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Mar
6
comment Enumerating double coset representatives in the symmetric group on a vector space
@DerekHolt: Thank you for this info. I should have thought of O'Nan-Scott myself.
Mar
5
comment Suppose $G$ is a group and $a\in G$ with $|a| = m$. Prove that $\langle a^k \rangle=\langle a \rangle \iff \gcd(k, m) = 1$.
I don't see a difference in the level of the answers for this problem, only in style. But anyway, let's agree to disagree. I'll keep liking (and upvoting) your answers to difficult problems, and disliking (but not downvoting) your answers to easy problems.
Mar
5
comment Suppose $G$ is a group and $a\in G$ with $|a| = m$. Prove that $\langle a^k \rangle=\langle a \rangle \iff \gcd(k, m) = 1$.
I like your posts about intermediate to difficult problems where an experienced mathematician (like you are) has to think a bit about a solution. There shortness is a plus. But a beginner knowing how to manipulate equivalences can prove just easy things. I think learning the proper techniques that work also for difficult problems is worth more than being able to solve a trivial problem more quickly. (Universal properties are a good thing to know about, but many - not you - on mse don't have the technique to apply them.)
Mar
5
comment Suppose $G$ is a group and $a\in G$ with $|a| = m$. Prove that $\langle a^k \rangle=\langle a \rangle \iff \gcd(k, m) = 1$.
+1 for the good style
Mar
5
comment Suppose $G$ is a group and $a\in G$ with $|a| = m$. Prove that $\langle a^k \rangle=\langle a \rangle \iff \gcd(k, m) = 1$.
You proof is nicely short, but I wouldn't teach math beginners this proof style. Matt's style is the way to go (it generalizes also to harder problem) at least for beginners' questions. For advanced questions style wouldn't matter so much, but equivalences might not work out anymore.
Mar
5
comment Subgroup structure of $\mathrm{SL}(2, p^2)$ and its irreducible characters
A semi-obvious subgroup is the quaternion group. Less obvious is $SL_2(3)$, see math.stackexchange.com/questions/1054917/….
Mar
2
comment Group presentation: How can we determine the group with the presentation below?
Forget my last comment. Just try to find a semi-direct product of two cyclic groups of order $4$ that is the image of your group $G$. This shows $|G|\ge 16$, and $|G|\le 16$ you know by my first comment. Hence $G$ is isomorphic to that semi-direct product.
Mar
2
comment Group presentation: How can we determine the group with the presentation below?
Next try to show that both $g^2$ and $h^2$ are in the center of $G$. (I'm not sure yet which group we will get, I just work comment to comment, I hope you don't mind. It might be that the given relations imply additional relations, but our group has at least $8$ elements.)
Mar
2
comment Can I recover a group by its homomorphisms?
@Turion: Yes, thanks, this should be $n_G$ (also in the line below the formula). Induction is on the order of the group $H$.
Mar
2
comment Group presentation: How can we determine the group with the presentation below?
As a first step you could try to prove that all elements of $G$ can be written as $g^ih^j$ with $0\le i\le 3$ and $0\le j\le 3$.
Feb
28
comment How many conjugates does a regular permutation group have?
The number of conjugates of a subgroup $G$ of $H$ is the index of its normalizer $N_H(G)$ in $H$. Now look at this question.
Feb
27
comment Can I recover a group by its homomorphisms?
@Turion: I hope to have some time for it at the weekend.
Feb
27
comment Structure of the semidirect product decomposition
One has always to worry a bit about the order of both elements (there are different conventions), but otherwise it is as easy as you think.
Feb
27
comment Structure of the semidirect product decomposition
ad 1) Try multiplication (if $G = N\rtimes H$ then you can write each element $g\in G$ uniquely as product $g = nh$ with $n\in N$ and $h\in H$.) ad 2) look at 1) again. Does this give you an idea?
Feb
25
comment Can I recover a group by its homomorphisms?
You get that number by induction on the subgroups of $H$.
Feb
25
comment Can I recover a group by its homomorphisms?
@anomaly: Modulo conjugation by $H$ the number of surjective homomorphism $G\to H$ should be number of possible kernels times the order of $\mathrm{Out}(H) = \mathrm{Aut}(H)/\mathrm{Inn}(H)$.
Feb
25
comment Can I recover a group by its homomorphisms?
@anomaly: What's $K$? My $U$ or my $H$? Anyway, I hope to solve this by looking at $N_H(K)$.
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
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).