Binary subgroups of direct products
نویسندگان
چکیده
We explore an elementary construction that produces finitely presented groups with diverse homological finiteness properties – the binary subgroups, $B(\Sigma,\mu)\<G\_1\times\dots\times G\_m$. These full subdirect products require strikingly few generators. If each $G\_i$ is presented, $B(\Sigma,\mu)$ presented. When are non-abelian limit (e.g. free or surface groups), provide new examples of residually-free do not have finite classifying spaces and Stallings--Bieri-type. settle a question Minasyan relating different notions rank for groups. Using we prove if $G\_1,\dots,G\_m$ perfect groups, requiring at most $r$ generators, then $G\_1\times\dots\times G\_m$ requires $r \lfloor \log\_2 m+1 \rfloor$
منابع مشابه
Maximal subgroups of direct products
A group G is simple if and only if the diagonal subgroup of G ×G is a maximal subgroup. This striking property is very easy to prove and raises the question of determining all the maximal subgroups of G , where G denotes the direct product of n copies of G . The first purpose of this note is to answer completely this question. We show in particular that if G is perfect, then any maximal subgrou...
متن کاملSubgroups of direct products of limit groups
If Γ1, . . . ,Γn are limit groups and S ⊂ Γ1 × · · · × Γn is of type FPn(Q) then S contains a subgroup of finite index that is itself a direct product of at most n limit groups. This answers a question of Sela.
متن کاملSubgroups of direct products closely approximated by direct sums
Let I be an infinite set, {Gi : i ∈ I} be a family of (topological) groups and G = ∏ i∈I Gi be its direct product. For J ⊆ I , pJ : G → ∏ j∈J Gj denotes the projection. We say that a subgroup H of G is: (i) uniformly controllable in G provided that for every finite set J ⊆ I there exists a finite set K ⊆ I such that pJ (H) = pJ (H ∩ ⊕ i∈K Gi); (ii) controllable in G provided that pJ(H) = pJ(H ∩...
متن کاملOn Normal Subgroups Which Are Direct Products
Now that the classification of finite simple groups is complete, it is logical to look at the extension problem. An important special case to consider is when M is a minimal normal subgroup of G and both G/M and M are known groups. If M is abelian, various techniques have been used to derive information about G. Indeed, almost the entire theory of finite solvable groups can be said to rest upon...
متن کاملTwo subgroups and semi-direct products
Proof. Clearly, the image of F is HK by definition. To see that F is injective, suppose that F (h1, k1) = F (h2, k2). Then by definition h1k1 = h2k2. Thus h −1 2 h1 = k2k −1 1 . Since H is a subgroup, h−1 2 h1 ∈ H, and since K is a subgroup, k2k −1 1 ∈ K. Thus h−1 2 h1 = k2k −1 1 ∈ H ∩ K = {1}, and so h −1 2 h1 = k2k −1 1 = 1. It follows that h−1 2 h1 = 1, so that h1 = h2, and similarly k2k −1 ...
متن کاملذخیره در منابع من
با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید
ژورنال
عنوان ژورنال: L'enseignement mathématique
سال: 2023
ISSN: ['0013-8584', '2309-4672']
DOI: https://doi.org/10.4171/lem/1057