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If G is abelian, then the holomorph of G is a reasonably nice group. If G is a finite elementary abelian p-group of order pn, then you can consider it to be a vector space over Z/pZ. The automorphis …

When n=2q is twice an odd prime power q, I think G = AGL(1,q) wr Sym(2) (amongst others) has the largest minimum orbit length on 2-sets. If G acts on Z = X ∪ Y, each of size q, then G has two orbits …

Here is a little explanation for why nilpotent groups are the only possible example for groups satisfying sufficient finiteness properties. Finitely presented is barely a finiteness property at all, …

BabAba = 1, where Bb=1 and Aa=1. SL(2,Z) does contain free subgroups, for instance on [1,2;0,1] and [1,0;2,1], so there is no reason it could not contain free subgroups, it just so happens that {a,b} …

If you have explicit finite presentations of G and H is small and finite, then you should be able to just ask GAP or magma for a presentation of the kernel of the projection. This is another disguise …

The paper's stated goal is to check that certain proposed ``subgroups'' are not actually subgroups at all. The proposed subgroups are groups that are direct products A × B. To show that A × B is not …

E = SmallGroup(32,28) is the first example. It has two central subgroups A1 and A2 isomorphic to A ≅ 2 with quotient isomorphic to SmallGroup(16,11), but A1 and A2 are not conjugate in Aut(E). Examp …

No. The simple group of order 60 is a counterexample to (1) and (3). If {x,y} is a generating set of G, call its signature 1/|x| + 1/|y| + 1/|xy|. It appears that the signature of most generati …

Yes. I believe (ab)^3 or so, at least (ab)^n for high enough n, never terminates. As far as I recall, only (ab)^1 and (ab)^2 terminate. If you start to collect them, I think you'll see the proof.

To help searching: ω(|G|) = |π(G)|, and I see the latter usually. A finite simple group G with |π(G)| = 1 must be cyclic of order p. By Burnside's paqb theorem, if |π(G)|=2, then G is not simple. T …

For algebras over non-prime fields, a good reference is Isaacs (1995). It proves a fundamental theorem establishing that these F-algebra groups behave like |F|-groups, not just like p-groups, for cha …

One phrase is just "normal product", but people who use such a phrase usually don't pay attention K∩L, so I don't think it will help you very much. These normal products are much weirder than direct …

Here is an approximation of an answer to "For what finite groups is Aut(G) simple?" As Daniel Miller mentioned, Inn(G) is a normal subgroup of Aut(G), so for Aut(G) to be simple either Inn(G) = 1, in …

If $G$ and $H$ are non-abelian simple groups and $N$ is a normal subgroup of $G×H$, then $N$ is equal (not just isomorphic) to $1×1$, $1×H$, $G×1$, or $G×H$. If $G$ and $H$ are simple groups but not …

It is difficult to find this number for arbitrary finite groups, but many families have been solved. A somewhat early paper that has motivated a lot of work in this area is: Johnson, D. L. "Minimal …