Difference between revisions of "Upper central series may be tight with respect to nilpotency class"

From Groupprops
Jump to: navigation, search
(Proof)
Line 17: Line 17:
 
==Proof==
 
==Proof==
  
{{further|[[particular example::Faithful semidirect product of cyclic p-groups]]}}
+
Let <math>H_1, H_2, \dots H_c</math> be [[group]]s such that each <math>H_k</math> is a [[nilpotent group]] of [[nilpotency class]] precisely <math>k</math>, i.e., it is ''not'' nilpotent of class smaller than <math>k</math>. Define <math>G</math> as the [[external direct product]]:
  
Let <math>p</math> be an odd prime. Let <math>H</math> be the cyclic group of order <math>p^c</math>, and <math>K</math> be the cyclic group of order <math>p^{c-1}</math>. Consider the action of <math>K</math> on <math>H</math>, where the generator acts via multiplication by <math>p + 1</math>. Let <math>G = H \rtimes K</math> be the semidirect product.
+
<math>G = H_1 \times H_2 \times \dots \times H_c</math>
  
The group <math>Z_k(G)</math> in this case is isomorphic to the semidirect product of the cyclic group of order <math>p^k</math> with the cyclic group of order <math>p^{k-1}</math>, with the generator acting via multiplication by <math>p + 1</math>. This tells us that the upper central series of <math>Z_k(G)</math> is precisely the first <math>k</math> terms of the upper central series of <math>G</math>, as required.
+
Now, for each <math>k</math>, we have:
 +
 
 +
<math>Z_k(G) = Z_k(H_1) \times Z_k(H_2) \times \dots \times Z_k(H_c)</math>
 +
 
 +
In particular, we obtain that:
 +
 
 +
<math>Z_k(G) = H_1 \times H_2 \times \dots \times H_k \times Z_k(H_{k+1}) \times \dots \times Z_k(H_c)</math>
 +
 
 +
From the given data, in particular the fact that <math>H_k</math> has nilpotency class exactly <math>k</math>, it is clear that <math>Z_k(G)</math> has nilpotency class exactly <math>k</math>.

Revision as of 23:19, 2 February 2012

Statement

Let c be any natural number. Then, we can construct a Nilpotent group (?) G with the following property.

Let Z_k(G) denote the k^{th} member of the Upper central series (?) of G: Z_1(G) = Z(G) is the center and Z_k(G)/Z_{k-1}(G) is the center of G/Z_{k-1}(G) for all k. By definition of Nilpotency class (?), Z_c(G) = G.

We can find a G with the property that for any k \le c, the upper central series of Z_k(G) is precisely the first k terms of the upper central series of G. In particular, we can find a G with the property that each Z_k(G) has nilpotence class precisely k: in other words, it is not a group of class k - 1.

Related facts

The corresponding statement is not true for the lower central series. Some related facts:

Proof

Let H_1, H_2, \dots H_c be groups such that each H_k is a nilpotent group of nilpotency class precisely k, i.e., it is not nilpotent of class smaller than k. Define G as the external direct product:

G = H_1 \times H_2 \times \dots \times H_c

Now, for each k, we have:

Z_k(G) = Z_k(H_1) \times Z_k(H_2) \times \dots \times Z_k(H_c)

In particular, we obtain that:

Z_k(G) = H_1 \times H_2 \times \dots \times H_k \times Z_k(H_{k+1}) \times \dots \times Z_k(H_c)

From the given data, in particular the fact that H_k has nilpotency class exactly k, it is clear that Z_k(G) has nilpotency class exactly k.