Let $G$ be a group of order $p^2$. Use the class equation to prove that $G$ is abelian.   The conjugacy class equation, at least how I remember it, is $$ |G| = |Z(G)|+\sum_{x\in I \backslash Z(g)} \frac{|C|}{C_G(x)} $$ where $|Z(G)|$ is the center of $G$, and $|C_G(x)|$ is the central subgroup of $G$.  I am tempted to start with using Lagrange's theorem, which gives possible orders $o(x,y)$ of $1,p,p^2$.    Now, to prove that $G$ is abelian, perhaps I need to prove that $|C_G(x)|=G$? ($C_G(x)$ being the central subgroup in which every element inside of it commutes)
I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements.   My problem is showing that $G$ is abelian, and I would be glad if someone could show me how.   I had two potential approaches in mind and I would prefer if one of these were used (especially the second one).   First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?  Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l $. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$).   Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine.
is the group of rational numbers the fundamental group of some space? I was thinking of the noted 1980 paper by Sidney Coleman and Frank de Luccia--"Gravitational effects of and on vacuum decay"-- about metastable vacuum states that could tunnel to a lower energy "true vacuum" with catastrophic results. I suspect the answer is that if our vacuum is a false one, we wouldn't "see" the true one til it hit us. disable clickable landmark on google map Ok. I was playing mcpe and I noticed that mossy cobblestone isn't in the crafting table. And I wanted to make a house that looks old. So if its possible to get mossy cobblestone without mods, how can I get it on survival? If it is possible can somebody please leave a step by step guide. Thank you. Geometric interpretation of $\frac {\partial^2} {\partial x \partial y} f(x,y)$ What are Click packages? I noticed that Ubuntu SDK recently had some updates and it installed a program named "Click". I would like to know what they are and how to use them.  Also, would they make Debian packaging easier? Many times it takes me a while to properly compose my question before posting. Sometimes it's just a jumble of ideas and concepts, and is not something I'd want seen immediately. However, there is no way to save a partial question unless you post it.  This should work exactly like it does in Gmail. There, once you start composing your question, it's autosaved as a draft. If your browser crashes or you need to come back later to complete it, it will be there for you to finish and post later.  Optionally, allow multiple draft questions and a page to edit/delete them. If needed, make them expire after some amount of time. Groups of order $56$ Let $G$ be a group of order $56$. (We do NOT assume the Sylow-$7$ subgroup to be normal.) Then either the Sylow-$2$ subgroup is normal or the Sylow-$7$ subgroup is normal. How to prove?  My idea: consider the case $n_2=7,n_7=8$. Then the number of elements in all Sylow $2$ -subgroups and Sylow $7$-subgroups is at most exactly $(7-1)*8+(2-1)*7+1=56$. This cannot give a contradiction. We need to prove that $G$ is simple. Then $|G||(8!/2)$. To prove $G$ simple, we need to prove there does not exist normal subgroups of order $2$ and order $4$. But I do not know how to prove. GET and POST methods with the same Action name in the same Controller How to derive variance-covariance matrix of coefficients in linear regression I am reading a book on linear regression and have some trouble understanding the variance-covariance matrix of $\mathbf{b}$:    The diagonal items are easy enough, but the off-diagonal ones are a bit more difficult, what puzzles me is that  $$ \sigma(b_0, b_1) = E(b_0 b_1) - E(b_0)E(b_1) = E(b_0 b_1) - \beta_0 \beta_1 $$  but there is no trace of $\beta_0$ and $\beta_1$ here. Why are all nonzero eigenvalues of the skew-symmetric real matrices pure imaginary? Assume that $A$ is an $n\times n$  real matrix, i.e. $$A^T=-A.$$  Since $\det(A-\lambda I)=\det(A^T-\lambda I)$, $A$ and $A^T$ have the same eigenvalues. On the other hand, $A^T$ and $-A$ also have the same eigenvalues. Thus if $\lambda$ is an eigenvalue of $A$, so is $-\lambda$. If $n$ is odd, $\lambda = 0 $ is an eigenvalue.  A curious search in Google returns  the nonzero eigenvalues of $A$ are all pure imaginary and thus are of the form $iλ_1, −iλ_1, iλ_2, −iλ_2,$ … where each of the $λ_k$ are real.   Here is my question:     How can I prove the fact that "the nonzero eigenvalues of $A$ are all pure imaginary"? Chromium on Ubuntu 12.04. Youtube says:     This video is currently unavailable   I don't know why but this is the case for some videos, but not all. Well, most videos. And I don't know what's wrong.  Anyone help? Given the Python function:  def a_method(arg1, arg2):     pass   How can I extract the number and names of the arguments. I.e., given that I have a reference to func, I want the func.[something] to return ("arg1", "arg2").  The usage scenario for this is that I have a decorator, and I wish to use the method arguments in the same order that they appear for the actual function as a key. I.e., how would the decorator look that printed "a,b" when I call a_method("a", "b")? The "Satellite Paradox": Twin paradox in orbiting satellites I want to simply eliminate the effect of acceleration, which is the cornerstone in resolving the traditional twin paradox, through the following thought experiment: Assume twins $A$ &amp; $B$ are located in the same orbital height, far away from the surface of a massive planet, on a long concrete column at rest WRT the planet. A rapid explosion that happens between the twins makes them orbit the earth with a high velocity, say, $0.6 \ c$, in the same orbital path but in different directions. According to relativity, since the orbiting satellites are completely inertial frames, each twin attributes a variable time dilation to the other one, and each twin expects the other one to be younger. Nonetheless, these twins are automatically reunited after traveling half of the orbit's circumference without being affected by additional accelerations. Which twin is younger in this case? Indeed, If each twin sees the other's clock as running slow all along the way, how can the clocks agree when they meet? For this configuration:  The acceleration due to the explosion does not affect the twins' clock rates since it occurs abruptly in the same place on the column where the twins are located back-to-back. The clocks have been synchronized just before the explosion.  The traditional gamma factor, attributed to one twin from the viewpoint of the other, is variable because the resultant velocity of one twin complying with the relativistic velocity addition is a variable measured from the viewpoint of the other one. However, whether or not the corresponding time dilation is varying, each twin anticipates the other one to be younger. There is no breaking of symmetry!  The $g$-field of the planet has nothing to do with orbiting satellites since they are inertial, and thus GR corrections are not inserted. (Remember the GPS settings!) Collinearity can pose certain problems in various kinds of regression problem. In particular, it can make the parameter estimates have high variance and be unstable.   Various methods have been proposed to deal with this including ridge regression, partial least squares regression, principal components regression, dropping variables and getting more data.   One method that is controversial is standardizing or scaling the independent variables, with various experts saying it is a good (e.g. Garcia) or bad (e.g. Belsley) idea. Belsley's issue seems to be (in non-technical terms) that changing the IVs just pushes the problem under the rug.  But other experts seem not to agree. And the authors tend to get rather heated in defending their own positions.   Back when I did my dissertation (on collinearity diagnostics) I found Belsley's arguments convincing, but that was long ago (I got my degree in 1999).  I'm looking either for expert guidance or for any current review article that is unbiased. Co-signed a loan and I need to have my name removed I was playing around with some integrals and noticed that some integrals of the form:  $$I(a,b,c)=\int_a^\infty \frac{\arctan(x+b)}{x^2+c}dx$$ Does have a closed form. I am trying to find for what constant $c$ will this work. In case you wonder why only $c$ is problematic, I will try to show by an example.  $$I=I(1,3,16)=\int_1^\infty \frac{\arctan(x+3)}{x^2+16}dx$$ Let's start by letting $x-1=t\,$ thus: $$I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})}}{t^2+2t+17}dt$$ With $\displaystyle{t=\frac{17}{y}\rightarrow dt=-\frac{17}{y^2}dy}$ $$I=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{y}+4}\right)}}{\left(\frac{17}{y}\right)^2 +\frac{34}{y}+17 }\frac{17}{y^2}dy\overset{y=t}=\int_0^\infty \frac{{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}{t^2+2t+17}dt$$ $$2I=\int_0^\infty \frac{{\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}}{t^2+2t+17}dt$$ $${\arctan(\color{blue}{t+4})+{\arctan\left(\color{red}{\frac{17}{t}+4}\right)}}=\arctan\left(\frac{\color{blue}{t+4}+\color{red}{\frac{17}{t}+4}}{1-(\color{blue}{t+4})\left(\color{red}{\frac{17}{t}+4}\right)}\right)$$ $$=\arctan\left(\frac{x^2+8x+17}{x}\frac{x}{-4(x^2+8x+17}\right)=\pi-\arctan\left(\frac14\right)$$ Above follows since the original integral is positive so we take $\arctan(-x)$ as $\pi-\arctan x $ and therefore getting a negative answer will not be an issue. $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\int_0^\infty \frac{1}{t^2+2t+17}dt$$ Well, now the inner integral is not hard to compute and the final answer happens to be: $$I=\frac12 \left(\pi -\arctan\left(\frac14\right)\right)\frac14\arctan\left(\frac{t+1}{4}\right)\bigg|_0^\infty =\frac{\pi^2}{16}-\frac{3\pi}{16}\arctan\left(\frac14\right)+\frac18\arctan^2\left(\frac14\right) $$ There are more examples that I found by checking and try such as:  $$I(1,2,9)=\int_1^\infty \frac{\arctan(x+2)}{x^2+9}dx$$ $$I(2,1,6)=\int_2^\infty \frac{\arctan(x+1)}{x^2+6}dx$$ $$I(2,2,13)=\int_2^\infty \frac{\arctan(x+2)}{x^2+13}dx$$ And so on... All those can be solved by the same method: First substitute $x-a=t$, then let $t=\frac{\alpha}{y}$, where $\alpha$ is the "free of x" coefficient from the denominator.   The problem is that I tried more than $100$ combinations to get those integrals which is not that nice. How can we "smartly" find $c$ so that $I(a,b,c)$ is evaluable by symmetry? Or put in other word what should be $c$ if one wants to compute by symmetry $I(7,13,c)$? What is a tainted kernel in Linux? How to find integer solutions to $M^2=5N^2+2N+1$? How does the Doctor know how old he is? I've often wondered, how does the Doctor know how old he is? He not only travels through time, but also through space, and visits planets where days can be shorter or longer than 24 hours. How does he 'keep score'?  Also, does he count his age in Earth years or Gallifrey years? They might be completely different in length.