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Oct
16 |
awarded | Custodian |
Oct
16 |
reviewed | No Action Needed Integer-distance sets |
Oct
16 |
reviewed | No Action Needed Integer-distance sets |
Jul
21 |
awarded | Autobiographer |
Jul
13 |
comment |
Cayley graphs with special subgraphs and some related problems
Infinite trees yield a somewhat trivial positive answer to 1). |
Jun
11 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
Isometries of some simple Cayley graphs
Sorry, I missed the "finite". Please disregard my comment. |
Mar
24 |
awarded | Yearling |
Mar
23 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 |