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Mar
12
comment Computing the order of a group element
@DerekHolt: What use does an order-finding algorithm have if you can get only elements of order $p-1$ or $\frac{p-1}{2}$?
Mar
12
comment Computing the order of a group element
I somehow doubt that your memory is correct: If $p = 2qr+1$ where $q$ and $r$ are primes both of about the same size, then one is unlikely to find elements of order $q$ without knowing $r$. Maybe you read the implication the other way round (which is true)?
Mar
12
comment If G is a group such that all of its proper subgroups are abelian, then G itself must be abelian
$Q_8\times\mathbb 1\times 1\le Q_8\times\mathbb Z_2\times\mathbb Z(3^{\infty})$ is abelian?
Mar
12
comment Computing the order of a group element
If you take $Z_n^\times$ for $n$ a product of two primes instead of $n$ prime, then being able to find the order of group elements quickly implies that you can break the RSA for modulus $n$.
Mar
12
comment Let $H,K$ be finite groups , if for any finite group $G$ , $h(G,K)=h(G,H)$ holds , then is it true that $i(G,H)=i(G,K)$ for any finite group $G$ ?
Rereading your question I wondered if you considered the implications of $i(G,H) = i(G,K)$ for $G$ being $H$ or $K$?
Mar
12
comment Can I recover a group by its homomorphisms?
@Turion: I think it is your question, so you deserve the credit for it being a good question.
Mar
12
comment Can I recover a group by its homomorphisms?
@Turion: You could post it as another question (maybe without the conjugation). I'd be interested in knowing the solution.
Mar
12
comment Let $H,K$ be finite groups , if for any finite group $G$ , $h(G,K)=h(G,H)$ holds , then is it true that $i(G,H)=i(G,K)$ for any finite group $G$ ?
$i(G,H) = h(G,H)-\sum_{N \lhd G, N\ne 1 } i (G/N , H)$, so induct on the order of $G/N$.
Mar
11
comment Show that under certain conditions the factors of direct product are isomorphic
@Stefan: You could look also at the projection from $G$ to the first coordinate, and restrict it to $D$.
Mar
11
comment Can I recover a group by its homomorphisms?
@Turion: Do you have a clue about the answer to your question for finitely generated residually finite groups?
Mar
10
comment How many conjugates does a regular permutation group have?
Using my last comment one should be able toarrive at the conclusion that the desired number is $\frac{(n-1)!}{|Aut(G)|}$.
Mar
10
comment Give an example where $A \subseteq B$ with $A \neq B,$ but $\left\langle A\right\rangle= \left\langle B\right\rangle.$
@Hayden: Why non-trivial? ;-)
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.)