Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.  My approach: Let's try to prove by induction on $o(G)$.  If $o(G)=1$ then the result is obvious. Suppose it is true for all groups with order less than $n$.  Suppose that $o(G)=n$ and $H\neq \{e\},G$ because otherwise the result is trivial.  Let $x\in H$ s.t. $x\neq e$ and consider subgroup generated by $x$, i.e. $\langle x\rangle$ then $H/\langle x\rangle\leq G/ \langle x\rangle$. Note that $(G/\langle x\rangle)/(H/\langle x\rangle)\cong G/H$. Since $o(G/\langle x\rangle)&lt;o(G)=n$ then by induction hypothesis there exists subgroup $\bar{K}\leq G/\langle x\rangle$ such that $\bar{K}\cong (G/\langle x\rangle)/(H/\langle x\rangle)$ $\Rightarrow$ $\bar{K}\cong G/H$.  Then I was trying to find subgroup of $G$ which is isomorphic to $G/H$ but no results. I've considered the surjective mapping $\pi: G\to G/\langle x\rangle$ which does give any useful results.  I would be very thankful if anyone can show how to complete my approach?
I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states     Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.   Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism $\phi\colon G\to K$ for $K\leq G$, and $\ker\phi=H$, and then just use the isomorphism theorems to get the result.  After a while of trying, I've failed to come up with a good map, since $H$ seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you.
I have just started learning Japanese Mahjong and I am a bit lost as to why this hand failed.  I did several pons and chis over the course of a hand, having a total of three open melds (one set and two sequences).  Apart from these I had an incomplete bamboo sequence of II and III as well as a pair of dragons, therefore I was on a single wait for I or IV bamboo stick tile. When the IV tile appeared in one of the discards of other players, I was given a choice to do another chi and declare another open meld, but I couldn't win with a ron. Why was that?  Am I wrong that winning by ron guarantees a single needed yaku to win a hand, just like Closed Tsumo does?  After declaring the chi I was offered with, my hand was left with a single pair of tiles, one of which I had to discard.  Eventually the round finished with a draw.  Why couldn't I win this hand with a ron on the IV bamboo tile, and why was I given only the choice of chi instead? Blade of the Ruined King Vs. The Bloodthirster Effects:  +25 attack damage +15% life steal +40% attack speed Passive  Unique: Your attacks deal 5% of the target's current health in physical damage (60 max vs minions and monsters). Active   Unique: Drains target champion, dealing 15% of the champion's maximum Health in physical damage and healing you by the same amount. Additionally you steal 30% of their movement speed for 4 seconds - 60 second cooldown. (500 range)  Cost: 3200g VS.   Effects:  +70 attack damage +12% life steal Passive: Gain +1 attack damage and +0.2% life steal per kill (maximum +30 attack damage and +6% life steal). Half of the bonuses are lost upon death.  Cost: 3200g What is better for what situations, and when should you build BT instead of BotRK and Vice Versa. Which provides more lifesteal, the only variable being the stats of the weapon? I find myself constantly questioning my decision on which weapon to get and would really love some in depth answers. Would be great if any examples given were given from the ADC side of the house. Connected graph with colored edges We have connected undirected graph with colored edges in two way (green or blue). And also each vertex have the same number of green and blue edges. How to prove that there are alternate colored (green-blue) path between any two nodes? Can I clear an area of my prison for everyone BUT max security prisoners? What is the fastest algorithm for finding the natural logarithm of a big number? Is it possible to save `bc` command line history? bash has a handy file .bash_history in which it saves the history of commands and on the next execution of bash the history is populated with saved commands. Is it possible to save bc command history to a file in the same way and then load it on startup so that bc history is preserved? I tried reading  and it mentions readline and libedit. From ldd /usr/bin/bc I see mine uses readline and readline has  and read_history functions. Is this functionality implemented in bc or to do it I'll need to patch bc? Due to my ignorance, I find that most of the references for mathematical analysis (real analysis or advanced calculus) I have read do not talk much about the "multivariate calculus". After dealing with the single variable calculus theoretically, it usually directly goes to the topic of measure theory.    After reading the wiki article "", I'd like to find the rigorous proof for this test. The first book comes to my mind is Courant's Introduction to Calculus and Analysis which includes the multivariate case in the second volume.   Motivated by this, I'd like to put the question here:     What are the usual references for the   theoretical treatment for   ? Is this correct for Rudin exercise 3.7? Prove the series is convergent This is Baby Rudin exercise 7 of Chapter 3. Prove that the convergence of $\sum{a_n}$ implies the convergence of $\sum{\sqrt {a_k} \over k}$ if $a_n \ge 0$.  Proof:  I will attempt to show that the cauchy criterion is met so $${\sum_{k=n}^m {\sqrt {a_k} \over k}} \le \epsilon $$ for all $n,m \le N_1$. Im disregarding the absolute value since its all positive terms.   Since $\sum{a_n}$ converges, we know that the $\lim \limits_{n \to \infty} a_n =0$ So there exists some $N_2$ such that for all $n \ge N_2$, $$   a_n \lt \ {{(m-n)}^2k^2(\epsilon)^2} $$ I am again disregarding the absolute values because its always positive Let N=max{$N_1,N_2$} then for all $n,m \ge N$ $$\sum_{k=n}^m {\sqrt {a_k} \over k} \le \sum_{k=n}^m {\sqrt {{(m-n)}^2k^2(\epsilon)^2} \over k} \le \sum_{k=n}^m{(m-n)(\epsilon)} = \epsilon $$  Maybe this is just crazy, should I just use Cauchy-Schwarz inequality and forget about this or could I possibly make this method work? How to show that $\int_{0}^{\pi}{\sin^5 x\over 1+\cos^3 x}\mathrm dx=\ln 3?$ Given the integral $(1)$     $$\int_{0}^{\pi}{\sin^5 x\over 1+\cos^3 x}\mathrm dx=\ln 3\tag1$$   This integral seems to be quite tough, I can't even know where to start. I have try substitution $u=1+\cos^3 x$ and $u=\cos^3 x$, can seem to remove the x variables.  We have the following identities   $$\sin^5a={5\over 8}\sin a-{5\over 16}\sin 3a+{1\over 16}\sin 5a\tag2$$  $$\cos^3 a={3\over 4}\cos a+{1\over 4}\cos 3a\tag3$$  We may write $(1)$ as  $$\int_{0}^{\pi}{\sin x(1-\cos^2 x)^2\over 1+\cos x(1-\sin ^2 x)}\mathrm dx\tag4$$  $u=1-\sin^2 x\implies \mathrm du=-2\sin x\cos x\mathrm dx$ then $(4)$ becomes  $${1\over 2}\int_{0}^{1}{(1-u)^2\over \sqrt{u}+u^2}\mathrm du\tag5$$  $u=v^2\implies du=2vdv$ then $(5)$ becomes  $$\int_{0}^{1}{(1-v^2)^2\over v+v^4}\cdot{v\mathrm dv}\tag6$$ Simplify to  $$\int_{0}^{1}{(1-v)(1-v^2)\over 1-v+v^2}\mathrm dv\tag7$$  I guess using partial decomposition of fraction and integrate term by term, seem lengthy process  How else can we prove $(1)$? A common mistake for beginning group theory students is the belief that a quotient of a group $G$ is necessarily isomorphic to a subgroup of $G$.  Is there a characterization of the groups in which this property holds?  If this question is too broad, I might ask if such a characterization exists for $p$-groups.    History: I originally posed the opposite question, regarding groups for which $\exists N\unlhd G\,:\, \not\exists H \unlhd G\, \text{  s.t. } H \cong G/N$, and crossposted this to .  I received an answer there to the (now omitted) peripheral question about probability, which shows that most finite groups probably have this property.  After this, I changed the question to its current state, as this smaller collection of groups is more likely to be characterizable. Let $L$ denote the set of all lines in the Euclidean plane $\Bbb R^2$. We say that two lines $L$ and $M$ are parallel if either $L = M$ or $L$ and $M$ have no points in common.  Using Euclid’s Parallel Postulate prove that parallelism is an equivalence relation on the set $L$.  I need some help determining if I have the right idea about this.  $L=L$ which means that $L$ is parallel to itself. So this is reflexive.  Suppose we have $L || M$. Then Either $L=M$ or $L$ and $M$ share no points. So Then $M=L$ or $M$ and $L$ share no points. Then $M||L$. So this is symmetric.  Suppose we have $L || M$ and $M || N$. So then either $L=M$ or $L$ and $M$ share no points and either $M=N$ or $M$ and $N$ share no points. So we have that either $L=N$ or $L$ and $N$ share no points. So $L || N$. So then this is transitive.  So from all of this, parallelism is an equivalence relation. I'm not sure if this is correct or not. Any help would be appreciated. How do I connect to the wifi in Ubuntu within my Virtual Machine? I have a virtual machine installed on my Windows 8.1 laptop with Ubuntu installed on it, but when I open the Network menu it doesn't show any wireless networks. How do I connect Ubuntu in my virtual machine to a wireless network that my host machine is detecting? Rename Multiple files in unix I have multiple files whose names look like this  2313_ABCDEAUG2014_PQRST_0_.pdf 3244_ABCDEAUG2014_PQRST_0_.pdf 4234_ABCDEAUG2014_PQRST_0_.pdf   t want to rename the file in the following format.  AZ-2014-08-2313.pdf AZ-2014-08-3244.pdf   Please suggest. Is posting someone's comment as an answer okay? I came across  and at the time there were no suggested answers. After reading Mark Hubbart's comment (at the time with four upvotes), which was the correct answer, and seeing that it had been over 30 minutes since his comment, I figured I would provide that answer in my own words, which I did. My answer is longer, and I think more rounded and more useful, but the core of the answer came from the comment. I have no idea if I'd have thought up the answer on my own--a definite maybe to that.  Now had I tried to be "honorable" or something, such as instead of answering, suggesting to the commenter that he make his comment an answer, I would only have accomplished giving away easy rep to some other user not as into gratuitous self-denial.  The commenter had merely to enter his comment into an answer box instead of a comment box! It was his choice! Whether he was too lazy or didn't think his answer worthwhile or some other reason is not my responsibility. Must I bend over backward to help him earn points that he so carelessly left lying around for another person to take? It just took a tiny bit of non-laziness on my part to get my text typed into the box that would reward me instead of the box that wouldn't.  If my perspective is incorrect and the community would rather see such answers-as-comments rewarded and supported, I will gladly accept that. In the meantime, I just want to shake off the tiny bit of guilt my overly-nice and overly-concerned-with-propriety self is feeling. :) Word for someone who is driven by morality I'm looking for a word to describe someone who is very driven by morality, constantly trying to do what they believe is morally right.  This word would be used in a sentence like "I really admire her; she truly is a (...) person." Deriving the Lagrangian for a free particle I'm a newbie in physics. Sorry, if the following questions are dumb. I began reading &quot;Mechanics&quot; by Landau and Lifshitz recently and hit a few roadblocks right away.  Proving that a free particle moves with a constant velocity in an inertial frame of reference ($\S$3. Galileo's relativity principle). The proof begins with explaining that the Lagrangian must only depend on the speed of the particle ($v^2={\bf v}^2$): $$L=L(v^2).$$ Hence the Lagrance's equations will be $$\frac{d}{dt}\left(\frac{\partial L}{\partial {\bf v}}\right)=0,$$ so $$\frac{\partial L}{\partial {\bf v}}=\text{constant}.$$ And this is where the authors say   Since $\partial L/\partial \bf v$ is a function of the velocity only, it follows that $${\bf v}=\text{constant}.$$  Why so? I can put $L=\|{\bf v}\|=\sqrt{v^2_x+v^2_y+v^2_z}$. Then $$\frac{\partial L}{\partial {\bf v}}=\frac{2}{\sqrt{v^2_x+v^2_y+v^2_z}}\begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix},$$ which will remain a constant vector $\begin{pmatrix} 2 \\ 0 \\ 0 \end{pmatrix}$ as the particle moves with an arbitrary non-constant positive $v_x$ and $v_y=v_z=0$. Where am I wrong here? If I am, how does one prove the quoted statement?  Proving that $L=\frac{m v^2}2$ ($\S$4. The Lagrangian for a free particle). The authors consider an inertial frame of reference $K$ moving with a velocity ${\bf\epsilon}$ relative to another frame of reference $K'$, so ${\bf v'=v+\epsilon}$. Here is what troubles me:   Since the equations of motion must have same form in every frame, the Lagrangian $L(v^2)$ must be converted by this transformation into a function $L'$ which differs from $L(v^2)$, if at all, only by the total time derivative of a function of coordinates and time (see the end of $\S$2).  First of all, what does same form mean? I think the equations should be the same, but if I'm right, why wouldn't the authors write so? Second, it was shown in $\S$2 that adding a total derivative will not change the equations. There was nothing about total derivatives of time and coordinates being the only functions, adding which does not change the equations (or their form, whatever it means). Where am I wrong now? If I'm not, how does one prove the quoted statement and why haven't the authors done it? Before the seasonal flu vaccine was recommended for all healthy adults, people in developed countries might get vaccinated against on the order of 10 different pathogens and develop immunity for some more via natural exposure.  Now the same person is expected to take a seasonal flu vaccine for a new strain every year - won't this put them in an uncharted territory as far as effects on the immune systems are concerned within a couple of decades? Cannot add custom launcher to Ubuntu Dock ('Add to Favorites' option does not show up) I have a custom .desktop file: [Desktop Entry] Type=Application Terminal=false Icon[en_US]=/home/alex/bin/idea-IC-173.4127.27/bin/idea.png Name[en_US]=IntelliJ Exec=/home/alex/bin/idea-IC-173.4127.27/bin/idea.sh Name=IntelliJ Icon=/home/alex/bin/idea-IC-173.4127.27/bin/idea.png  Output of ls -lha .local/share/applications/intellij.desktop: -rwx--x--x 1 alex alex 248 Dez 30 10:54 .local/share/applications/intellij.desktop  Answers posted to a previous question () don't work for me. I can find this launcher in Activities and I can double click on that file and IntelliJ will start. Nevertheless, I cannot add this launcher as a favorite to the dock. The option does not show up. Search across Stack Exchange sites in Android app Is it possible to search for a question across all Stack Exchange sites in the Android app?   Right now I can search in only one specific site at a time.   If not, can you add that feature? I need an approach to allow delegated admins (i.e., non-Administators) to add members to Public Groups and Collaboration Groups (the GroupMember and CollaborationGroupMember objects).  I added a custom object (UserGroupRequest__c) to capture which Users should be added/removed to which Groups.    I've also built a batchable class to process the requests.  However, GroupMember is a Setup object, and CollaborationGroupMember is not.  So this doesn't work if I combine the operations in a single batchable class.  Should I instead be using 2 batch classes, calling one from the finish method of the other?    Any recommendations on how best to proceed?  Am I way off base?    Would a trigger on the Request object perhaps be better? I'm guessing that there may be a permission problem if this is attempted via a trigger as the delegated admins won't be able to add users to public Groups.  Also, I'm guessing the trigger approach would run into the same mixed DML issue - on Insert of the Custom object, I would need to Insert/Delete from GroupMember and from CollaborationGroupMember, and then perform a 2nd update on the Custom object to indicate that processing was completed...