How do I show that both the additive and multiplicative groups of an infinite field are non-cyclic?
Why must a field whose a group of units is cyclic be finite?
How to colorize an image as in the example? I have seen such an effect being used in many places however I haven't figured out how to replicate it via Photoshop.  I have tried desaturating the image, placing it under a layer of solid colour and then lowering the opacity of the colour. I have also tried a couple of blend modes, to no avail. The effect is nowhere near the same.  Can someone please tell me the effect used on these images? In chapter 1, of Landau-Lifshitz Mechanics' book, Landau through isotropy and homogeneity of space and homogeneity of time proves that the Lagrangian must depend of magnitude of velocity of the particle.  This seemed fine. But he didn't give a reason as to why it must depend of $|v|^2$ and not on some $|v|^n$ where $n\neq2$. PreparedStatement IN clause alternatives? What are the best workarounds for using a SQL IN clause with instances of java.sql.PreparedStatement, which is not supported for multiple values due to SQL injection attack security issues: One ? placeholder represents one value, rather than a list of values.  Consider the following SQL statement:  SELECT my_column FROM my_table where search_column IN (?)   Using preparedStatement.setString( 1, "'A', 'B', 'C'" ); is essentially a non-working attempt at a workaround of the reasons for using ? in the first place.    What workarounds are available? How to show only date after the date of today in JCalendar How do I restrict my kids' computing time? How do I find this limit: $\lim_{n\to\infty} \left[ n-{n\over e}\left(1+{1\over n}\right)^n \right]$? How can I set desktop defaults for new users? How can I set up desktop defaults, much like most distributions do, for every new user? (ex. default wallpaper, apps, etc) I have a question on Holder space.Is the space $C^2(\mathbb{R}^n)$ embedded in $C^{1,\alpha}(\mathbb{R}^n)$? I think in the bounded domain case, this should be true. But what if in the case where the domain is not bounded? Thanks for any hint! I have got a specific folder located in "/home/.../reboot/".  I have a series of users that require read-only access to the ever changing files within.  My problem is that the same users are not authorized to view any of the parent directory files.  Right now, the only thing I have in place is that their starting folder is at "/home/.../reboot/" by changing the etc/passwd file; and I need to block all access to other folders.  How could I only give them access to this directory? Load Excel data sheet to Oracle database I am looking for a free tool to load Excel data sheet into an Oracle database. I tried the Oracle SQL developer, but it keeps throwing a NullPointerException. Any ideas? Prove that the dihedral group $D_4$ can not be written as a direct product of two groups I like to know why the dihedral group $D_4$ can't be written as a direct product of two groups. It is a school assignment that I've been trying to solve all day and now I'm more confused then ever, even thinking that the teacher might have missed writing out that he means normal subgroups.  On another thread it was stated (as the answer to this question) that the direct product of two abelian groups is again abelian. If we consider the direct product of abelian subgroups $H$,$K\in G$ where $HK=G$ (for all $g \in G$, $g=hk$ $h \in H$, $k \in K$.) I can't understand why this would imply $g=kh$? It is not stated anywhere that $H$,$K$ has to be normal! But if this implication is correct I do understand why $D_4$ (that is non-abelian) can't be written as a direct product of two groups. But if it's not, as I suspect, I need some help!  We know that all groups of order 4 and 2 are abelian, (since $4=p^2$), but only 4 of the subgroups of $D_4$ are normal:  Therefor I can easily show that $D_4$ can't be a direct product of normal subgroups: The only normal subgroups of $D_4$ is the three subgroups of order 4, (index 2 theorem): $\{e,a^2,b,a^2b\}$, $\langle a\rangle$, $\{e,a^2,ab,a^3b\}$ and the center of $D_4=\{e,a^2\}$ We can see that these are not disjoint. So $D_4$ can't be a direct product of normal subgroups. The reason for this being that the center is non-trivial. But why can't $D_4$ be a direct product of any two groups?  If we write the elements of $D_4$ as generated by $a$ and $b$, $a^4=e$, $b^2=e$, $ba=a^3b$ why isn't $D_4=\langle b\rangle\langle a\rangle$ ? I calculated the products of the elements of these two groups according to the rule given above and ended up with $D_4$, and also $\langle a\rangle$, $\langle b\rangle$ is disjoint...? Why is this wrong? Very thankful for an answer! How to align two images side by side if one label has more lines I have two same sized images and want them side by side with a caption under them.  My code looks like this:   \begin{figure}[H] \centering \begin{minipage}{.5\textwidth}   \centering     \includegraphics[width=\textwidth]{3}     \caption{label1}   \label{fig:test1} \end{minipage}% \begin{minipage}{.5\textwidth}   \centering     \includegraphics[width=\textwidth]{4}     \caption{label2}   \label{fig:test2} \end{minipage} \end{figure}   which works fine until one of the labels has 2 lines, than it looks like this:    and I woud like to prevent it to look something like this:    How can it be done?  Thanks Earth Engine convert list with coordinates and values into a feature collection for export I need to create a feature collection from a list containing coordinates and band values.  These are extracted from a landsat image as shown in this linked script.    I'm unable to find a solution in creating a feature collection.  Here is the tail end code.  var valuesList = joinedImage.reduceRegion({   reducer: ee.Reducer.toList(4),   geometry: myGeometry }).values().get(0);  var feature = ee.Feature(null, valuesList);   // Wrap the Feature in a FeatureCollection for export. var myFeatures = ee.FeatureCollection([feature]);   // Export the image, specifying scale and region. Export.table.toDrive(myFeatures, "data", "myData", "B4-B5",  "CSV");   This doesn't work, because it is clearly the wrong way to create the feature collection.  I've tried to create it similar to Rodrigo E. Principe's solution of mapping a Feature Collection but I can't seem to create the initial feature collection to start with. Problem from iOS 10.1.1 to 10.2.1.  I have no photos at all on my iPhone yet in settings it says my photo library is taking up ~200MB of space. The number of space being taken up keeps changing constantly(in the settings app on my phone). In iTunes, the storage bar at the bottom does(when in the sync menu) not show photos.           What is taking up this storage, and how can I reduce it?  Troubleshooting:   I disabled iCloud photo library and iCloud photo stream, still showing the same thing. Tried doing what @User9125 said, didn't work. I had set the date all the way back when I first got my phone, nothing showed up. Rebooting I had hundreds of photos in Whatsapp, I deleted them but I didn't get the storage back. It didn't even make a dent in the amount of storage being used by Photos so I don't think a third-party app could be taking up this storage. I don't think a system app is taking this space either, because I cleared the pictures in mail and messages. I tried using (through sideloading) in Secure Mode and Fast Mode, but it did not clear up the photos. Setup restarts the laptop and then restarts the installation all over and over again? Tips to extend battery life for laptops and notebooks I use the Stack Exchange sites very often to find solutions to problems and help me with decision-making.  In some sites I have asked question and received answers, and I have more than 15 reputation. On Stack Overflow I have always found answers to my questions and never needed to post a question (avoiding duplicate questions). However, I would like to vote for the answers that helped me but I do not have the ability to do that, due to low reputation.  Posting a question or answering an answer on each site seems like the only way to get enough reputation.  Since I have no input to give in these areas unless they will be counted as duplications, many answers that I have already used are missing my vote.  I suggest to take into account all of the reputations from all Stack Exchange sites for voting, or thinking about a different way to allow votes. Understanding the length extension attack I have been trying to understand exactly how a length extension attack works on SHA-1. I'll detail below what I've understood so far, so that I can convey my understanding of the same and hopefully get advice on where I'm going wrong.  Lets assume that for a message $m$, a secret $s$ is appended to it and the SHA-1 hash is calculated. If $\textrm{len}(m) + \textrm{len}(s)$ is not a multiple of the block size, a certain amount of padding is added to it to make it a multiple, and the hash is generated using what is essentially $s || m || \textrm{padding}$.  Now SHA-1 internally uses 5 registers — it takes in an block, generates new register values, turns the crank, takes in another block, generates new register values, turns the crank ... and so on till the end. When all blocks are exhausted, the register values are concatenated and spit out.  Assume that an attacker knows a certain value of $m$, for which he also knows the hashed key. In such a case, he knows the state of the 5 registers when $(s || m || \textrm{padding}$ was processed. Lets also assume that the attacker knows the length of $s$, so that he can predict how much padding was added.  Here is where I get confused. From what I've read so far, the attack involves taking the known hash, adding a certain amount of padding and then the value that we want. However, isn't the padding already used up in creating the hash? Why are we using padding in our attack? A local IP address (RFC 1918) on my network that is not … local I've noticed an IP address today on my network (192.168.15.1) which when visited actually goes to a page that has a router webconfig page (Not my router)  The puzzling thing is that i tried traceroute 192.168.15.1 which said that the distance between me and that host is 9 hops, whereas the distance between me and any other computer on my home network is 2 hops away (My computer --> Router --> Host)   Can someone please explain what's going on here ?  tracert output:  Tracing route to 192.168.15.1 over a maximum of 30 hops    1    &lt;1 ms    &lt;1 ms    &lt;1 ms  192.168.1.1   2    47 ms    51 ms    49 ms  [163.121.171.123]   3     *        *        *     Request timed out.   4   128 ms     *        *     163.121.211.134.dnet.net [163.121.211.134 ]   5     *      134 ms   126 ms  10.37.242.18   6    94 ms    90 ms    95 ms  10.35.2.157   7   178 ms   254 ms   206 ms  10.32.8.107   8    85 ms    57 ms   318 ms  10.35.0.186   9   215 ms   241 ms   230 ms  192.168.15.1  Trace complete. Where to start with Oracle? This question only needs one answer: a list of great resources for getting started with Oracle, including Database Administration, PL/SQL and SQL.   You may find your question has been closed as a duplicate of this one, if so, please don't be offended: we are trying to be as helpful as we can without . Closing a question as a duplicate of this one is us saying "What you really need is to spend some time reading some of the best resources available rather than have your specific question answered right now. If you do get through these resources, you'll soon be back with more !