It is the "free" group of nilpotency class at most $2$, exponent dividing $4$, on three generators. (In other words, every group in that class is a quotient of it.) According to https://etd....

Almost all $n$ are insipid. In fact, the number of non-insipid numbers at most $n$ grows like $2n/\log n$. See the paper Cameron, Peter J.; Neumann, Peter M.; Teague, David N. On the degrees of ...

It is the only finite group with exactly two conjugacy classes and it is the only non-trivial group with trivial automorphism group. EDIT: For the second one, it turns out you need the axiom of ...

After some googling, one finds a few references. Most point to Cameron, Peter J.; Solomon, Ron; Turull, Alexandre, Chains of subgroups in symmetric groups. J. Algebra 127 (1989), no. 2, 340–352. ...

EDIT : Cleaned up answer, added more info. 20 is small enough that it is possible to find ALL the symmetric 6-valent graphs on 20 vertices. In fact, all the vertex-transitive graphs up to 32 ...

For $x\geq 2$, and $m$ odd, there is a regular digraph of order $P=(x-1)m$ and of out-valency $K=(x-2)m+\frac{m-1}{2}=\frac{(2x-3)m}{2}-\frac{1}{2}$ with no cliques of size $x$: start with the ...

Have a look at : S. P. Glasby, P. P. Pálfy, Csaba Schneider, p-Groups with a unique proper non-trivial characteristic subgroup, Journal of Algebra, Volume 348,Pages 85-109 http://www.sciencedirect....

A graph is called semisymmetric if it is regular, edge-transitive but not vertex-transitive. Semisymmetric graphs are walk-regular hence they provide example of graphs that are regular and walk-...

I claim the answer to your question is yes. This is my first time posting on mathoverflow. I hope my latex goes ok. Given $\Gamma(S_n,A)$, build an auxiliary graph $X(\Gamma(S_n,A))$, with vertex set ...

The smallest counterexample is the dicyclic group of order $36$: $G=C_9\rtimes C_4$, with the generator of $C_4$ acting by inversion on $C_9$. In this case, $\Phi_{-}(G)\cong C_6$, while $G/\Phi_{-}(... View answer Accepted answer 7 votes First, some terminology: if$G\cong\mathrm{Aut}(\Gamma)$then$\Gamma$is often called a GRR (for graphical regular representation). This may help in looking for references. Determining whether a ... View answer Accepted answer 7 votes A graph is Cayley if and only if its automorphism group has a regular subgroup. A computer algebra system such as Magma or GAP (I think) usually can determine if a permutation group of order 288000 ... View answer Accepted answer 6 votes An example was constructed by Pablo Spiga: https://arxiv.org/abs/2102.13614 "A generalization of Sims conjecture for finite primitive groups and two point stabilizers in primitive groups" View answer 6 votes The answer to Q2 is no. Let$H$be an elementary abelian$2$-group, in its regular action, so$n=2^a$for some$a$. Clearly, to generate$H$, we need$a$elements with full support, so the total ... View answer Accepted answer 6 votes Everything that is not obviously forbidden can happen. For example, a single group can have two Cayley graphs that are isomorphic (as graphs) but which are not of the same type. Such examples will ... View answer Accepted answer 5 votes Let$\Omega=\{1,2,3,4,5,6,7,8,9\}$and let$G\leq\mathrm{Sym}(\Omega)$be the group generated by the following permutations:$(1, 2, 9)(4, 5)(7, 8)(1, 4, 7)(2, 5, 8)(3, 6, 9)(3, 6)(4, 7)(5, ...

The most obvious family of examples is $AGL(1,q)$ for $q$ a prime power. As spin said in the comments, finite $2$-transitive groups are classified. They are all almost simple or of affine type (...

$\mathrm{P\Sigma L}(2,8)$ is a counterexample. All its cyclic groups of order $6$ are self-normalising. (They fall into two conjugacy classes.) (This the automorphism group of $\mathrm{PSL}(2,8)$, it ...

$\mathrm{P\Sigma L}(2,8)$ is a counter-example. A corrected version of the statement is that, under the hypothesis, either a) $G\cong Z_p$ for some prime $p$, b) $G\cong Z_p\times Z_q$ for some ...

According to : http://arxiv.org/pdf/1009.5795v3.pdf this is known for all n up to 120 except 72, 96, 108 and 120.

Updated: In the finite case (as in the reference), the only examples are $\mathrm{AGL}(1,q)$ for $q$ a prime power. This follows, for example, by "Lucchini, Mainardis, Stellmacher, Transitive ...
Since nothing else came up, I'm posting this as an answer. For $n$ a power of a prime $p$, the answer is any group of exponent $p$, acting regularly. Indeed, if $n$ is a power of $p$, then a Sylow $... View answer 4 votes In SmallGroup(16,3), the derived subgroup has order 2, is central, and its generator is not a square. (See groupprops.subwiki.org/wiki/SmallGroup(16,3)#Subgroups) View answer Accepted answer 4 votes For$k=0$, take a complete graph and for$k=1$, a star. For$k\geq 2$, take a path with$k$vertices, and a complete graph with$n-k$vertices, then join one end of the path with all the vertices in ... View answer 3 votes Miller, G.A. Groups of order$g$containing$g/2-1$involutions. Tôhuku math. J. 17, (1920) 88-102 (https://www.jstage.jst.go.jp/article/tmj1911/17/0/17_0_88/_pdf ) is on exactly this question. The ... View answer Accepted answer 3 votes 1.No, this is a hard question in general. It could maybe be done for special classes of groups, say nilpotent groups. 2.The only (finitely-generated) groups which have a unique minimal generating ... View answer 3 votes You are asking for a sharply$2$-transitive set of degree$n$. These objects are closely linked to projective planes of order$n$. In particular, they exist if$n$is a prime power, and no example is ... View answer Accepted answer 3 votes Pablo Spiga found a proof a few weeks ago. Together, we then proved a slightly more general result which is now on the arxiv: http://arxiv.org/abs/1501.05046 It is more general in two ways: it deals ... View answer Accepted answer 3 votes This problem is known as ''pancake sorting'' or ''sorting by prefix reversal''. Imagine you have a stack of pancakes numbered$1,2,3,\ldots$starting from the top. Then$t_k$corresponds to taking the ... View answer Accepted answer 3 votes Here is an example. Adjacency matrix:$ \left[ \begin{array}{cccccccccccccccc} 0&1&1&0&1&1&1&0&0&0&0&0&0&0&0&0\\1&0&0&1&1&...