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 Yearling
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Feb
25
comment Can I recover a group by its homomorphisms?
@AlexWertheim: For $G$ finite the answer should be yes. If you can determine for every finite group $H$ whether it is a quotient of $G$, then the biggest quotient will be $G$ itself. I think you can get the number of surjective homomorphisms $G \to H$ modulo conjugation by induction: the number of all homomorphisms is given and every non-surjective homomorphism has a proper subgroup $U$ of $H$ as image, so you know their number by induction (you'll have to correct this number considering $N_H(U)$ and the conjugates $U^h$).
Feb
24
revised A group whose automorphism group is cyclic
added warning that answer is wrong
Feb
24
suggested approved edit on A group whose automorphism group is cyclic
Feb
24
comment A group whose automorphism group is cyclic
In case you didn't see it: the accepted answer is wrong! Can you maybe un-accept it? Thanks!
Feb
23
comment A group whose automorphism group is cyclic
I thought about it, but couldn't come up with alternatives (I tried variations of Jack Smith's example for an older question by W4cco). I tend to believe that the answer is no, and that the question could be asked at mathoverflow.net (but infinite abelian groups ain't my strength).
Feb
23
comment What's so special about the group axioms?
@Shaun: The distinction between groups and inverse semigroups depending on whether they are distance-preserving or not in your first comment lacks a bit of context (I guess one finds it in the book you linked to).
Feb
23
accepted How many non-isomorphic central extensions of a cyclic group of order $2$ by the Lamplighter group exist?
Feb
23
comment A group whose automorphism group is cyclic
A hint in case you don't find any other automorphism of $(\mathbb{R}/\mathbb{Z})^2$: Look for vector space automorphisms of $\mathbb{R}$ seen as vector space over $\mathbb{Q}$ fixing $\mathbb{Q}\le\mathbb{R}$.
Feb
23
comment What's so special about the group axioms?
You are not the only one not recognizing the importance of these axioms at once. Historically seen it took more than a century before they were generally considered important in mathematics. Shortly ago I attended a lecture by Bernd Fischer (discoverer of three sporadic finite simple groups) who spoke about his first years in university as student. Several objects (like the sympletic group) were introduced to him without even mentioning that they were groups.
Feb
23
comment A group whose automorphism group is cyclic
I'd read the original question as speaking of automorphisms as an abelian group. In your example the two automorphisms are the only ones that additionally preserve the structure as an abelian variety. There are surely more group automorphisms that do not preserve this structure.
Feb
20
comment What does this statement mean?
$\psi\circ\phi$ does not have to be onto, so if you want to identify its image (which is a set of cosets in $H$) with some set of cosets in $G$ you have to restrict the image, not the domain of the map. The "is" in the mysterious last sentence shouldn't be taken literally, but instead as "can be identified with".
Feb
20
comment How many non-isomorphic central extensions of a cyclic group of order $2$ by the Lamplighter group exist?
Very nice, thank you. I'm still struggling to understand if the extra-special $2$-groups $N_n$ are isomorphic or not (if you happen to know, a simple yes/no would be appreciated a lot), but the $G_n$ are different.
Feb
19
comment How many non-isomorphic central extensions of a cyclic group of order $2$ by the Lamplighter group exist?
@pjs36: Sorry for driving you nuts. Earlier I didn't have time to look up "\wr" and just copied and pasted from the other question.
Feb
19
asked How many non-isomorphic central extensions of a cyclic group of order $2$ by the Lamplighter group exist?
Feb
19
comment Finitely many group extensions?
@DerekHolt: You're right. As you still seem to think that your construction works, I'll post it as a question.
Feb
19
comment Can a normal subgroup of a finite nonabelian group be nonabelian?
$D_{4n}$ has $D_{2n}$ as subgroup of index $2$ and subgroups of order $2$ are normal. Even easier is to take the direct product of two non-abelian groups.
Feb
19
comment Finitely many group extensions?
@DerekHolt: As the extensions proposed in your comment have to be central, how do you hope to put a central element of order $2$ "below" the Lamplighter Group? (I have a hard time finding a second possible $G$.)
Feb
18
awarded  Civic Duty
Feb
18
comment Finitely many group extensions?
@DerekHolt: You can distinguish the groups by checking if all elements of order $2^n$ are central or not.
Feb
18
comment Finitely many group extensions?
If you take $F=Z_3$ and pick different homomorphisms from $H$ to $Aut(F)$ (only the $k$th summand of $H$ being not in the kernel) then the proof that the groups are not isomorphic follows easily from looking at the central elements of $G$.