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comment 
$2$group with two isomorphic normal subgroups of index $4$ with nonisomorphic quotients
I'm not sure I understand your example. Could you maybe be more explicit? (Keep in mind that I am only considering finite groups.) 
Apr 14 
awarded  Yearling 
Apr 14 
awarded  Student 
Apr 14 
asked  $2$group with two isomorphic normal subgroups of index $4$ with nonisomorphic quotients 
Mar 26 
comment 
Max order for which connected Cayley Graphs are known to be Hamiltonian
It has been checked to 1280 that, in the cubic vertextransitive case, only the four wellknown exceptions occur. The question asked for Cayley graphs of arbitrary valency. 
Mar 21 
answered  Max order for which connected Cayley Graphs are known to be Hamiltonian 
Feb 24 
comment 
On the Groups of Order $(p^2+1)/2$
ADDENDUM: there are three more candidates for $p$ between 3 and 4 million: $p=3319597,3456127,3636443$, and then none up to 10 million. 
Feb 24 
comment 
On the Groups of Order $(p^2+1)/2$
A small observation : since the group has odd order, it is soluble. In particular, it has SOME (minimal) abelian normal subgroup. Moreover, by "Groups of CubeFree Odd Order", by Curran, we may assume that the group is not cubefree. Anyway, I checked the conjecture up to $p=3000000$. I was only checking that $n=(p^2+1)/2$ was not squarefree and that Sylow's theorem would not force a normal $q$Sylow subgroup of order at most $q^2$ for some prime $q$. Up to $p=3000000$, the only exceptions are for $p=239$, when we get n=$13^4$ and $p=2905807$ when we get $n=5^4∗13∗61∗97∗137∗641$. 
Jan 23 
answered  Upper bound on the number of vertex transitive graphs 
Nov 15 
answered  Fantastic properties of Z/2Z 
Sep 15 
revised 
Presentation of the Monster Group
deleted 344 characters in body 
Sep 15 
answered  Presentation of the Monster Group 
Jul 17 
comment 
Maximum automorphism group for a 3connected cubic graph
In the cubic vertextransitive case and n twice an odd number, it immediately follows from the same paper that you get a polynomial rather than exponential upper bound. For example Corollary 4 yields that, for large enough n, we have G<n^2. This is not best possible, but is not far off, at least for some values of n. If you need more precise estimates, I can show you a few more references that deal with this. (By the way, the link in your "added" section has a typo.) 
Jul 17 
comment 
Maximum automorphism group for a 3connected cubic graph
The family of vertextransitive graphs you have in mind are in fact best possible among large enough cubic vertextransitive graphs. (n=100 or so should already suffice.) This is shown in the paper : "Bounding the order of the vertexstabiliser in 3valent vertextransitive and 4valent arctransitive graphs", arxiv.org/abs/1010.2546. Therefore, a counterexample to your conjecture will necessarily be not vertextransitive. 
May 26 
comment 
frobenius group
Actually, the Wikipedia page answers both queries (it gives more examples and mentions the odd dihedral groups) and also point towards the fact that being a semidirect product is not enough. It's clear that the author has not even glanced at that page and I would close this question as not being research level. 
May 26 
comment 
frobenius group
en.wikipedia.org/wiki/Frobenius_group Not all semidirect products are Frobenius. 
Mar 29 
awarded  Citizen Patrol 
Mar 25 
awarded  Yearling 
Mar 21 
comment 
Smallest nonisomorphic strongly regular graphs
The links appear to be broken. In the meantime, win.tue.nl/~aeb/graphs/srg/srgtab.html has some information. 