5,704 reputation
1342
bio website boolesrings.org/nickgill
location San Jose, Costa Rica
age 37
visits member for 5 years
seen 9 hours ago

I'm a visiting professor at the Universidad de Costa Rica.


Oct
20
awarded  Yearling
Oct
10
comment Equivalence classes of (2,3)-pairs in PSL(2,q)
My mistake! Alexander's answer below shows my error - the triples fuse in $S_5$ (explaining Robin's calculation in the question) but, of course, there is no $S_5$ in $PGL_2(q)$...
Oct
2
comment Equivalence classes of (2,3)-pairs in PSL(2,q)
One further thing: For products of order equal to $2$, $3$ and $4$, I guess the problem reduces to studying the number of orbits in $Aut(G)$ of subgroups isomorphic to $S_3$, $A_4$ and $S_4$. I'm guessing there's only one such subgroup in the latter two cases, not sure about more generally. (This should be a tractable problem though.) For product $6$ the pair commutes so this is easy. For product $7$ one is studying Hurwitz groups and so the pair will generate $PSL_2(p)$ where $p$ is the prime. Now one can check how many non-conjugate triples generate any given $PSL_2(p)$.
Oct
2
comment Equivalence classes of (2,3)-pairs in PSL(2,q)
.. Thus I think the two orbits found by Stefan will fuse in $PGL_2(11)$. And, in general, there will be one orbit of $(2,3)$-pairs with product equal to $5$ in the automorphism group of $G$. For other product orders things will be different I guess.
Oct
2
comment Equivalence classes of (2,3)-pairs in PSL(2,q)
I'm slightly confused - I think this question neglects the OP's consideration of the automorphism group of $G$. In the case where you have a $(2,3)$-pair whose product is $5$ they will generate a group isomorphic to $A_5$. Subgroups of this form exist in $PSL_2(q)$ exactly when $q=-1,0,1\pmod 5$. What is more if one considers the automorphism group of $G$, then all subgroups isomorphic to $A_5$ are conjugate. Now Robin's calculations for $q=5$ suggest that all such pairs should be conjugate in the automorphism group of $PSL_2(q)$.
Sep
30
awarded  Explainer
Sep
24
comment A generalization of real characters on a group
+1 for the first sentence.
Sep
17
comment Normal subgroup of classical groups
@M.B Sorry, no idea. I don't know anything about the fields you describe...
Sep
10
comment Ends of Coxeter Groups
@YCor, of course your comment is entirely reasonable. There are plenty more results in the cited text that give more explicit information (see especially Thm. 8.7.3)... but I don't want to write them all out!
Sep
10
answered Ends of Coxeter Groups
Sep
10
comment Ends of Coxeter Groups
You can find some information on ends of Coxeter groups in this paper by Mihalik: sciencedirect.com/science/article/pii/0022404995001174
Sep
5
comment existence of a finite group which is the union of self normalizing subgroups
That's a cool fact about Carter subgroups. I didn't know about those.
Sep
5
awarded  Nice Answer
Sep
4
comment (Connected) Cayley graphs of PSL(2,q) from (2,3,n)-triples
It might help to give a little motivation also - why is this particular triple of elements of interest??
Sep
4
comment (Connected) Cayley graphs of PSL(2,q) from (2,3,n)-triples
what do you mean "there are elements of order $n$ in $G$, but not in any of its proper subgroups"? This will only be true when $G$ is cyclic of order $n$, but I guess you're still assuming $G=PSL(2,q)$... Perhaps you mean subspace subgroups??
Sep
4
reviewed Approve suggested edit on Trace inequality for matrices with determinant 1
Sep
4
revised Which finite simple groups can be characterized by their action on a small set?
Added a discussion of the general setting.
Sep
4
answered existence of a finite group which is the union of self normalizing subgroups
Sep
2
answered Which finite simple groups can be characterized by their action on a small set?
Sep
2
comment Which finite simple groups can be characterized by their action on a small set?
... By the way I have an e-copy of the LPS-memoir - email me if you want it....