List of centers of finite groups - MathOverflow

List of centers of finite groups

2010-06-26T20:25:04Z

What I'm really looking for is a good set of examples for a semi-direct product of two (coprime) cyclic groups, with a non-trivial (not $1$ and not everything) center. What resource would you recommend for that purpose?

Answer by Johannes Hahn for List of centers of finite groups

2010-06-26T20:49:47Z

One can calculate the center of a semidirect product $U\ltimes V$ very explicit. Let $\phi: V\to Aut(U)$ the corresponding homomorphism. Then the following holds: $$(u,v)\in Z(U\ltimes V) \iff v\in Z(V), u\in C_U(V), \kappa_u=\phi(v^{-1})$$ where $\kappa_u$ is the conjugation with $u$. In particular $\phi(v)\in Inn(U)$. If $U$ is abelian, this gives $Z(U\ltimes V)=Z(V) \cap \ker(\phi)$. If $V$ is abelian too, then $Z(U\ltimes V)=\ker(\phi)$. Now all examples you're looking for can be easily written down.

Answer by Jack Schmidt for List of centers of finite groups

2010-06-27T01:10:38Z

I don't know of a resource with such a list, though you could easily make one with programs such as GAP or Magma. Since it is reasonably easy and interesting to analyze this problem, I'll outline how I suggest thinking about the examples:

Let ⟨a⟩ act on the normal subgroup ⟨b⟩ in the semi-direct product ⟨a,b⟩. Then z = aⁱb^j is in the center of ⟨a,b⟩ if and only if it commutes with both a and b. Just writing it out you get a⋅(aⁱb^j) = aⁱ⁺¹b^j and (aⁱb^j)⋅a = aⁱ⁺¹(b^j)^a, so b^j has to be centralized by a. Similarly to commute with b, b has to be centralized by aⁱ.

You could try to use this in two directions: make ⟨a⟩ centralize part of ⟨b⟩, or let the kernel of the action of ⟨a⟩ on ⟨b⟩ be non-trivial (and proper).

The first direction runs into some trouble, but does let you produce pretty obvious examples. As a concrete family, consider:

Let b have order 3k for k an integer coprime to 6, and let a of order 2 act on b as b^a = b^(k²+1). Then ⟨a,b⟩ has center ⟨b^k⟩.

This example is not particularly satisfying since it is just the direct product S₃×k. For the verification, check that (b³)^a = b^(3k²+3) = b³ since 3 ≡ 3k²+3 mod 3k, but (b^k)^a = b^k(k²+1) = (b^k)² since 2 ≡ k²+1 mod 3. This might make a good example to show that "a" and "b" are not uniquely defined by the abstract isomorphism type of the group meeting your requirements (since the "k" could just as easily have been part of the "a" as part of the "b").

Unfortunately you cannot do much better than this using the first direction, since if ⟨a⟩ centralizes a subgroup of ⟨b⟩ of prime order p, then ⟨a⟩ centralizes the entire Sylow p-subgroup of ⟨b⟩, and you get a direct product. This follows from your coprime hypothesis.

The other direction looks a bit different. Take any example semi-direct product ⟨A,b⟩ that has ⟨b⟩ normal, |A| and |b| are coprime, and Z(⟨A,b⟩) a proper subgroup. Let ⟨a⟩ be a cyclic group of order k|A| where k is coprime to ⟨b⟩, and define b^a = b^A. Then the center of ⟨a,b⟩ is still a proper subgroup and contains ⟨a^|A|⟩ of order k. More concretely:

Let b have odd order m and let a have order 2ⁿ⁺¹ and let a act on b as b^a = b⁻¹. Then ⟨a,b⟩ has center ⟨a²⟩ of order 2ⁿ.

You can combine the two directions, but again because the first direction trivializes, the combined version looks pretty silly. Just take the direct product any example of the second direction and a cyclic group of order coprime to its |a|. This exhausts all possibilities, but I suspect most people would be irritated with the direct product aspect and only consider the second direction valid.

If you relax the coprime hypothesis, then you get another type of example that is reasonably interesting: take the Sylow p-subgroup of the holomorph of a cyclic p-group. The center will have order exactly p. If ⟨b⟩ has order pⁿ⁺¹, then ⟨a⟩ will have order pⁿ, ⟨a,b⟩ will have order p²ⁿ⁺¹, and the center of ⟨a,b⟩ will be ⟨b^pⁿ⟩ of order p. The action of a on b is easy to write down: b^a = b^1+p.

Answer by SandeepJ for List of centers of finite groups

2010-06-27T18:20:55Z

In Humphrey's Group theory, Chapter 21 gives the construction of Central Extensions. [A group G is a central extension of N by H with Z(G)=N.] You could use that construction to build appropriate examples.

Wreath products are also semidirect products. In Weinstein's Examples of Groups, Sec 4.4 on wreath products, Theorem 4.4.7 constructs and proves the restricted wreath product W=(G rwr A) such that Z(W) = Z (diag($G^A$) x {1}) where diag is the diagonal subgroup (result applies only if A is finite)

Meldrum's book on Wreath Products Chap 1 might provide more constructions.

Answer by Tim Dokchitser for List of centers of finite groups

2010-11-19T17:09:57Z

If you would like a complete classification following Jack's strategy, say your group is $G=C_n\rtimes C_m$ with $n,m$ coprime, $C_n=\langle c\rangle$, $C_m=\langle x\rangle$ and $xcx^{-1}=c^j$. The order of $j$ in $({\mathbb Z}/n{\mathbb Z})^\times$ has to divide $m$, and every such triple $(n,m,j)$ gives a valid semi-direct product $G$. Now

$G$ has a non-trivial centre if and only if either

The order of $j$ is less than $m$, or

$j$ is congruent to 1 modulo a prime $p|n$.

In the first case, $C_m$ does not act faithfully on $C_n$, and the elements in the kernel commute with both $C_n$ and $C_m$, so they are in the centre. In the second case, $C_p\lt C_n$ is central. If neither condition holds, then the centre must be contained in $C_n$ (by faithfullness of the action) and cannot contain any $C_p$ for $p|n$ (by the failure of the second condition), so it has to be trivial.