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Upper central series may be tight with respect to nilpotency class

Revision as of 23:19, 2 February 2012 by Vipul (talk | contribs) (Proof)


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


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.