I need a counterexample to the following statement: if A, B are subgroups of a group G, and A is a normal subgroup of B, B is a normal subgroup of G, then A is a normal subgroup of G. I've tried using groups such as $Z/nZ$ and $Z_n^{*}$, but neither seem to work. Thanks for your help.
Suppose $G$ is a group and $K\lhd H\lhd G$ are normal subgroups of $G$. Is $K$ a normal subgroup of $G$, i.e. $K\lhd G$? If not, what extra conditions on $G$ or $H$ make this possible?  Applying the definitions, we know $\{ghg^{-1}\mid h\in H\}=H$ and $\{hkh^{-1}\mid k\in K\}=K$, and want $\{gkg^{-1}\mid k\in K\}=K$. Clearly, the best avenue for a counterexample is if $gkg^{-1}\not\in K$ for some $k\in K$ and $g\in G-H$.  If no such element exists, $\{gkg^{-1}\mid k\in K\}\subseteq K$ implies $\{gkg^{-1}\mid k\in K\}=K$ because if $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
How to add csv file in QGIS 2.0 using Delimited Text Layer Plugin? I used to import csv files in previous versions of QGIS using the format below, but when I upgraded to QGIS 2.0 the delimited text layer plugin was changed and I can't import the file anymore. The plugin can't also detect the x,y or long lat field unlike in previous versions. Any ideas how to add the file? Thanks      Heres what it looks in Text Editor: How do I effectively defeat the Wall of Flesh? Is there any method to close connection similar to  for  in WordPress?  Is it not necessary to close connection for global $wpdb? Laggy viewport in 2.9 when model has a lot vertices I've taken up self-study of math. (How smart can that be?) I've just about finished a course in real analysis which spent a lot of time on metric spaces and some time revisiting calculus.  I was thinking of trying abstract algebra. I would appreciate any book recommendations.  Thanks in advance.  Andrew int t=-1'd1; //trying to assign -1 to t; bit vb[7:0]; // simple declaration  bit matrix[0:1][0:7]='{'{0,1,0,0,0,0,0,0},'{0,0,1,0,0,0,0,0},'{1,0,0,1,0,0,0,0},'{0,0,0,0,1,0,0,0},'{0,0,0,0,0,1,0,1},'{0,0,0,0,0,0,1,0},'{0,0,0,0,1,0,0,1},'{0,0,0,0,0,0,0,0}};// initialising multi-dimensional array.  I am using QuestaSim to compile and simulate. this is the report: # -- Compiling module test # ** Error: (vlog-13032) test.v(8): near &quot;=&quot;: Syntax error. # ** Error: (vlog-13032) test.v(9): near &quot;;&quot;: Syntax error.  I am somewhat New to the Verilog. Please help me in fixing these errors. multiline command chunks in bash history into multiple lines In Bash (as well as Bash in WSL), multiline written commands can call-back with up/down arrows correctly also built-in history command report them correctly (until I'm still logged-in in bash). But after exiting from bash and press up/down arrows, multiline commands appears each as separate entries and that's because cat ~/.bash_history has chucked the command on multiple separate entries and writing problem here; the history command result messed too. I have set these options in ~/.bashrc shopt -s cmdhist lithist HISTTIMEFORMAT='%F %T '  and now writing problem fixed but history command still report in multiple line and separate entries (reading problem now). Bash version is 4.3, same issue on v4.4 there is  since a long years but no solution yet. Update: it's also to mention that below multiline command as an example is correctly displaying even after exit Bash: for i in {1..10} do echo $i done  but this is failing: awk ' { print $1}' &lt;&lt;&lt;'first last'  Update 2: the issue mentioned above is fixed in Bash v5.0.0+ that changed to recognize entries between &quot;timestamp as delimiter&quot; as single entry of command (HISTTIMEFORMAT, cmdhist and lithist must be set). Do I need a visa for Paris, if my connecting flight stops for 24 hours in Paris? Do I need visa for Paris, if my connecting flight stops at Paris for 24hrs and I want to visit Paris during that 24hrs? I have US B1 visa. I'll be travelling from San Francisco to Bangalore (India) via Paris. I hold Indian passport. If yes, what is the procedure for that visa? What is the method to make files in a USB Flash Drive undeletable? Does these method work with USB 3.0 interface?  (I have surveyed some Flash Drive manufacturers, and I only see undeletable files for USB 2.0 drives, and not 3.0 drives. One manufacturer claimed that USB 3.0 cannot do undeletable file, so I am curious whether the claim is correct...) show that if $a | c$ and $b | c$, then $ab | c$ when $a$ is coprime to $b$. Given two numbers $a$ and $b$, where $a$ is co-prime to $b$,     Show that for any number $c$, if $a|c$ and $b | c$ then $ab| c$. Is the reverse also true? In other words, if $ab |c$ and $a$ is co-prime to $b$, then do we have $a | c$ as well as $b|c$? What causes the expansion of space? Which is more idiomatic in a bash script: `|| true` or `|| :`? I don't do terribly much shell scripting, so I was a little surprised when I was reading  and I saw the syntax they used in this documentation:     A non-zero return from the command in any submodule causes the processing to terminate. This can be overridden by adding || : to the end of the command.   I had to look up that || : was a shorthand for forcing a command to exit successfully.  Anytime I've had to make a command exit successfully, I used || true.  Is || : considered to be more idiomatic? Why don't we allow low reputation users to insert more than two images in their posts? I was searching for privileges one can get after a certain amount of reputation, and I ended up with this question in my mind: Why don't we allow users with low reputation to insert more than two images or links in their posts?  Is it good to make restrictions for them? Let's say someone is going to post a good question and he/she needs to insert three images to explain it properly; because this can't be done the question may become vague or incomplete to answer, and ultimately be flagged by others. Is this a rational decision for those users?  Yes, I believe that restrictions should be there to make a control over new users like for voting privileges but is it good to make restrictions on Questions and Answers?  Here is my question:  Is it decent to restrict new users in such a manner? So this homework question is in the context of $\mathbb{R}$ only, and we are using Lebesgue measure.  The sum $A+B$ is defined to be $A+B=\{a+b|a\in A,b\in B\}$. The question is: If $m(A),m(B)&gt;0$, is it necessarily the case that $A+B$ contains an open interval?  This question is quite annoying, since it is quite difficult to calculate the sum even after you construct an example. However, it seems like no matter how I constructed a set of positive measure, there is a region that have a lot of point in it (yet this region is not necessarily dense). This suggest that when summed up, the holes should be covered up somehow and thus there would be an open interval, and so the answer is yes. Sorry for a rather vague idea, but I'm just not sure what to do with this problem. I have a feeling that somehow topology would have a way to capture precisely the idea above about a region with a lot of point.  (also, I think my middle school teacher mentioned this once, and I remembered the answer to be yes; but that is quite a while ago though)  Thank you. Any help will be appreciated.  EDIT2: So Camilo's answer is gone. I am looking at the convolution method right now. Because the convolution is continuous, if it is nonzero I am done. But then there is no guarantee that the convolution is nonzero even if $x\in A+B$. I still need help with this direction. Can someone give me some hints?  EDIT: Camilo's answer below essentially suggest the method of getting 2 open interval from regularity theorem, one cover a subset of A and one cover a subset of B such that the sum of the measure of these two subset is bigger than the measure of the larger interval. user99680's answer point to a wiki which show a method that essentially simply acquire an open set. After some thinking, I believe that these method simply can't be made to work. I constructed this example in the hope that it would convince you that it is the case:  Let $I_{0},I_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is nonzero and $m(I_{n})=9^{-n}$ and all of them are inside $[0,2]$. Let $J_{0},J_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is at least $3$ and $m(J_{3n})=m(J_{3n+1})=m(J_{3n+2})=\frac{m(I_{n})}{3}$. Let $O=\bigcup I_{n}$ and $H=\bigcup J_{n}$. Construct $A\subset O$ such that $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$. Construct $B\subset H$ such that $m(B\bigcap J_{n})=\frac{3m(J_{n})}{4}$. Now if you apply Camilo's method, it is possible that your $I$ would be one of the $I_{n}$ above, and your $J$ is one of the $J_{m}$ above. No matter which $I_{n},J_{m}$ you have, the argument does not work. Rewrite $m=3q+r$ where $0\leq r\leq 2$. If $q&lt;n$ then $m(J_{m})&gt;m(I_{n})$ so the longer interval have length $\frac{m(I_{q})}{3}$; $m(A\bigcap I_{n})$ is at most $\frac{3m(I_{q+1})}{4}=\frac{m(I_{q})}{12}$ and $m(B\bigcap J_{m})=\frac{3m(J_{m})}{4}=\frac{m(I_{q})}{4}$. Thus $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. If $n\geq q$ then $m(I_{n})&gt;m(J_{m})$ so the larger interval is of length $m(I_{n})$; $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$ and $m(B\bigcap J_{m})$ is at most $\frac{3m(J_{3n})}{4}=\frac{m(I_{n})}{4}$. Once again $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. Hence the Camilo's argument does not work. user99680's method does not work either, since you get $O$ and $H$ which is even bigger than any interval. Need help with a question regarding the Big Oh In class we are currently covering upper/lower bounds, big Oh and omega and the like. I am pretty good on the "typical" functions one would do, but at a complete loss at "general" statements. This ons function, for example:  $$\forall f\in\mathcal{F}: \lfloor \sqrt{\lfloor f(n)\rfloor }\rfloor \in O(\sqrt{f(n)}).$$  the domain F by the same is all natural numbers that return a real positive number when inputted. My first instinct with something like this is to remove the floors, the general strategy I though to solve this is remove the outer floor, then get rid of the square root, and then the inner floor. But I have no idea how to go about this or if this is even the right way to do it.  Help please? Thanks in advance. What do you recommend as an optimal point and orientation for mounting a rear-view mirror? I'm open to handlebar-, glasses-, or helmet-mounted.  Some of the considerations, as I understand it, are:   Ease of mounting. View provided. Vibration. Likelihood of breaking. Robustness for repeated off and on (for glasses-mounted).   I've got a hybrid bike with flat handlebars.  I'd also welcome specific product recommendations and recommendations about other aspects to look for in a mirror. The Ubuntu installer does not allow me to install on NTFS partitions, but certain circumstances requires me to do so. Is it possible? Examples and further results about the order of the product of two elements in a group Let $G$ be a group and let $a,b$ be two elements of $G$. What can we say about the order of their product $ab$?   says "not much":     There is no general formula relating the order of a product $ab$ to the orders of $a$ and $b$. In fact, it is possible that both $a$ and $b$ have finite order while $ab$ has infinite order, or that both $a$ and $b$ have infinite order while $ab$ has finite order.   On the other hand no examples are provided. $(\mathbb{Z},+), 1$ and $-1$ give an example of elements of infinite order with product of finite order. I can't think of any example of the other kind! So:   What's an example of a group $G$ and two elements $a,b$ both of finite order such that their product has infinite order?     Wikipedia then states:     If $ab = ba$, we can at least say that $\mathrm{ord}(ab)$ divides $\mathrm{lcm}(\mathrm{ord}(a), \mathrm{ord}(b))$   which is easy to prove, but not very effective. So:   What are some similar results about the order of a product, perhaps with some additional hypotheses? Multiple addresses in JSON-LD Schema.org "With who" vs. "with whom" Is this correct?     The person with whom I'm doing the project should be here soon.   If it is, is with always a dative preposition (like mit in German)?