Lower central series is fastest descending central series

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Statement

Suppose G is a Nilpotent group (?) with a Central series (?):

G = H_1 \ge H_2 \ge \dots H_n = \{ e \}

Then, if we consider the Lower central series (?) of G:

G_1 = G, G_{i+1} = [G_i,G]

Then, for every i, we have:

H_i \ge G_i

In particular, if G has Nilpotence class (?) c, then:

n \ge c + 1

Related facts

Facts used

We use the following fact:

  • If H_1 \le K_1 \le G and H_2 \le K_2 \le G, then [H_1,H_2] \le [K_1,K_2].

Proof

Given': G is a nilpotent group with a central series:

G = H_1 \ge H_2 \ge \dots H_n = \{ e \}

Consider the lower central series of G:

G_1 = G, G_{i+1} = [G_i,G]

To prove: For every i, we have:

H_i \ge G_i

In particular, if G has nilpotence class c, then:

n \ge c + 1

Proof: We prove this by induction on i.

Base case for induction: For i = 1, H_1 = G_1 = G so it's true.

Induction step: Suppose H_i \ge G_i. Then by the definition of the lower central series:

[G,G_i] = G_{i+1}

Because H_i \ge G_i, we have, by the fact mentioned above:

[G,H_i] \ge [G,G_i]

Finally, by the definition of central series, we have:

H_{i+1} \ge [G,H_i]

Combining these facts, we see that:

H_{i+1} \ge G_{i+1}

Thus, if H_n is trivial, G_n must be trivial. Since the smallest n for which G_n is trivial is c + 1, we have n \ge c + 1.