bio | website | |
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location | Perth | |
age | ||
visits | member for | 3 years, 4 months |
seen | Jul 31 at 11:14 | |
stats | profile views | 284 |
Researcher in Graph and Group Theory
Jul 25 |
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Counting the k-factors of the complete graph on n vertices
I thought your final way was fine : how many $3$-factors does $K_{10}$ have. By the way, the (connected) 3-regular graphs on at most 10 vertices are given here for example : en.wikipedia.org/wiki/Table_of_simple_cubic_graphs together with the size of their automorphism groups. Given this, it is a small computation taking only a few minutes to find the answer you are looking for. (In the connected case, each graph $G$ contributes $\frac{10!}{|Aut(G)|}$, the only disconnected possibility is two components of size 4 and 6, which is also not hard.) I don't think this is research level. |
Jul 25 |
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Counting the k-factors of the complete graph on n vertices
@DKal: to be more precise, I think you would need to add "on a given vertex-set" (to avoid confusion with counting up to isomorphism, for example). |
Jul 21 |
awarded | Autobiographer |
Jul 13 |
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Cayley graphs with special subgraphs and some related problems
Infinite trees yield a somewhat trivial positive answer to 1). |
Jun 11 |
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Is there a big solvable subgroup in every finite group?
No, for $p\geq 3$, this subgroup always contains a $p$-cycle and is thus always transitive and primitive. It is not necessarily maximal. (Note that the answer you linked says that all maximals are of these types, not that all groups of these types are maximal. Sometimes, one of the groups listed there is contained in another, as happens for $AGL(1,23)\cap A_{23}$ which seems to be contained in $M_{23}$.) |
Jun 11 |
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Is there a big solvable subgroup in every finite group?
In fact, it seems that $H\cap A_p$ is maximal in $A_p$ for $p$ greater than 23. (The small exceptions are due to things like the Matthieu groups.) Again, this should be provable (it's probably already known in fact), and it shows that every large alternating group of prime degree has a big (maximal even) solvable subgroup. |
Jun 11 |
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Is there a big solvable subgroup in every finite group?
$S_p$ always contains $H:=AGL(1,p)$ so I consider $H\cap A_p$. This is primitive soluble (and as large as possible with respect to this property) but it is not always maximal in $A_p$. In fact, I am not sure how to tell whether it is maximal in general (but some people probably do), but I used magma to determine which small alternating groups had maximal soluble groups... (For p>16, it is easy to see that such a group must be as above.) |
Jun 11 |
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Is there a big solvable subgroup in every finite group?
By the way, there is a sequence of primes $\{p\}$ such that $A_p$ contains a maximal soluble (primitive) subgroup. It includes: {13,19,29,31,37,41,43,47,53,59,61,67,71,73,79...}. I am not sure if this sequence is infinite but I guess so. |
Jun 10 |
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Is there a big solvable subgroup in every finite group?
$A_6$, $A_7$ and $A_8$ all have maximal soluble subgroups. For example, $(S_4\times S_4)\cap A_8$ is a maximal intransitive subgroup of $A_8$. You can play a similar trick in $A_9$ by taking a maximal imprimitive group. $A_{10}$ is the smallest alternating group that doesn't have a maximal soluble subgroup. |
May 1 |
awarded | Revival |
Apr 29 |
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Guess that group via product queries
What do you mean by "determine $G$"? Do you mean find the isomorphism type, or find the full multiplication table. (I'm not sure if the answer is different, I just want to clarify). In any case, the phrase "black-box group" will probably help in looking for references. See for example: math.uzh.ch/fileadmin/user/rosen/publikation/zu08p.pdf |
Mar 25 |
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Isometries of some simple Cayley graphs
Sorry, I missed the "finite". Please disregard my comment. |
Mar 24 |
awarded | Yearling |
Mar 23 |
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Isometries of some simple Cayley graphs
Suppose that we take as connecting set the whole group except for the identity. Then the graph is complete and its automorphism group is the full symmetric group. This is always bigger than the group generated by translations and group automorphisms once |G|>4. In other words, except for a few small exceptions, the answer cannot be yes for all generating sets of a given group. |
Mar 11 |
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Generalization of Hamiltonian cycle
Sure, the equivalences don't necessarily hold for infinite graphs, I was simply answering the remark in the OP that there isn't an obvious way to generalise HC. |
Mar 11 |
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Generalization of Hamiltonian cycle
Moreover, I think Hamiltonian cycles generalise in a straightforward way to infinite graphs. After all, a HC is simply a 2-valent spanning subgraph. This works in the infinite case as well and is well-accepted I think. |
Mar 11 |
answered | Generalization of Hamiltonian cycle |
Mar 11 |
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Section of Cayley graphs
Another way to say what Dave said is that, since $S$ is the full preimage of $S_1$, as a graph, $\mathrm{Cay}(G,S)$ is the lexicographic product of $\mathrm{Cay}(G_1,S_1)$ with the edgeless graph on $G/G_1$ (and thus any choice of one vertex per preimage will induce an embedding). |
Feb 21 |
awarded | Revival |
Feb 6 |
answered | Condition(s) for the full autormophism group $\operatorname{Aut}(C(G, S))$ of the Cayley graph of $G$ to be isomorphic to $G$ |