Let $a\in G$, where $G$ is a group. Prove that $a$ commutes with each of its conjugates in $G$ if and only if $a$ belongs to an abelian normal subgroup of $G$.   This is what I did:  $"⟹"$   First, I thought  of $&lt;a&gt;$ to be that normal subgroup containing a. But if it is true, I am not sure how to prove it. Then I thought of another subgroup, namely $C_G(a)$.   I conclude from the hypothesis given that $Cl_G(a) ⊂ C_G(a)$  And I know that  $|Cl_G(a)| = |G :  C_G(a)|$ by G acting transitively on $Cl(a)$ by conjugation. But I don't know how to get that $ C_G(a)$ is normal in $G$.  I am thinking also of  a normal subgroup of $G$ , say $K$ containing $a$, so  $&lt;a&gt;$ $\cap$ $K$ and  $ C_G(a)$ $\cap$ $K$ are subgroups of G inheriting abelianity from   $&lt;a&gt;$  and $ C_G(a)$ respectively, and normality from $K$.  $"\Leftarrow"$   Let $a ∈ H$ s.t. $H$ is an abelian normal subgroup of G. So $ah=ha$, ∀ $h∈H$.  Since $H$ is normal in $G$, hence $g^{-1}hg ∈ H$, ∀ $g∈G$   So $g^{-1}ag ∈ H$, ∀ $g∈G$   Which means     $Cl_G(a)$ $\subseteq$ $ H$  Therefore, a commutes with each of its conjugates in G.  Could someone correct me?
Let $a$ be an element of a group G. Prove that $a$ commutes with each of its conjugates in $G$ iff $a$ belongs to an abelian normal subgroup of $G$.   My Try:  I proved the backward direction. For forward one, given $g\in G, a(gag^{-1})=(gag^{-1})a$. I was trying to prove that $a\in Z(G)$. In order to prove that, I must show that $ag=ga$. But failed. Am I on the right track? Can anybody please give me a hint? I want to try it myself.
Alphabetically display the items in itemize I am wondering whether it is possible to display the items alphabetically regardless of the order we type on the backend? For example I am using itemize as follows:  \begin{itemize}     \item ISDYNSTP:  Is dynamic time step used?     \item ISCDCA:     \item MVAR     \item IS2TL_ \end{itemize}   But I want to display the items in the alphabetical order. I don't know whether it is possible or not. I have like hundreds of itemize items. Actually I am trying to create a glossary for the different parameters. On every boot up I get this:    For about 10 seconds (or more) before the "normal" Ubuntu Gnome loading screen and the login screen comes up.  I've read some questions and a bug report about this already, but this problem was dismissed as a minor inconvenience or there were no real solutions (that worked for me). For me, this problem leads to an unacceptable slow boot compared to Ubuntu (Unity) or Windows.  Does anybody know how to fix this or is this a "feature" of Ubuntu Gnome and not a bug? I don't understand this. These identities are given in the online notes for MIT's 18.01 calculus class. It's related to taking the sum of two trig functions and transforming them into a single trig function. I can use the formulas to do this, but I am having trouble finding anything on the internet to use a proof for my understanding.  Thanks for any responses!  EDITED (added after the first answer):  Sorry, the identity goes like this:  $A\sin{k(x-c)}=a\sin{kx}+b\cos{kx}$ ,  with the relationships  $a=A\cos{kc}$ , $b=-A\sin{kc}$ , $A=\sqrt{a^2+b^2}$ , and $\tan{kc}=-b/a$ This option seems to have gone away after upgrading from Ubuntu 16.04. I tried installing nautilus-open-terminal but it is unable to locate the package for 18.04. Also trying to install nautilus-actions gives:  Package nautilus-actions is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source     How to fix this? How to programmatically create a WPF window in a WinForm application Column line overshooting the horizontal line in table \begin{tabular}{|m{3.5cm}|m{5.5cm}|m{4.5cm}|} \hline \emph{Category of Benefits} &amp; \emph{General Benefit} &amp; \emph{Operational Advantages} \\ \hline \multirow{3}{*}{\pbox{3.5cm}{Robust operational performance}} &amp; \multirow{3}{*}{\pbox{5.5cm}{One sensor can contribute information while other are unavailable, denied(jammed), or lack coverage of a target or event}} &amp; -Allows continued operation despite jamming \\ &amp;  &amp; -Graceful degradation \\ &amp;  &amp; -Increased probability of detection \\ \hline  \multirow{2}{*}{\pbox{3.5cm}{Extended spatial coverage}} &amp; \multirow{2}{*}{\pbox{5.5cm}{One sensor can look where another cannot}} &amp; -Increased survivability \\ &amp;  &amp; -Probability of detection increased  \\ \hline \label{tab1.1} \end{tabular}   For the LateX code mentioned above the table which comes looks like this ![enter image description here][1]   In the resulting table, column 1 vertical line overshoots the last horizontal line.  What is the problem? Outlook from time to time makes an operation 'Send and Receive.' This can sometimes be quite slow and Outlook hangs while this is done.    Is is possible to set different intervals for different mail accounts (if user has multiple accounts configured within Outlook, e.g., main Exchange account and one or more IMAP accounts) If yes (or even if only for some [e.g., IMAP accounts], where in the settings it can be configured?  (e.g., Outlook 2007) What makes induction (over natural numbers) a valid proof technique? Is $$ \dfrac{ P(0) \quad \forall i \in \mathbb{N}. P(i) \Rightarrow P(i+1) }{ \forall n \in \mathbb{N}.  P(n)} $$ just taken for granted as a proof rule, or can it be derived from more foundational axioms?  Similarly, can the principle of induction over well-founded sets be derived from something more foundational, or does it have to be assumed to be a valid inference rule? Are there any rulesets or laws for Q or Q continuum? Anything, that would prevent them from doing... anything? Or at least limit them from doing something really bad?  In "" episode Q explains to  that they can do, have, get, become anything they want or appear anywhere they want, with just a snap of fingers. If that is true and if there are absolutely no borders or limits, they can't pass, then how can universe work in general? Or, how did it manage to exist so far?  I mean -- any Q could wipe out Borg, entire human population or even entire universe... with just a snap of fingers. What could prevent Q from doing that? When you're logged in to a Stack Exchange account, it's possible to add and remove logins (associated OpenID/OAuth accounts) without performing any form of reauthentication.  This is not a full vulnerability in itself, but it provides an extremely easy way for an attacker who's found a temporary XSS vulnerability to turn that into a permanent account hijack. An XSS exploit on its own may only provide a minute of indirect access to a user's account. But with this login configuration, the attacker could use that minute to automatically:   Add a new attacker-controlled login to the account. For example:   The XSS JavaScript would generate a random identifier, and use that to construct a login URL of the form http://hacker.example/openid/1234567. The hacker.example server would automatically authenticate the first OpenID login request made with a given identifier, allowing it to be added to the account without user interaction. All subsequent login attempts would be automatically rejected for the user. The hacker.example server would notify the operator with the account name and the generated identifier, and would allow them to login to it.  Remove all existing logins from the account. Log the user out of their account, with no way to get back in.   If this exploit were used to hijack moderator and staff accounts it could makes it more difficult to quickly respond to an attack, creating a greater opportunity for extended abuse.  This is a broader instance of the same fundamental vulnerability which was previously-reported as . That instance is a little more straightforward to fix: just ask for the password again when changing it. However, safe reauthentication with OpenID/OAuth logins is more complicated. Most users' OpenID providers will be configured to automatically re-authenticate upon request, so there would be minimal security gain from requesting that.  Instead or as well, my suggestion is that for users that have a verified email address, require an emailed confirmation link to be clicked before committing any login changes.    I'm not a real security expert, but I've just found that fellow Stack Overflow user , who is, previously described this vulnerability in his 2012 blog post :     If an attacker can add additional login accounts then they can create a persistent mechanism to get back into that Stack Overflow account at any time without needing to hijack the session. Now of course the victim can always go into their account settings and identify the presence of the new login, but it’s a question of how much damage is done before then.      How would an attacker mitigate the risk of having their newly added account removed? Just remove the victim’s original account(s) instead.      [...]      Finally, this is also a lesson for those considering using third party authentication providers. Yes, it’s great that you can farm off this work to someone who has a solid implementation, but you still have responsibility for securing the authenticated session on your site. If you choose not to, be aware of the risks because certainly we’ve seen them exploited before. Triggering actions when a variable is set Visa for European countries for US passport holder I'm a US passport holder. What documents do I need when I fly from the U.S. to Germany?  Assume I'm going for a few weeks of vacation.  Obviously I need my passport.  What about a "visa"?  Anything else? Suppose $f$ and $g$ are entire functions with $|f(z)|\leq|g(z)|$ for all $z$.  How can we show that $f=cg$ for some complex constant $c$?  Thanks for any help :) Currently there is not enough space on by /boot partition/folder and the sofware update can not be performed.   Question: how should I correctly free-up some space in that directory?  Here is the listing:  root@mindaugas-ubuntu-14:/boot# ls -la total 156607 drwxr-xr-x  4 root root     3072 Kov 12 09:37 . drwxr-xr-x 24 root root     4096 Kov 23 23:36 .. -rw-r--r--  1 root root  1163858 Rgs  4  2014 abi-3.13.0-36-generic -rw-r--r--  1 root root  1164720 Grd 16 03:17 abi-3.13.0-44-generic -rw-r--r--  1 root root  1164967 Sau 13 22:12 abi-3.13.0-45-generic -rw-r--r--  1 root root  1164852 Kov 10 22:43 abi-3.13.0-46-generic -rw-r--r--  1 root root   165671 Rgs  4  2014 config-3.13.0-36-generic -rw-r--r--  1 root root   165748 Grd 16 03:17 config-3.13.0-44-generic -rw-r--r--  1 root root   165748 Sau 13 22:12 config-3.13.0-45-generic -rw-r--r--  1 root root   165748 Kov 10 22:43 config-3.13.0-46-generic drwxr-xr-x  5 root root     1024 Kov 12 09:37 grub -rw-r--r--  1 root root 29225220 Rgs 25  2014 initrd.img-3.13.0-36-generic -rw-r--r--  1 root root 29264463 Sau 24 20:22 initrd.img-3.13.0-44-generic -rw-r--r--  1 root root 29267419 Vas 23 20:30 initrd.img-3.13.0-45-generic -rw-r--r--  1 root root 29268175 Kov 12 09:37 initrd.img-3.13.0-46-generic drwx------  2 root root    12288 Rgs 14  2014 lost+found -rw-r--r--  1 root root   176500 Kov 12  2014 memtest86+.bin -rw-r--r--  1 root root   178176 Kov 12  2014 memtest86+.elf -rw-r--r--  1 root root   178680 Kov 12  2014 memtest86+_multiboot.bin -rw-------  1 root root  3386479 Rgs  4  2014 System.map-3.13.0-36-generic -rw-------  1 root root  3388834 Grd 16 03:17 System.map-3.13.0-44-generic -rw-------  1 root root  3389258 Sau 13 22:12 System.map-3.13.0-45-generic -rw-------  1 root root  3389458 Kov 10 22:43 System.map-3.13.0-46-generic -rw-------  1 root root  5806848 Rgs  4  2014 vmlinuz-3.13.0-36-generic -rw-------  1 root root  5814496 Grd 16 03:17 vmlinuz-3.13.0-44-generic -rw-------  1 root root  5814112 Sau 13 22:12 vmlinuz-3.13.0-45-generic -rw-------  1 root root  5814592 Kov 10 22:43 vmlinuz-3.13.0-46-generic   There was a mention of it here:      $ sudo apt-get remove linux-image-2.6.31-21-generic   but will it delete all the files? Want to copy the file name quickly in Windows. I always use rename, copy , but feel it is quite slow. Any better way? For me at least,  has answers that are not sorted by score:    Is this normal?  I've tried refreshing the page a few times, but the order doesn't change... How to select even or odd elements based on class name if you create html layout like so  &lt;ul&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt;      &lt;li class='a'&gt;&lt;/li&gt;      &lt;li class='b'&gt;&lt;/li&gt; &lt;/ul&gt;   and try to select odd elements with 'a' class like so $$('.a:odd') you will get empty array, and if you do $$('.a:even') you will get all four li elements with 'a' class.. It really strange.. But im new to mootools, maybe im doing something wrong..    So my question is how to select first and third li elements with a class. I know that i can do it with function like this  $$('.a').filter(function(item, index) { return index%2; }  but its too complicated for such small task as selecting odd or even elements.. Replacing some characters in the previous command Can bounties be awarded multiple times by the same person on the same question? Is there any restriction to prevent multiple bounties ( not simultaneously) being awarded to different answers on the same question ?  i.e., on a question  -  I want to set multiple bounties since I've received 2 awesome answers. Can I set award a bounty on answer A first, and then set the bounty and award it to answer B ? Sum of (example: 0+1+2+3 = 6 , 0+1+2+3+4+5+6+7 = 28 and so on)