There are certainly legitimate reasons when it is useful to copy text of a comment. As Emil Jeřábek said, it is not clear whether they are frequent enough to be integrated in Stack Exchange software.
All right, there are surely exceptions, but the question is if they are frequent enough to warrant a special software support. -- Emil Jeřábek
If you think that such feature can be useful you should probably upvote the feature request on Meta Stack Exchange: Is there a way to view a comment's source?
I will summarize some of the methods how to copy comments. There are certainly many possibilities how to do this, I'll mention the ones I have at least tried.
I will also refer to the answers here: Is there a way to view a comment's source? on Meta Stack Exchange and How to copy mixture of text and latex formulas in a comment? on Mathematics Meta.
Clearly, if the comment contains only plaintext, it can be copied directly from the browse. It is more problematic what to do if the comment also uses some MarkDown and MathJax. We can try whether various methods work on this comment which contains both.
There is a bookmarklet for this available among NormalHuman's Stackmarklets (Wayback Machine). In Google Chrome I had no problems with using it - all I needed to do was to drag the bookmarklet to the bookmark bar and then I was able to start using it.
The disadvantage is that in this way you can only copy the comments under the question. Also, it won't work if the question is closed, since the comment is copied into the answer box.
In this way I get:
In addition to [@PaulBroussous](https://mathoverflow.net/a/295954)'s nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ <i>has</i> no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$). -- [LSpice](https://mathoverflow.net/questions/295897/contragredient-of-a-cuspidal-representation#comment735801_295897)
In addition to @PaulBroussous's nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ has no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$). -- LSpice
You can also simply view the source of the page (Ctrl+U in Google Chrome) and view the text of the comment of the comment in this way. If there are many comments, you can use StackPrinter to see the version of the question with all comments displayed. Notice that you get things in HTML, while the bookmarklet mentioned above changes some of it to MarkDown.
In addition to <a href="https://mathoverflow.net/a/295954">@PaulBroussous</a>'s nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ <i>has</i> no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$).
In addition to @PaulBroussous's nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ has no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$).
It is also possible to use this SEDE query with comment id or comment link. See this answer for more details.
In addition to [@PaulBroussous](https://mathoverflow.net/a/295954)'s nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ *has* no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$).
In addition to @PaulBroussous's nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ has no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$).
Here both url and italics have been converted to MarkDown.
