Skip to main content
Tom De Medts's user avatar
Tom De Medts's user avatar
Tom De Medts's user avatar
Tom De Medts
  • Member for 13 years, 10 months
  • Last seen this week
awarded
comment
Subgroups of groups of Square-free order
@Martin: This is just a way of rephrasing your original question in the case $r=3$; or am I missing something?
awarded
comment
Subgroups of groups of Square-free order
The fact that every such group has the structure of the form that I've explained in my answer, is due to Zassenhaus (hence the name Zassenhaus metacyclic groups or $Z$-groups for short). The statement that for every divisor of $|G|$ there is a corresponding subgroup, follows easily from this description (write this divisor as $m'n'$ with $m' \mid m$ and $n' \mid n$, and take $r' = r^{n / n'}$). Alternatively, in your specific case you can simply invoke Hall's theorem saying that a solvable group $G$ has a Hall $\pi$-subgroup for every set $\pi$ of primes.
comment
Subgroups of groups of Square-free order
In fact, any group $G$ for which all Sylow subgroups are cyclic, contains a subgroup of order $m$ for any divisor $m$ of $|G|$.
comment
Subgroups of groups of Square-free order
The relations are: $\gcd(m,n) = 1$, $\gcd(m, r-1) = 1$ and $r^n \equiv 1 \pmod{m}$. A possible reference is Huppert's "Endliche Gruppen I", Chapter IV, Satz 2.11 (p. 420).
answered
Loading…
answered
Loading…
comment
Local complementation group of simple graphs
The local complementation w.r.t. a given vertex v is the operation that changes the subgraph induced on the neighbors of v (not including v) into its complement. In particular, it is an involution, so the "local complementation group" is a transitive group generated by involutions.
accepted
comment
Recovering n from sigma(n)/n
By the way, the "$b$" in my previous comment is really the "$b$" from the comment below my question, so $q=a/b$ with $a,b$ coprime. So only looking at $n=bc$ with $b,c$ coprime really is a restriction.
comment
Recovering n from sigma(n)/n
@Francois: What I'm doing at the moment is a combination of recursively writing $n = bc$ with $b,c$ coprime (maybe the coprimeness is too restrictive, but it makes the computation easier), together with a fast lookup-table containing the values for $\delta(n)$ for $n$ going from $1$ to $10^7$. But I find this somehow rather down-to-the-earth, and I was wondering whether there exist more advanced methods instead.
comment
Recovering n from sigma(n)/n
Thanks a lot! Are there more recent developments since Pomerance's paper (which is from 1977)? Do you know of any kind of efficient algorithms to try to compute, for example, the smallest element in $\delta^{-1}(q)$ for given $q$ (up to some given upper bound)?
comment
Recovering n from sigma(n)/n
@Francois: Something more general is true indeed; namely if $m \mid n$, then $\delta(n) \geq \delta(m)$. Your observation is the case $m=6$.
awarded
comment
Recovering n from sigma(n)/n
@Charles: good point; this part of the question is indeed too ambitious.
Loading…
asked
Loading…
Loading…
comment
A problem in Finite Group Theory
Yes: if $A$ is a $\pi$-group (where $\pi$ is a set of primes), then the fact that $A \cap F(G) > 1$ gives $A \cap O_\pi(G) > 1$. Now proceed as in the previous problem with $F(G)$ replaced by $O_\pi(G)$.