[This post is admittedly a development of  of mine, based on therein accepted answer.]     Here I kindly ask for a "solution verification" of my final interpretation.     For $n$ positive integer, let be:   $I_n:=\{1,\dots,n\}$; $G$, $\overline G$ groups of order $n$; $\psi\colon G \rightarrow \overline G$ isomorphism; $f$ bijection and $\bar f = \psi f$; $\theta$, $\bar \theta$ Cayley embeddings; in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets; $S_n$ the symmetric group of degree $n$.   Visually:    Then:  $$\varphi^{(f)} \theta f = \varphi^{(\bar f)} \bar \theta \bar f \tag 1$$  Proof. Since $\varphi^{(\bar f)}=\varphi^{(f)}(\varphi^{(\psi)})^{-1}$ and (*) $\bar\theta=\varphi^{(\psi)}\theta\psi^{-1}$, we get:  $$\varphi^{(\bar f)} \bar\theta \bar f = \varphi^{(f)}(\varphi^{(\psi)})^{-1}\varphi^{(\psi)}\theta\psi^{-1}\psi f = \varphi^{(f)}\theta f$$  $\Box$  My interpretation of $(1)$ is as follows:     If two (finite) groups are isomorphic (via $\psi$), then, for a given labelling of the elements of the first one (via $f^{-1}$), a labelling of the elements of the second one exists (given by $f^{-1}\psi^{-1}$) such that the two groups' structures manifest in $S_n$ as one same set (by $(1)$, actually).     Edit. As pointed out in the comments, (*) needs not to be true, unless the embeddings are Cayley's. Anyway, since these latter always exist, the conclusion keeps unaltered.    Edit #2. If we'd define structure of $G$ (under a given labelling $f^{-1}$) the map $\varphi^{(f)}\theta f$, where $\theta$ is Cayley embedding of $G$ into $S_G$, then $(1)$ would precisely state that isomorphic groups have the same structure.    Edit #3. I think we can upgrade this result to an "iff" one.  Let:    where, for $n$ positive integer:   $I_n:=\{1,\dots,n\}$; $G$, $\overline G$ are groups of order $n$; $f$, $\bar f$ are bijections; $\theta$, $\bar \theta$ are Cayley embeddings; in general, $\varphi^{(\alpha)}$ is the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets; $S_n$ is the symmetric group of degree $n$.   Claim. Let $f$, $\bar f$, $G$, $\overline G$, $\theta$ and $\bar \theta$ be as above. Then $G \cong \overline G$ if and only if:  $$\exists  f,\bar f \mid \varphi^{(f)}\theta f = \varphi^{(\bar f)}\bar \theta\bar f \tag 1$$  Proof. $\Rightarrow$) Let $\psi$ be the isomorphism between $G$ and $\overline G$; then, for any $f$, the bijection $\bar f := \psi f$ does the job (see  for the proof). $\Leftarrow$) Let $(1)$ hold for some $f$ and $\bar f$; then $\bar f f^{-1}$ is an isomorphism from $G$ to $\overline G$ (see  for the proof). $\quad \Box$
I was trying to give the wording "isomorphic groups have the same structure" a precise interpretation, at least for the finite order case. A positive answer to my final question would reach the goal, as it would translate a "$\cong$" between groups into a "$=$" between sets.    For $n$ positive integer, let be:   $I_n:=\{1,\dots,n\}$; $G$, $\overline G$ groups of order $n$; $\psi\colon G \rightarrow \overline G$ isomorphism; $f$, $\bar f$ bijections; $\theta$, $\bar \theta$ embeddings; in general, $\varphi^{(\alpha)}$ the isomorphism between symmetric groups on sets of the same cardinality, defined by $\sigma \mapsto (g \mapsto (\alpha\sigma\alpha^{-1})(g))$, where $\alpha$ is a bijection between the sets; $S_n$ the symmetric group of degree $n$.   Visually:  $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c}    I_n     &amp;   &amp;    I_n      \\     \da{f}         &amp;           &amp;     \da{\bar f}          \\    G &amp; \ras{\psi} &amp;    \overline G  \\     \da{\theta}         &amp;           &amp;     \da{\bar \theta}          \\    \operatorname{Sym}(G) &amp; \ras{\varphi^{(\psi)}} &amp;    \operatorname{Sym}(\overline G)  \\     \da{\varphi^{(f)}}         &amp;           &amp;     \da{\varphi^{(\bar f)}}          \\    S_n &amp;  &amp;    S_n  \\ \end{array} $$     Is it $(\varphi^{(f)}\circ\theta)(G)=(\varphi^{(\bar f)}\circ\bar \theta)(\overline G)$?
How to enable projection transformation in QGIS My QGIS project has two layers, one layer is retrieved by Google layer plugin which fetches the static Google map image whose CRS is +proj=merc +lon_0=0 +lat_ts=0 +x_0=0 +y_0=0 +a=6378137 +b=6378137 +units=m +no_defs and another vector layer is a postgis layer whose CRS is +proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs.  I digitalized some polygon features from Google Layers and added it to the vector layer. However, QGIS seems do not make the automatic CRS transformation. How can I change its projection?   Thanks NEWID() In Joined Virtual Table Causes Unintended Cross Apply Behavior My actual work query was an inner join, but this simple example with cross join seems to nearly always reproduce the problem.  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT NEWID() TEST_ID ) BB ( B )   With my inner join I had many rows for which I added to each a GUID using the NEWID() function, and for about 9 out of 10 such rows the multiplication with the 2-row virtual table produced the expected results, just 2 copies of the same GUID, while 1 out of 10 would produce different results. This was unexpected to say the least and gave me a really hard time trying to find this bug in my test data generation script.  If you take a look at the following queries using as well non-deterministic getdate and sysdatetime functions, you won't see this, I don't anyway-I always see the same datetime value in both final result rows.  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT GETDATE() TEST_ID ) BB ( B )  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT SYSDATETIME() TEST_ID ) BB ( B )   I'm currently using SQL Server 2008 and my work around for now is to load my rows with GUIDs into a table variable before finishing out my random data generation script. Once I have them as values in a table as opposed to virtual table, the problem goes away.  I have a workaround, but I'm looking for the ways to workaround without actual tables or table variables.  While writing this I tried without success these possibilities: 1) placing the newid() into a nested virtual table:  SELECT * FROM (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A ) CROSS JOIN (     SELECT TEST_ID     FROM (         SELECT NEWID() TEST_ID     ) TT ) BB ( B )   2) wrapping the newid() within a cast expression such as:  SELECT CAST(NEWID() AS VARCHAR(100)) TEST_ID   3) reversing the order of appearance of the virtual tables within the join expression  SELECT * FROM (     SELECT NEWID() TEST_ID ) BB ( B ) CROSS JOIN (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A )   4) using uncorrelated cross apply  SELECT * FROM (     SELECT NEWID() TEST_ID ) BB ( B ) CROSS APPLY (     SELECT 1 UNION ALL     SELECT 2 ) AA ( A )   Just before finally posting this question, now I tried this with success it seems, a correlated cross apply:  SELECT * FROM (     SELECT NEWID() TEST_ID ) BB ( B ) CROSS APPLY (     SELECT A     FROM (         SELECT 1 UNION ALL         SELECT 2     ) TT ( A )     WHERE BB.B IS NOT NULL ) AA ( A )   Anyone have any other more elegant, simpler workaround? I really don't want to use cross apply or correlation for a simple row multiplication if I don't have to. How will the super massive black hole affect our galaxy? In Ubuntu 18.04, when moving the windows there is a resize effect that occurs when accidentally touching one of the screen edges.  This control was available in CCSM and Ubuntu Tweak in previous versions with Unity.  I'm not finding it in GNOME Tweaks.  I also searched the repository with:  apt search ubuntu.*tweak Unsure whether to use "enquire" or "inquire" of Stack Overflow, all content posted on Stack Exchange sites by their users (i.e. you wonderful people) has been provided to the whole universe under . For my fellow non-lawyers, that license basically means:   Anyone can use any Stack Exchange posts at any time without having to ask for permission Making money off of the copied content is permitted You don't even have to copy stuff from here verbatim; you can just use it as a starting point and make whatever edits you want There are just two rules you have to follow:   You have to provide . Simple links to the original post and author info are just fine. You have to  and allow other people to use your content, as long as they follow these very same rules. How meta!    (If you ever forget any of that, and want to refresh your memory, the license info is linked to in the footer of every page.)  There are, in fact, a lot of people who republish varying amounts of our content for assorted reasons. Unfortunately, there are some Stack Content Republishers Attributing Poorly and/or Excelling at Ranking (SCRAPERs, for short).  In this context, "attributing poorly" means any use that doesn't follow our attribution rules or make any other reasonable attempt at give credit. This can get pretty egregious; I've seen SCRAPERs that not only don't link back to SE originals, but also use fake author info and post dates to make it harder to find originals. By "excelling at ranking," I'm referring to copycat sites that end up higher in Google results than the original SE sites do for the same content. That's not necessarily wrong, but in some cases, it indicates inappropriate SEO hackery.  So, the question is: what can you do if you spot a SCRAPER? Inserting the value from Datetime field form I have a form where the user can input datetime and then I want to store it in the database. Here's my schema :     'start' => array(           'type' => 'varchar', 'mysql_type' => 'datetime',           'not null' => FALSE,         ),   The form :     $form['add']['start'] = array(         '#type' => 'datetime',   '#default_value' => $date,       );   When I hit submit button, it shows me :     db_insert failed. Message = SQLSTATE[22007]: Invalid datetime format: 1292 Incorrect datetime value:...   What I got out of it was that the DateTime in Drupal is incompatible with MySQL's. Any suggestions or how should I process the value from the form so I can insert it into the db?  Alternatively, any way to convert DrupalDatetime into string?  (For what it's worth I also tried 'timestamp' as the MySQL type but it throws me the same error). When applying dropout in artificial neural networks, one needs to compensate for the fact that at training time a portion of the neurons were deactivated. To do so, there exist two common strategies:   scaling the activation at test time inverting the dropout during the training phase   The two strategies are summarized in the  below, taken from .  Which strategy is preferable, and why?    Scaling the activation at test time:    Inverting the dropout during the training phase: I want to find the probability that mean is less than 5 given that poisson distribution states it is 6 ie find p(x&lt;5|x~po(6))  Here is the actual question: Two grocers agree that the daily demand for a particular item has Poisson distribution. However, grocer A claims that the mean demand is 3 items per day, while grocer B claims that the mean demand is 6 items per day. They agree to resolve the disagreement by observing the demand on one particular day: B agrees to accept A’s claim if the observed demand is 4 or less, and A agrees to accept B’s claim if the observed demand is 5 or more. (a) Calculate the probability that A’s claim is accepted when, in fact, B’s claim is correct Get value of R, G, B by color Simple problem with Manipulate and Plot What is the difference between "information on", "information of", "information about" "Information on something", "information of something", "information about something" — these three usages all sound having the same meaning for me. Are there any differences? What does $(command) & do? I saw the following syntax in a bash script:   &gt;$(command) &amp;   I know that the ampersand &amp; at the end makes a command run in the background, but I have never seen it in combination with the dollar sign plus the brackets around the command.  What does it do? Can somebody help me to prove this identity?     $$\prod_{k=1}^{n-1} \left(\;\cos\theta-\cos\frac{k\pi}{n}\;\right) = 2^{1-n}\sin(n\theta)\csc\theta$$   I was thinking of using the properties of the complex numbers and the binomial theorem, because I found this problem in the Complex Variables book from Schaum's, but it did not solve my problem.  Does anyone have an idea?  Thanks in advance! Define the Lie algebra $\mathfrak{g}$ of a Lie group G to be the set of all left-invariant vector fields of $G$. I want to prove that   $$          f: \mathfrak{g} \rightarrow T_eG \\              X \mapsto X(e) $$ is a linear isomorphism.  The only thing that I have left to prove is surjectivity.Letting $v\in T_eG$, we set $X\in \mathcal{X}(G)$ such that $\forall x\in G,\,X(x)=L_{x_{*,e}}(v)$ (I still have to show that such an $X$ is indeed a vector field on $G$). We get that  $$ f(X)=X(e)=L_{e_{*,e}}(v)=v $$  where the last equality holds because  $$       L_{e_{*,e}}(v)f = v(f \circ L_e) = v(f \circ id_G) = v(f) $$ for any $f\in C^\infty(G)$. (I am thinking of the tangent vectors as derivations).  Now all we have left to do is prove that such an $X$ is left-invariant, and it is therefore an element of $\mathfrak{g}$, but I do not know how. I tried to prove that  $$            \forall f\in C^\infty(G),\, L_{{x}_*,y}(X(y))f=X(L_x(y))f $$ but with no success. Backup and restore settings in Texstudio Does someone know, how to save all the settings of Texstudio and restore them in a simple way?  Thank you in advance! Necromancy In Harry Potter How is Necromancy accomplished in the Harry Potter universe?  I thought that people cannot be raised from the dead in Harry Potter.  Were bodies just use as puppets that can only do simple tasks, for instance when Dumbledore and Harry were getting the locket, the  were there to guard the locket.  Also, besides Inferi were there any Zombies in Harry Potter? Configure tab-switch hotkey in gnome-terminal I used to use Alt+number to switch between gnome-terminal tabs. In my recent 19.10 installation, this does not work. When I press Alt+1, instead of switching tab, the command line shows this: (arg: 1)   and if I press Alt+2, it becomes: (arg: 12)   I checked the options in gnome-terminal, it is set to use Alt+number to switch between tabs. I would like to know which application is hijacking the hotkey, and how to fix it. PS. I do NOT want to use Ctrl+PgUp/PgDn for this. Thanks. Is there a word to describe a person who likes chaos? I am wondering if there is a word to describe a person who likes chaos. By this I mean a few specific things:   The person is pleased to hear when chaos is created, or confusion emerges The person will take opportunities to create chaos or confusion The person enjoys taking part in chaotic or disorganized situations.   I can't seem to think of a word to describe the type of person who would act in this way.  I hope this question is clear.  EDIT: Below gbutters sums up nicely:     What I wonder is if there is a word for a generally good-natured person who just likes some good old-fashioned chaos. It seems like any person that thrives on chaos would have to have some extra baggage. Can not update Bash on Debian 6.0 (Squeeze) I can not update  on a Debian 6.0 (Squeeze) server to get rid of the discovered vulnerability:  bash --version GNU bash, version 4.1.5(1)-release (x86_64-pc-linux-gnu)  apt-get update apt-get install bash Reading package lists... Done Building dependency tree Reading state information... Done bash is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 7 not upgraded.   Can I use Squeeze-LTS for this server just to update Bash? After one week I will be on another server, so I will not make any other updates.  uname -m x86_64  lsb_release -a No LSB modules are available. Distributor ID: Debian Description:    Debian GNU/Linux 6.0.5 (squeeze) Release:        6.0.5 Codename:       squeeze