Questions like this have been asked before, but I just want some quick clarification on something.    Suppose that $G$ is the sum of its Sylow $p$-subgroups.  Then each Sylow subgroup must be normal, and of course we have the other direction.  If each Sylow subgroup is normal, then $G$ is the product of its Sylow subgroups.    My question that I need clarified is, "if $G$ is a sum of its Sylow subgroups, when will this be abelian, non-abelian?"
Proposition II.7.5 of Hungerford's Algebra goes     Proposition 7.5. A finite group is nilpotent if and only if it is the direct product of its Sylow subgroups.   Let $G = (\mathbb{Z}_6, +)$. $G$ is abelian, so it is nilpotent and thus by the proposition is the direct product of its Sylow subgroups.  The Sylow subgroups of $G$ are $H = (\mathbb{Z}_3, +)$ and $K = (\mathbb{Z}_2, +)$, so the proposition says $G = H \times K$. In particular, $0 \in H$ and $0 \in K$, so $(0,0) \in H \times K = G$... but that's not true, since $G = \{0,1,2,3,4,5\}$.  Where is/are my misunderstanding(s)?
Automatically login and lock Height of colorbox Can't SSH in as root I'm using the default installation of openssh-server for my Ubuntu 14.04 server. I can connect through the console as root like normal, with my username and password. When I try to SSH with the same username/password, however, I repeatedly get Access Denied errors. /var/log/auth.log reports Failed password for root from &lt;ip address&gt; port &lt;port&gt; ssh2, but I'm entering the correct password.  Why can't I connect to the server via SSH, even though the username and password ARE correct? Why is conservation of angular momentum considered a law? We know that angular momentum is defined as the cross product of position and linear momentum. By taking the time derivative, we can deduce that the time rate of change of the angular momentum equals the net torque.  Then why is the conservation of angular momentum considered a law, when we can easily show it is conserved when no net torque is applied, by using Newton's laws?  It truly confuses me. I am trying to find the formula for the number $f(n,k)$ which is defined as the number of $k$-subsets of $[n]$ in which no two elements are consecutive numbers in $[n]$.  From the simple thinking I had made up following recursion:  $$f(n,k) = f(n-1,k) + f(n-2, k-1) $$  This is true since I can construct without last element of $[n]$ then make up $k$-subsets, then otherwise must contain the last element of $[n]$ then construct remaining $k-1$ elements among $[n-2]$ since with the last one, I can't use $n-1$.  However, I had never learned any techniques regarding changing those recursion into a well-summarized formula in terms of $n$ and $k$.  Any guidance would be appreciate. SC2272-M4 Behaviour when using relay SRD-05VDC-SL-C 5V DC I'm currently encountering a problem when using a relay switch with a 5v receiver. When using the same receiver with a row of LEDS it works flawlessly. I've noticed that the relay keeps making the switching noise over and over like I'm pressing the transmitter really fast apart from the fact I'm not. Also, when the transmitter is moved closer to the receiver from 20" to around 1" the problem stops. Below is some circuits to help show you my setup.  Any help appreciated thanks.  Setup Specs   USB 5v DC (2AMPS) Receiver (XY-DJM-5V - SC2272-M4) Relay (SRD-05VDC-SL-C 5V DC) Transmitter (XY-04) Diode (1N4001) Transistor (2N2222A) Resistor (1.8K Ohms)          &ndash; Schematic created using   Antenna Signal Workaround or Booster  I believe this issue is related to wireless signal on D0. Basically above 20" inches the output to the relay drops from 5.05v to 0.5v and is obviously causing the problem. I think I need to boost the signal or replace the transistor with something that triggers at low and high current but have no clue. Any help appreciated, I'm new at this.  Board Setup Lie bracket for Lie algebra of $SO(n,m)$ Is it possible to set a DNS record to redirect all example.com/* traffic to www.example.com/*?  I'm running a  app on Heroku's .  I set up , and the documentation discusses . I also read about how to do this with . However, I'm wondering if there's a way to do this via DNS, i.e., without hitting my app. I'm using Jabref to manage my Bibliography.  Sometimes, I put directly:     First Author et al.   when adding the article author's name in Jabref.  When compiling my .tex file, LaTeX doesn't recognize the et. al as more names. It takes it as a name/surname.  This is an example of a Bibliography entry in Jabref.  @ARTICLE{Wolf2003,     author = {Wolf, K.et al.},     title = {Compensation mechanism in tumor cell migration: mesenchymal–amoeboid     transition after blocking of pericellular proteolysis},     journal = {J. Cell Biol.},     year = {2003},     volume = {160},     pages = {267–277},    }   This is a .texMWE with the bib entry mentioned:  \documentclass[11pt]{book}   \usepackage{cite}   \bibliographystyle{unsrt} \begin{document}   Here is the bib entry mentioned\cite{Wolf2003}    \bibliography{bibfile}  \end{document}   Here's the output:   How can I solve this problem? After my suspension period, my reputation was not returned On Stack Overflow, I have a suspension period of 7 days which ends today. But my actual reputation has not been returned to me as one of the moderators said it would. I have a message like this by @Andrew Barber:     Once your suspension period is over, you will be returned to your   correct reputation level, minus the votes from the accounts which have   been deleted. At that time, you are more than welcome to return, and   continue contributing your knowledge, and gaining legitimate   reputation points. Just please keep our rules in mind when you do.      Regards      Andrew Barber, Stack Overflow moderator   Can any moderator here help me or tell me what the rules are of getting back my reputation?  How can I get the attention of a moderator so that they can resolve my problem? What overpowered combinations would be available if I allow a bonus action to be used in place of a standard action? It has come up in game a couple of times that a player might want to cast a spell that has a casting time of 1 bonus action using their "main" action (if they have another bonus action they also want to take on that turn, such as giving bardic inspiration, or controlling a Bigby's hand, etc.)  On the face of it, it seems obvious that something (a bonus action) that is usually much faster than a full action could be done as your full action.  Although the question comes up most often with respect to spellcasting, if I house rule this, I would rule that any bonus action can be taken as a regular action instead; however, I would not allow the same type of bonus action to be taken twice (so no giving bardic inspiration to two allies on the same turn, for instance).   Are there any abusive or overpowered combinations I should be wary of if I were to allow a character to take 2 bonus actions instead of one regular action and one bonus action on a turn?   The issue of casting two bonus-action spells would not come up because the rule against casting 2 spells on your turn unless one of them is a cantrip with a casting time of 1 action would still be in effect:  PHB p. 203 (under Bonus Action casting time)     You can't cast another spell during the same turn, except for a cantrip with a casting time of 1 action.   I know it's hard to prove/justify a negative answer to a question like this, but I'd be happy to get answers that say you don't think there would be any issues if you describe how you came to that conclusion. Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$ I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$   $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.      I think the first step should look something like this:      $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$   From here I tried to proceed with two cases. 1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$ 2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$  My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.  Any help is appreciated. Thx in advance. Cherio Woltan Customizing titlesec-etoc (further question) One thing I've noticed about the relatively new sitewide layout is that I can't find a quick way to get back to www.stackexchange.com without finding StackExchange in the sites list. As a result, I am typing stackexchange in my toolbar a whole lot more often now. In particular, I tend to enjoy browsing the hot questions list and my own personal feed quite often so visit these pages on a daily basis. Is there an easy link to them that I'm missing?  (On the android app, there is a feed in the menu that I really appreciate, something like that on the desktop would be ideal!) Is there any direct relation among first, third, fourth order moments? Does Android hide some amount of RAM from the User? I bought a new tablet, but Android (2.3.1) shows that it has only 150MB free of 256MB, and it should have 512MB.  I checked the file /proc/meminfo, MemTotal information. It only shows 256MB. Is there a possibility that the Android "hid" 256MB of RAM from the user? Is there an official list of Disney Princesses? This is what my GRUB 2 menu looks like so far:  I want to do a few things here:  I want to rename &quot;Windows Boot Manager (on /dev/sda1)&quot; to just &quot;Windows 8.1&quot;. I'd like to rename &quot;Ubuntu&quot; to &quot;Lubuntu 14.04.1&quot;. I'd also like to switch the order of the options to this:  Windows 8.1 Lubuntu 14.04.1 Advanced options for Lubuntu    (Windows will be the default OS) If it helps in any way, I'll explain the steps I took prior to this, since I am very new to Linux and don't know what information is required.  I bought a laptop on Amazon (ASUS X551MAV-EB01-B) pre-installed with Windows 8.1 64-bit I downloaded the proper Lubuntu .iso file for my system (Lubuntu 14.04.1 amd64) and put it on my flash drive (via UNetBootin) I went to Disk Management, and shrunk the C:/ drive down 32Gb so that 32Gb will be free for Lubuntu I used my flash drive (32Gb Sandisk Cruzer) as a LiveUSB for Lubuntu 14.04.1 I went through the install process. It didn't recognize Windows 8.1 so I had to do the &quot;Something Else&quot; option. I highlighted the free space, and pressed the + button I created a 10,000mb ext4 partition and mount point is / (is now dev/sda7) I created a 4,000mb swap area partition (is now dev/sda8) I created an ext4 partition using the rest of the free space and the mount point is /home (is now dev/sda9) I selected dev/sda7 to install Lubuntu and selected dev/sda7 for the bootloader installation It installed just fine, booted up just fine, and I'm happy to know the GRUB menu works. Show that the set $\mathbb{Q}(\sqrt{p})=\{a+b\sqrt{p}; a,b,p\in\mathbb{Q},\sqrt{p}\notin \mathbb{Q}\}$ is a field  I am having a little trouble proving that for each non-zero $b\in\mathbb{Q}(\sqrt{p})$, there is $d\in\mathbb{Q}$ such that $b\cdot d=1$.  My attempt:  Let $b=a_1+b_1\sqrt{p}$  Let $d=a_2+b_2\sqrt{p}$  We want $$b\cdot d=(a_1+b_1\sqrt{p})(a_2+b_2\sqrt{p})$$ $$=a_1a_2+a_1b_2\sqrt{p}+a_2b_1\sqrt{p}+b_1b_2p$$  $$=(a_1a_2+b_1b_2p)+(a_1b_2+a_2b_1)\sqrt{p}$$ $$=1$$  Am I following the right approach? What should be my next step An example of a lattice that's not totally ordered