Let $H \triangleleft G$ such that $[G:H]=n$. Show that $g^n \in H$ for all $g \in G$.   So I've been trying this for a while and got to nowhere. So I tried to solve it by arriving to a contradiction. So take $g \in G$ such that $g^n \notin H$. Then if  $G=\bigcup_{i=0}^{n-1}Hg^i$,  I was able to arrive to a contradiction. However I wasn't able to prove that $Hg^j \cap Hg^k = \emptyset$ for every $0 \leq k,j \leq n-1$ as to be able to write $G$ as that disjoint union of coclases.  Am I on the right track? Is that true? Any hint if not?  Edit:     The idea is not to use the quotient group. It's an exercise of a chapter of Rotman's Introduction to Theory of Groups which is previous to the introduction to the quotient group.
I need some help in relation to this exercise    "Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in H$ for all $g \in G$."  I'm not allowed to use quotient groups in the proof, because the exercise is in the chapter before.   I tried by induction on $n$. The case $n=1,n=2$ are obvious, but even  the case $n=3$ is giving me trouble so I give up studying the general case of the inductive step.  My other approach was studying left or right coset of $G$. But I only proved that if $g \in aH$ then $g^2 \notin aH$ if $a \notin H$, and I can't find a way to demonstrate that $g^n \in H$. (my starting idea was to prove that every power of $g$ is in a different coset but then I realize that in this way I don't handle several case, for example $g$ has period strictly lesser than $n$ and in conclusion it doesn't prove the exercise)  Maybe I'm missing something about indexes, and this is why I asked here for some help,  (I can't use quotient groups because they are introduced later than this exercise, forgot to add this info in the beginning) Thanks in advance :)
I normally write using the American English forms, but for some particular words, I tend to naturally write it in the British manner (like with the word favourite rather than favorite).  Is it wrong to mix both styles? If not *wrong*, does it looks weird? Why do people not upvote, or accept answers that worked for them? I am regular Stack Exchange user, especially at Stack Overflow.  One of the things I really dislike about new and sometimes veteran users is that:   Either they do not accept an answer that works for them, Or they accept the answer but do not vote it up   I mean they can really both accept the answer and vote it up but somehow they wouldn't =(.  How many of you feel the same way? Do Wine Viruses only work while Wine is running? I heard Linux can get viruses through Wine,and I was curious if the viruses work only while Wine is running.  In other words, could I stop a virus from doing it's thing by just quitting Wine? Let $$ f(x) =    \begin{cases}\begin{align}     1\quad&amp;\text{ if }\; 0\leq x \leq 1 \\     2 \quad&amp;\text{ if }\; 1&lt;x \leq 2 \\   \end{align}\end{cases}$$  Then $\int^2_0 f(x)dx=3$ and an anti derivative of f(x) is $$F(x)=\begin{cases}\begin{align}     x\quad&amp;\text{ if }\; 0\leq x \leq 1 \\     2x \quad&amp;\text{ if }\; 1&lt;x \leq 2 \\   \end{align}\end{cases}$$ But, $F(2)-F(0)=4-0=4 \neq 3.$ Why has this happened?  Find an anti-derivative $G(x)$ of $f(x)$ such that $G(2)-G(0)=3$, the correct answer.  My attempt: $F(2)-F(0)=4 \neq 3$ because we need to break it up into two parts like this: $\int^2_1 2x dx + \int^1_0 x dx = x^2|^2_0 + \frac{1}{2} x^2|^1_0 =4-1+\frac{1}{2}-0=3.5$ but this is not right either. :(  For the "Find an anti-derivative $G(x)$..."I'm completely lost but perhaps if we solve the first part then I'll understand what it's asking for there.  Any help would be appreciated. Installing debian package through Synaptic I want to install a .deb file. After going through various questions, I came across these three options: dpkg, GDebi and Ubuntu Software Center. However, I want to install the. Deb file via Synaptic Package Manager only. Is there any method by which I can install. Deb file as well as all it's dependencies via Synaptic Package Manager? Does Overhand Chop override the damage bonus from wielding a two handed weapon? I was showing a GM friend of mine a build I was making that uses the Two-Handed Fighter Archetype and we are both unsure of how it interacts with the damage bonus you get from two-handed weapons (which is 1-1/2 times your STR.) It reads:     Overhand Chop (Ex): At 3rd level, when a two-handed fighter makes a single attack (with the attack action or a charge) with a two-handed weapon, he adds double his Strength bonus on damage rolls. This ability replaces armor training 1.   So, does Overhand Chop turn the 1-1/2 times multiplier from two-handed weapons into a 2 times multiplier, or do the two add up? Or, do they interact in some other way?   This assumes a Medium sized character with 20 STR and a greatsword with 3 levels. Compile error: package javax.servlet does not exist Let $(a_n)$ be a strictly increasing sequence of positive integers such that: $a_2 = 2$, $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Let $(a_n)$ be a strictly increasing sequence of positive integers such   that: $a_2 = 2$ and $a_{mn} = a_m a_n$ for $m, n$ relatively prime. Show that $a_n = n$, for every positive integer $n$.   This is a result apparently due to Paul Erdős, and  has a proof by induction.  I tried like this, $a_{10}=a_2a_5$. After this what we can do?    [Editor's comment] Possibly due to the apparent simplicity of the conditions it may be difficult to appreciate the subtleties of this question. If we try and construct a counterexample like $a_3=4$, $a_4=5$, $a_5=6$, then the requirements dictate $a_6=8$, $a_{10}=12$, $a_{15}=24$. At this point we realize that we have been speeding. For monotonicity forces $a_9\le 11$, and therefore $a_{18}\le22&lt;a_{15}$, violating the requirements. It is not obvious why something similar ruins all the modifications to the sequence $a_n=n$. [/comment, JL] No sound in Ubuntu 13.04, only Dummy output device listed My sounds suddenly stopped playing on Ubuntu 13.04 64bit. I can only see Dummy output device in the Sound Settings and no sound is played during testing.  I can get the sound temporarily working by killing one of the strange processes, but I have to repeat the same after reboot  jan@jan-HP-Desktop:~$ ps -ef|grep pulseaudio 112       1405     1  0 09:38 ?        00:00:00 /usr/bin/pulseaudio --start --log-target=syslog 112       2132  1405  0 09:38 ?        00:00:00 /usr/lib/pulseaudio/pulse/gconf-helper jan       2575     1  0 09:39 ?        00:00:00 /usr/bin/pulseaudio --start --log-target=syslog jan       2597  2575  0 09:39 ?        00:00:00 /usr/lib/pulseaudio/pulse/gconf-helper jan       3160  2975  0 09:40 pts/2    00:00:00 grep --color=auto pulseaudio  jan@jan-HP-Desktop:~$ sudo kill 1405   Is there a permanent solution for this problem? Intuition on the Kullback–Leibler (KL) Divergence I have learned about the intuition behind the KL Divergence as how much a model distribution function differs from the theoretical/true distribution of the data. The source I am reading goes on to say that the intuitive understanding of 'distance' between these two distributions is helpful, but should not be taken literally because for two distributions $P$ and $Q$, the KL Divergence is not symmetric in $P$ and $Q$.   I am not sure how to understand the last statement, or is this where the intuition of 'distance' breaks down?   I would appreciate a simple, but insightful example. Entries I can safely exclude doing backups What is a fragment in a fragment shader? Wikipedia says that:     In general, a fragment can be thought of as the data needed to shade the pixel, plus the data needed to test whether the fragment survives to become a pixel (depth, alpha, stencil, scissor, window ID, etc.)   So is it textures, vertices or something else? Recursive bash script to collect information about each file in a directory structure How do I work recursively through a directory tree and execute a specific command on each file, and output the path, filename, extension, filesize and some other specific text to a single file in bash. While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$  Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so? What kind of passport do I need when entering the USA? I heard that you need a biometric passport if you want to enter USA. However, my friend has a biometric passport, but no fingerprints in it. Is this enough? And how long has the passport to be valid? Is it okay if it expires on the day after his return date? Is there an expression to indicate the strategy of wearing someone down with numerous small irritations? I would use rope-a-dope, but it's got connotations of pretending to lose that I don't need. I'm trying to describe the behavior of someone who sends twenty detailed emails a day about various projects, and the recipient of his emails has finally gotten tired of trying to parse them, so the recipient just agrees to whatever he proposes. I'd like an expression that evokes both outward earnestness and hidden creepiness, like Uriah Heep, Dickens's "Humble servant" with veiled ambition, who overdoes it with professions of humility until he's in power. Water torture would work, but comes too close to a more racist expression that I'm not comfortable with. Essentially, I would like to set my phone so that all my outgoing messages to a specific friend, with whom I currently text back and forth over iMessage, are sent out over SMS, never iMessage. Additionally, I would like to block any incoming iMessage texts of his and (hopefully) they'll fail and be resent via SMS.  Let me restate that: I would like to force SMS for both parties, going both ways, for a specific individual contact.   Is this possible? Blocking any usage of iMessage between myself and this other individual, so they must either send and receive any texts to or from me via SMS (or the texts fail/never arrive). Why does an unhinged body rotate about its centre of mass? Movie/TV series, which bullets carried DNA of the shooter How to make a GIS inventory? My office will be seeing a big change in its GIS section. This section has been operational since the 1980's and has a huge collection of GIS data (i.e., shapefiles, raster files, data, etc) but never been through any inventory. Now it will happen.   Is there any automated way to extract all the information about the GIS data (i.e., shapefile, arc-info coverage, layer file, *.mxd, gdb, raster file, and more) from a PC to an Excel file? The information may include date of creation, date last edited, folder or container name, etc.