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 Mar 15 comment computing the orbits for a group action Do you know the number of fixed points and the order of your group? Mar 13 comment Computing the order of a group element My first comment shows that my other comments are wrong: To factor $p=2qr+1$ just break the RSA with modulus $qr$. The order of elements in $\mathbb{Z}/qr\mathbb{Z}^\times$ give divisors of $\lambda(qr)$ that allow factoring $qr$. Mar 13 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 $G$ ? @SaunDev: So you can rephrase your question slightly simpler in the style of math.stackexchange.com/questions/1161733/… Mar 12 comment $G$ is a simple group of order $60$.Then $G$ contains a subgroup of order 12 $n_2 = 3$ gives you a non-trivial homomorphism to $S_3$. Mar 12 comment Can I recover a group by its homomorphisms? @Turion: Five minutes wikipedia (en.wikipedia.org/wiki/Residually_finite_group) and you'll know enough about it (e.g., that you should have included this condition from the beginning in your question). It is as essential as "finitely generated": f.g. prevents that your numbers are infinite and r.f. prevents that they are all zero. Mar 12 comment Computing the order of a group element @DRF: We seem to agree. My comment with $p=2qr+1$ was about the fact, that you won't get lucky often enough. 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.