This is an exercise from M.Isaac's book on character theory of finite groups (exercise 4.6, Pg 60):  Let $n \in {\mathbb{N}}$ and assume that $\chi^{(n)} \in {\mathrm{Irr}}(G)$ for every $\chi \in {\mathrm{Irr}}(G)$. Show that $G = H \times A$, where $A$ is abelian and ${\mathrm{g.c.d.}}(|H|, n) = 1$. Here $\chi^{(n)}(g) := \chi(g^n)$.  The first hint was: Let $d = {\mathrm{g.c.d.}}(|G|, n)$. Show that it is no loss to assume that ${\mathrm{g.c.d.}}(|G|/d, n) = 1$. [The conclusion is sort of obvious. Surely a priori it is difficult to find a subgroup $H$ with ${\mathrm{g.c.d.}}(|H|, n) = 1$]   The second hint was: Let $A = \bigcap_{\chi \in {\mathrm{Irr}}(G)} {\mathrm{ker}}(\chi^{(n)})$. Show that $A = \{ g \in G ~:~ g^n = 1 \}$ and $|A| = d$.     The first statement follows from the definition of $\chi^{(n)}$ itself. ${\textbf{ It is not so clear why $A$ must be abelian}}$. It is easy to realise $A$ being kernel of $\rho^{(n)}$, where $\rho$ is the regular character. The problem is many of the $\chi^{(n)}$ will coincide for different $\chi \in {\mathrm{Irr}}(G)$.  The third hint was: Let $H = \bigcap \left\{ {\mathrm{Ker}}(\chi) ~:~ \chi \in {\mathrm{Irr}}(G), \chi^{(n)} = 1_{G} \right\}$. Show that $[G : H] = d$.  Assuming the previous step, one can apply Schur-Zassenhaus theorem to conclude this. ${\textbf{Is there any direct way?}}$
Let $n&gt;0$ and assume that $\chi^{(n)} \in \mathrm{Irr}(G)$ for every $\chi \in \mathrm{Irr}(G)$. Show that $G = H \times A$, where $A$ is abelian and $(|H|, n) = 1$.  $\chi^{(n)}$ is defined by $\chi^{(n)}(g) = \chi(g^n)$. $\mathrm{Irr}(G)$ is the set of all irreducible characters of $G$.  Here are the hints in the book:   Let $d = (|G|,n)$. Show that it is no loss to assume that $(|G|/d,n) = 1$. Let $A = \bigcap_{\chi \in \mathrm{Irr}(G)} \mathrm{ker} \chi^{(n)}$. Show that $A = \{g \in G | g^n=1\}$ and $|A| = d$. Let $H = \bigcap \{\mathrm{ker} \chi | \chi \in \mathrm{Irr}(G), \chi^{(n)} = 1_G\}$. Show $|G : H| = d$.   Hint 1 and 2 are easy, and it is easy to deduce the result from the hints. As for hint 3, I have proven that $\{ g \in G | (o(g), n) = 1 \} \subseteq H$, and that $|G : H| \mid d$. And I am stuck here.  Can anyone help me?
How can I find the product key of Office 2011 after restoring from backup? I had to restore my iMac from a Time Machine backup.  Everything seemed fine until I tried to start Office 2011 and it is asking me to re-enter my product key. I am trying to locate my key (I believe I downloaded this version originally and it should be in my email but I can't find it)  I tried a program that is supposed to "find the keys for products installed on your computer" (Mac Product Key Finder) and it found my very old Office 2008 key, but not 2011.  I am surprised that if I had a totally working install before recovering from back-up that this wasn't restored as well - do I have any options to find it hiding somewhere on my drive?  Note: The recovery was after a hard drive was replaced by the Apple store - if that makes a difference, so this is a new internal drive. How to add the name of the supervisor in a @thesis field? In the biblatex doc, it says ( Version 1.7, p.11):     thesis      A thesis written for an educational institution to satisfy the   requirements for a degree. Use the type field to specify the type of   thesis.      Required fields: author, title, type, institution, year/date      Optional fields: subtitle, titleaddon, language, note, location,   month, isbn, chapter, pages, pagetotal, addendum, pubstate, doi,   eprint, eprintclass, eprinttype, url, urldate   I would like to know if there is any "canonical" way to add the name of the supervisor(s) in a field of the .bib file (no .bbl manipulation please, I want my bibliography to be usable "as it" in several documents).  The expect result could be something like  :     Meand Myself. “A brilliant result, explained”. PhD thesis**, under the   joint supervision of Pr. Toto and Pr. Tata**. Université Paris 24,   Nov. 2013.   The "under the (joint) supervision" could be a string that would be declined in four ways:   under the supervision of  under the joint supervision of and the abridged forms Show that $g\circ f\equiv 0$ implies $\det Df\equiv 0$, where $g(x_1,...,x_n)=x_1^5+...+x_n^5$ Let $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be a smooth function and let $g:\mathbb{R}^n \rightarrow \mathbb{R}$ be defined by $g(x_1,...,x_n)=x_1^5+...+x_n^5$. Suppose $g\circ f\equiv 0$. Show that $\det Df\equiv 0$.  I was going to start the solution from $n=2$ and then use induction on $n$.  So, for $n=2$, $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a smooth function and $g:\mathbb{R}^2 \rightarrow \mathbb{R}$ where $g(x_1,x_2)=x_1^5+x_2^5$. Then $Df=\begin{pmatrix}   f_{x_1x_1} &amp; f_{x_1x_2}  \\   f_{x_2x_1} &amp; f_{x_2x_2}   \end{pmatrix}$  But now I am stuck in proving the result. Am I in the correct track? Can anyone plese give me a hint to continue?  Ok so since $g\circ f$ is $0$, the image of $f$ is the line $x_1+x_2=0$ which has no min, max or saddle.   If det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&gt;0$ OR det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)&lt;0$, we have contradiction.   But if det$Df(x_0)&gt;0$ and $f_{x_1x_1}(x_0)=0$, then det$Df(x_0)=0$ or, det$Df(x_0)&lt;0$(Since $f_{x_1x_2}=f_{x_2x_1}$). So this is impossible.  If det$Df(x_0)&lt;0$ we have saddle and it is impossible.   So finally we conclude that det$Df(x)=0$ for all $x$. Am I correct? How to parse a string to an int in C++? What's the C++ way of parsing a string (given as char *) into an int? Robust and clear error handling is a plus (instead of ). Investment options for f1 visa students in USA Make bibliography appear in toc as a chapter, not section Operators can be cyclically interchanged inside a trace: $${\rm Tr} (AB)~=~{\rm Tr} (BA).$$ This means the trace of a commutator of any two operators is zero: $${\rm Tr} ([A,B])~=~0.$$ But what about the commutator of the position and momentum operators for a quantum particle? On the one hand: $${\rm Tr}([x,p])~=~0,$$ while on the other hand: $$[x,p]~=~i\hbar.$$ How does this work out? How does one change the frame size and colors in the frame option of titlesec as in MWE? Jawa asked me this  and I didn't know the answer. So I present it to the meta SO community:     [To get the Yearling badge] Do you need to earn the 200 rep within that one year or is it enough to get 200 at some point after being a member for a year? And does 'a year' mean the very first 12 months or the last (sliding window of) 12 months?    I would assume that if you hit 200 reputation after you've already been around for a year, then you would get the Yearling badge. But I don't know. If n is such that every element $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a root of $x^2-1$. Prove that $n$ divides 24. I have a hard time formulating proofs.  For this problem, I  can see that if $n$ is equal to $8,$ this statement is true.  $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements:  $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$  And obviously $8$ divides $24.$  But how do I prove this without depending on number calculations and only using theorems?  Help Please?  I need a step by step walk through of how to do this proof and what theorems would be appropriate to use. pgfplots - barplot with multiple bars on same coordinate, but labelled differently I'm showing several data sets next to each other using multiple \addplot commands within a pgfplots axis environment. These are then further grouped into categories. The output of the graph per se is fine, however, I need each bar to be labelled separately - basically, I want the legend to be displayed directly underneath its corresponding element. I presume a picture will illustrate my problem more clearly:     I prepared an MWE:  \documentclass{minimal} \usepackage{pgfplots}  \begin{document}  \pgfplotstableread{ 0 35569 8842    134984   1 30428 4689    34077 2 32920 6207    73787    3 16462 7562    23496    4 12315 8572    16565    5 76572 19572   26030    }\dataset \begin{tikzpicture} \begin{axis}[ybar,         width=.9\textwidth,         ymin=0,         ymax=150000,                 ylabel={Y-Label},         xtick=data,         xticklabels = {             Category 1,             Category 2,             Category 3,             Category 4,             Category 5,             Category 6         },         major x tick style = {opacity=0},         minor x tick num = 1,         minor tick length=2ex,         ] \addplot[draw=black,fill=black!20] table[x index=0,y index=1] \dataset; %Data1 \addplot[draw=black,fill=black!40] table[x index=0,y index=2] \dataset; %Data2 \addplot[draw=black,fill=black!60] table[x index=0,y index=3] \dataset; %Data3 \legend{Data1,Data2,Data3} \end{axis}  \end{tikzpicture}  \end{document} How do I check the health of a SSD? Two figures side by side How to prove that $P \rightarrow Q$ is equivalent with $\neg P \lor Q $? In my book about Logic, which is called 'Language, Proof and Logic', by the way, there is explained that the conditional $ P \rightarrow Q $ is equivalent with $\neg P \lor Q$.   that gives as answer: 'just compare the truth tables and you can see that they are equivalent'. However, I would like to know how to prove this using formal propositional logic. This should be possible, right? I found an asnwered question concerning this topic already here:  Copying \captionsetup[subfigure]{justification=centering} doesn't help me thought for I am using different packages.  I would like my Captions and Subcaptions to be justified as the text below. Just under my pictures. Any suggestions?  Thanks a lot  \documentclass{article} \usepackage{subfig} \usepackage[labelfont=bf, justification=justified]{caption} \usepackage[demo]{graphicx}% Remove demo option in real document  \begin{document}     \begin{figure} %   \captionsetup[subfigure]{justification=centering}     \centering       \subfloat[Subcaption one \newline including a linebreak ]   {     \includegraphics[width=0.2\textwidth]{image1}     }     \hspace{1cm}     \subfloat[Subcaption two \newline including a linebreak]     {     \includegraphics[width=0.2\textwidth]{image2}     }        \caption{Figure caption \newline with additional text right here} \end{figure}  \begin{center} This is correctly justified Text \\ including a longer linebreak than blabla. \end{center} \end{document} Do's and Don't's of Undergraduate Research I was fortunate enough to get a position as a researcher for the Mayo Clinic's SURF Program this year. My PI's lab focus is on the the immune system's role in CNS axonal and neuronal injury, specifically through the lens of how innate and adaptive immune effectors interact w/ infected neurons.  Although I do research under a professor at my college and I volunteered for a state university lab during the previous summer, this is my first REU/SURF opportunity, and I REALLY want to make a good impression. Here are my questions:  What are the do's and don't's in terms of being a skilled and efficient researcher?  Since I am still an undergrad, I know that I will be a less useful asset to the lab than a grad student or post-doc, but what can I do as an undergrad to not burden my colleagues and PI?  Thank you all for your help! Wishing you all the very best! Nuclear Binding energy The nuclear binding energy, is the energy that is needed to seperate the nucleons in a nucleus. And binding energy is also defined as the energy given out when a nucleus forms from nucleons.   Also the larger the nucleus is, the more energy is required to break it apart, so why doesn't that mean that larger nuclei are more stable? I mean Uranium has a lower binding energy per nucleon than Iron, but there are many many more nucleons in Uranium that Iron so the total binding energy is going to be much greater.   Basically I don't understand why whether an element gives out energy by fusion or fission (why the lighter element provide energy by fusion not fission and vice versa for heavy elements) depends on binding energy per nucleon and not "total" binding energy Since Ubuntu 11.10 use a new kernel, it's very difficult to have a decent support for virtualization.   VirtualBox doesn't support guest additions for ubuntu 11.10, so I can't copy to and from my ubuntu desktop and windows, which I absolutely require, plus FreeBSD seems not to be able to use DHCP without guest additions.  Virt-manager instead gives an error on launch:  Unable to open a connection to the libvirt management daemon.  Libvirt URI is: qemu:///system  Verify that:  - The 'libvirt-bin' package is installed  - The 'libvirtd' daemon has been started  - You are member of the 'libvirtd' group   unable to connect to '/var/run/libvirt/libvirt-sock', libvirtd may need to be started: Permission denied  Traceback (most recent call last):   File "/usr/share/virt-manager/virtManager/connection.py", line 1146, in _open_thread     self.vmm = self._try_open()   File "/usr/share/virt-manager/virtManager/connection.py", line 1130, in _try_open     flags)   File "/usr/lib/python2.7/dist-packages/libvirt.py", line 102, in openAuth     if ret is None:raise libvirtError('virConnectOpenAuth() failed') libvirtError: unable to connect to '/var/run/libvirt/libvirt-sock', libvirtd may need to be started: Permission denied   The problem is solved by running virt-manager as root, but I don't like that. How do I change permissions to run Virt-Manager as user?  Is there a way to install guest additions on Ubuntu 11.10? I used the prcomp() function to perform a PCA (principal component analysis) in R. However, there's a bug in that function such that the na.action parameter does not work. ; two users there offered two different ways of dealing with NA values. However, the problem with both solutions is that when there is an NA value, that row is dropped and not considered in the PCA analysis. My real data set is a matrix of 100 x 100 and I do not want to lose a whole row just because it contains a single NA value.  The following example shows that the prcomp() function does not return any principal components for row 5 as it contains a NA value.  d       &lt;- data.frame(V1 = sample(1:100, 10), V2 = sample(1:100, 10),                        V3 = sample(1:100, 10)) result  &lt;- prcomp(d, center = TRUE, scale = TRUE, na.action = na.omit) result$x                                # $ d$V1[5] &lt;- NA                           # $ result  &lt;- prcomp(~V1+V2, data=d, center = TRUE, scale = TRUE, na.action = na.omit) result$x   I was wondering if I can set the NA values to a specific numerical value when center and scale are set to TRUE so that the prcomp() function works and does not remove rows containing NA's, but also does not influence the outcome of the PCA analysis.  I thought about replacing NA values with the median value across a single column, or with a value very close to 0. However, I am not sure how that influences the PCA analysis.  Can anybody think of a good way of solving that problem? We know Euler formula:  $$ e^{ix}=\cos(x)+i\sin(x). $$  Does this formula work if we replace real number $x$ with complex number $z$?