By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$\langle a, y\mid y^2, y^{-1}a^2=a^{-2}y\rangle,$$
where $y=ab$. This presentation is equivalent to
$$\langle a, y\mid y^2, y^{-1}a^2y=a^{-2}\rangle,\tag{P}$$
which defines...
@XanderHenderson This last change in handle was a mod-determined change (deleting the temporary "joke", and replacing it with his default userid username). Likely in 28 days or so, RF can reclaim his typical handle. (One can change one's username only once per 30 days, I believe.)
@MartinSleziak Closed in the opposite direction: older of two posts is now closed as a dupe of the newer question, as I agree with your assessment about the better quality of the more recent question.
Perhaps a merge of the answers might be considered, as I saw on the older question, an upvoted answer with 10 net upvotes. But that depends on how varied the answers are on the newer question.
Thanks! I should probably work a bit harder - to get one of those shiny gold hammers myself.
Merging would lead to duplicate answers. (At least Jack D'Aurizio's answer on both questions seems almost the same to me. I did not look at other answer, but this one caught my eye.)
Andrei's answer on the newer question is similar to Deepak's answer on the older one.
I do not see a way of merging them without either having (almost) duplicate answers on one question or removing of some answers. So probably it's better not to merg in this case.
Maybe this math.stackexchange.com/questions/2658464/… can be encouraged to re-word in a form like "what tools do we have for assessing wagstaff primes / eliminating the possibility it's a wagstaff prime"?
@José Carlos: I don't think that just showing something that doesn't work counts as effort. But I would like to know the source and motivation of the problem.
I mean, I could post any PSQ and then add a couple sentences about some failed attempt - that is not what I think this site should count as "research effort".
@Lord_Farin I have mentioned one possibility for a dupe target about $|\mathbb R|=2^{\aleph_0}$ in the searching chatroom. Maybe someone else notices my message there and finds some other possibilities.
@JoséCarlosSantos Echoing @CarlMummert's comment, I take a the stance that work or effort shown is, essentially, no context at all 99% of the time. That particular question falls into that realm. The question itself is not intrinsically interesting (to me), there is no source or motivation, and there is no reason to believe that the question will be useful to anyone else in the future.
Anyone interested in speaking here, please simply request access, and most will be allowed in by me or any other room owner here.
@CarlMummert I added you to the list of room owners of CRUDE. That doesn't mean a whole lot, but it allows you to pin and unpin, with a few other privileges. Check the "room" upsidedown triangle immediately above the line of icons in the room, on the right hand side. You'll see some options on the drop down menu for room owners.
@CarlMummert Well, I've been wanting to add you to the room owners for awhile. The Gallery status won't change much at all, and is only temporary. It is essentially still public.
@Lord_Farin It's temporary. Seems a couple users want to be able to speak everywhere, but shut out others from speaking in their chats. Just demonstrating how nice open chatrooms everywhere is.
I thought "shutting out" to be peculiar, requested access to one of those chats, and flagged a chat mod. But chat mods are near worthless. (Always try to deal with chat issues through an MSE mod. Chat mods may come from who knows where, and have no clue about this site.) No one at all except the couple of wanna be private users will be affected here.
@XanderHenderson Hey... I didn't do anything to gain it, other than answer the question over four and a half years ago!
It seemed straightforward to me. $a > b$ if $a -b > 0$. I may have been naive at the time thinking the question was sincere, but, heck, naivete isn't punishable by death, or at least that was my belief, until yesterday!!
@XanderHenderson Yes. It's not my proudest moment!
By @user2345215's comment, the group $G$, defined by the presentation in question, is also defined up to isomorphism by
$$\langle a, y\mid y^2, y^{-1}a^2=a^{-2}y\rangle,$$
where $y=ab$. This presentation is equivalent to
$$\langle a, y\mid y^2, y^{-1}a^2y=a^{-2}\rangle,\tag{P}$$
which defines...
