Conversation started Aug 22, 2018 at 17:57.
user131753
Aug 22, 2018 17:57
Let $(H,\circ)$ and $(H,\ast)$ be two groups and $S\subseteq H$. Let $[S]$ and $(S)$ be the subgroups generated by $S$ in $(H,\circ)$ and $(H,\ast)$ respectively. If $[S]=(S)$ for all $S\subseteq H$, are they isomorphic @TobiasKildetoft?
Tobias Kildetoft
Tobias Kildetoft
This is a very weird question
Usually we don't care about having two group structures on the same set
Rudi_Birnbaum
Rudi_Birnbaum
@user170039 one way to identify two groups is "isomorphism"
user131753
@Rudi_Birnbaum Yeah I know that. But that's not my point.
Rudi_Birnbaum
Rudi_Birnbaum
if you don't want "isomorphism" to be how you identifiy two groups then you have to exactly define how else
Tobias Kildetoft
Tobias Kildetoft
But this does seem like it might be an even stronger condition that in the linked question (as I don't think the bijection given there can be extended to be defined on the elements)
Rudi_Birnbaum
Rudi_Birnbaum
Aug 22, 2018 18:01
also what if the sets of groups are the same (isomorphic) but one occurs in one group "more often"?
user131753
@Rudi_Birnbaum For example?
Rudi_Birnbaum
Rudi_Birnbaum
no idea, just trying to make you get the piont that your question is not really formulated very good
Tobias Kildetoft
Tobias Kildetoft
@Rudi_Birnbaum I understand the question now. But it feels very unnatural, so I have no idea what the answer might be
Rudi_Birnbaum
Rudi_Birnbaum
@TobiasKildetoft OK! sorry then @user170039
user131753
@TobiasKildetoft I think that $\ast=\circ$ (as functions) and that's what I was trying to prove but couldn't.
Semiclassical
Semiclassical
Aug 22, 2018 18:05
Do you have an example in mind?
user131753
@Semiclassical Was this asked to me?
Tobias Kildetoft
Tobias Kildetoft
@user170039 Now that sounds much stronger than I would expect
Semiclassical
Semiclassical
yeah
user131753
@TobiasKildetoft Yeah indeed.
Semiclassical
Semiclassical
I guess one way to pose it: Can two different operations on a given group produce the same subgroup?
I would imagine the answer is a definite yes.
Tobias Kildetoft
Tobias Kildetoft
Aug 22, 2018 18:09
@user170039 Right, the operations will not be equal as functions (just take a prime number of elements and permute the non-identity elements)
Rudi_Birnbaum
Rudi_Birnbaum
Hi @MatheinBoulomenos! How are you?
MatheinBoulomenos
MatheinBoulomenos
Hi @Rudi_Birnbaum
I'm well, thanks. I'm on vacation right now, that's why I haven't been online for a while
user131753
@TobiasKildetoft So then we are left with isomorphisms.
Tobias Kildetoft
Tobias Kildetoft
right
 
Conversation ended Aug 22, 2018 at 18:12.