash; Schematic created using   Will this work as a four-bit shift register? (And by extension a shift register of any size?) Each vertical PMOS / NMOS transistor pair acts as a multiplexer, and the PMOS transistor to its left amplifies the voltage back up to close to V_high. I'd be interested in hearing people's feedback on this. Right now I have only three transistors (and a resistor) per bit. Can I really get it that small a ratio, or are there reasons why I need more transistors per bit?  It's probably worth noting that I don't have master/slave here, so I can only assert (Shift) long enough to move one bit toward the right and no more.  Furthermore, I'm not trying to say that I need a resistor with 100 ohms. That appeared to be the only resistor I could use to make this schematic! What I want is a resistor that has significantly less resistance than the PMOS it's next to when it's at high impedance, but enough resistance to prevent a short circuit from V_high to ground when that transistor is conducting.  Okay, now I've changed each of the BJT's to MOSFET's, and reduced the resistance of each resistor from 100 ohms to 10 ohms. I'm still going on the assumption that (shift) will only be asserted long enough for the gate delays of M4 and M5, and then of M6, so the value of (dataIn) will only make it to M6's output, and won't move to M8 until the next time (shift) is asserted. Is this enough to get this circuit behaving like a four-bit shift register? Clarification on Clausius Inequality Consider a heat reservoir which gains heat $Q$ irreversibly at temperature $T$ from the surroundings which is at temperature $T_0$. The entropy change of reservoir is then given by $\frac{Q}{T}$, while that of the surroundings is $-\frac{Q}{T_0}$.  My question is, how is this possible? According to the Clausius inequality, the entropy change of a irreversible process is greater than that due to heat transfer. Please help, thank you! I tried to find an answer to this a few years ago, but it seemed it wasn't possible then (I suspect it isn't now either! - but what the hell!)   I have a three monitor system, that I mostly use for remote desktop. However occasionally it would be useful to have remote desktop on two monitors and use the other monitor for the local machine.   I know it is possible to have a single monitor full screen, or to stretch an RDP session across monitors, but this loses things like snapping etc.   P.S. this is a duplicate of  thread, but that was a couple of years ago, I'm hoping things have changed. Force JSF to process, validate and update readonly/disabled input components anyway MacBook Pro stuck in gray screen when booting I've got a Late-2011 15" MacBook Pro stuck in the gray screen when booting AFTER login. I've tried removing the drive (it's non-retina) and booting from a working bootable Yosemite installer, no luck. (it sees the startup USB drive, the progress bar progresses, but stuck at the gray screen again even without a drive.  Tried NVRAM reset (cmd option P R), it has reset (the login sound's volume changed) but nothing changed.  Tried holding D on startup, AHT doesn't start, instead, it tries to boot into the OS.  Tried Cmd V, it sometimes works and displays the verbose output, sometimes, it acts like I haven't even pressed. Again, stuck after login.  Tried booting again from USB, I select USB (after seeing both drives correctly), but no luck, it's also stuck.  I've stumbled upon this:    I've tried booting into single user mode with Cmd S, it worked ONCE, I've disabled the graphics extensions just as shown at that post, restarted, but again, nothing changed.  More weirdly, now the Cmd S doesn't work either. It just takes me to login screen without showing the terminal, just as I haven't pressed anything.  Why would this happen? Why would a Mac refuse to enter hardware test mode, single user mode, or refuse to boot from a working bootable Yosemite drive both with or without the main drive installed?  UPDATE: I've run the AHT and got an error:  4MEM/66/40000000: 0x84e0ec98  It looks like a memory issue, but I'm not sure.   UPDATE 2: I've tried reinserting the RAMs and the the problem is not reported anymore. However, the Mac still won't boot. How do I remove all loose edges? I have cut a hole in a plane, but I am left with some loose edges.  Is there an easy way to select and delete them all?  See this screenshot:    See this edges siting out with no face? Console freezes on auto-complete of /mnt/folder (CIFS) share I have a Ubuntu 14.04 server that will lock up after an undetermined amount of time when you attempt to tab-auto-complete the path of a CIFS share in the /mnt directory. I mount the share and it seems to work fine for a uncertain amount of time. I can do ls or run du on folders within the mounted share. Then I will go to tab out a folder path and my console locks up. I can start a new session but it will lock as soon as I try to mess with the mounted share; this even happens to other users if I have them try as well. I have no idea why this is happening? I've never seen this issue. A set $A \subset l_1$ is compact if and only if closed, bounded, and one other condition A set $A \subset \ell_1$ is compact if and only if $A$ is closed and bounded and given any $\epsilon &gt;0$, there exists $n_0$ such that $\sum_{k=n}^{\infty} |x_k| &lt; \epsilon$ for all $n&gt; n_0$ and $x \in A$.   It's easy to prove the part assuming that $A \subset l_1$ is compact, but I am finding difficulty to prove the other part.