How could I show that a normal subgroup of a normal subgroup of a group is not always a normal subgroup of a group?  Here is what I did. I know the Algebra system GAP, so I wrote the following function:  performTheProve:= function()     local g, gDash, AFour, NormalOfAFour, NormalK;     AFour:=AlternatingGroup(4);     NormalOfAFour:=NormalSubgroups(AFour);     for g in  NormalOfAFour do         for gDash in NormalSubgroups(g) do             if(not gDash in NormalOfAFour) then                 return [gDash, g];             fi;                  od;     od; end;    This function produces the following result:  [ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]) ]   So, now I am stating that the Group([ (1,3)(2,4), (1,2)(3,4) ]) is a normal subgroup of a ${A}_{4}$, Group([ (1,2)(3,4) ]) is a normal subgroup of a Group([ (1,3)(2,4), (1,2)(3,4) ]) and not a normal subgroup of a ${A}_{4}$.  Given this example I am proving that normal subgroup of a normal subgroup of a group is not always a normal subgroup of a group. Am I right?
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 $k'\in K$, $gk'g^{-1}=k\in K\Rightarrow k'=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}$.
I see the  on the user page.  It seems to edit posts sometimes.  What is it? And why has it downvoted over 1800 times? See also: Avoid if for chained objects Is there a reason why we gain ghost reputation on the meta of an SE website? 80s-90s anime (movie or series) in which an armored girl exchanges age with a dying man on a beach If a friend wants to transfer $7000 to me, what is the best way to do it, especially if we don't share the same banks? We could do wire transfer, but that comes with its own fees. Could we also do direct deposit? Paypal doesn't allow such large lump sum transfers.  This would be US-to-US transfer. Will reseting my device to factory setting unroot and remove superuser? I just upgraded to Ubuntu 18.04 but the UI did not change at all. For example I don't have the full screen app launcher; it's the same as it was back in 16 and 17. No notification center etc.   Do I need to somehow opt in? Changes to "close as duplicate" (part deux) Recently I've noticed more and more sign-in / sign-up workflows defaulting to the sign-up option first and providing a minimal "Sign-In / Login" action somewhere else on the page. What's the reason behind this practice?  Examples where I've seen this recently: My ubuntu partition was deleted and I can't boot from either a DVD or USB For what natural numbers is $n^3 < 2^n$? Prove by induction Problem  For what natural numbers is $n^3 &lt; 2^n$?  Attempt @ Solution   For $n=1$, $1 &lt; 2$ Suppose $n^3 &lt; 2^n$ for some $n = k \ge 1$ It looks like the inequality is true for $n = 0$, $n = 1$ and $n\ge10$ But, how can I prove this through induction? Female main character in an ice world earning gems/runes Should I activate the additional driver: Processor microcode firmware for Intel CPUs for intel-microcode I have recently noticed that there is an additional driver called Unknown: Unknown here which was not there before:    I have already read , and understand why it is there, but my question is, should I activate it? Will this make a difference? Or should I just leave it as it is as it sounds rather useful?    OS Information:  Description:    Ubuntu 15.04 Release:    15.04 Image of the union is the union of the images $$ f\left(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}\right) = \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$$  Let $b \in f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda})$ $\rightarrow b=f(a)$ for some $a \in (\bigcup\limits_{\lambda \in \wedge} A_{\lambda})$.  Since $a \in (\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) \rightarrow a \in A_{\lambda}$ for some $\lambda \in \wedge$.  Since $b=f(a)$, then $b \in f(A_{\lambda})$ for some $\lambda \in \wedge$.  $\rightarrow b \in \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$ $\rightarrow f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) \subset \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$.  I used a similar argument to prove $\bigcup\limits_{\lambda \in \wedge} f(A_\lambda) \subset f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda})$.  Which shows $ f(\bigcup\limits_{\lambda \in \wedge} A_{\lambda}) = \bigcup\limits_{\lambda \in \wedge} f(A_\lambda)$.  Please let me know if this is valid. Subdomain versus subdirectory I recently moved from having a main site like www.example.com with a subdomain associated site yyy.example.com, to moving everything off the subdomain to a subdirectory like www.example.com/yyy/, with the same hierarchy underneath the subdirectory, and now yyy.example.com is just a redirect page to the subdirectory, so yyy.example.com/abc/page-x is redirected to www.example.com/yyy/abc/page-x and so on.  The effect has been that the traffic increased by several times (the total page views are more than ten times at the present rate) what it was to the yyy subdomain. Previously almost all of the traffic to the yyy subdomain was from the www site. This is from monitoring the old yyy site for a year or more.  I'm curious to know if there's any research or results which would indicate whether this is universal (subdirectories beat subdomains) or "just me". Avoiding NullPointerException in Java My friend Evan,is griefing me. When I went on his world to get revenge, he banned me. The screen said " HerobineYoung has banned enderpizza you will be kicked out in 5 minutes." I begged him to tell me how but he wouldn't tell me.   How can I ban people in Minecraft PE? How can I customize a system locale? Windows 7 / 8 tool for amplifying sound / equivalent to pavucontrol? I'm watching live streamed videos, but the volume sometimes is barely loud enough.  With Windows, I can only turn the system volume and the flash volume up to maximum, and then bad luck if that's still not loud enough.  On Linux, on the same notebook, I have the same problem, but I can use pavucontrol (pulseaudio volume control) to amplify the sound volume beyond 100%.  Unfortunately, I can't always work on Linux, since Visual Studio and SQL-Server only run on Windows.  Is there any tool or way to amplify volume on Windows 7 / 8 , without having to resort to run Linux in a virtual machine? 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