Equivalence of definitions of subnormal subgroup

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This article gives a proof/explanation of the equivalence of multiple definitions for the term subnormal subgroup
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

  1. There exists an ascending chain H = H_0 \le H_1 \dots H_n = G such that each H_i is normal in H_{i+1}. The smallest possible n for which such a chain exists is termed the subnormal depth of H.
  2. Consider the descending chain G_i defined as follows: G_0 = G and G_{i+1} is the normal closure of H in G_i. Then, there exists an n for which G_n = H. The smallest such n is termed the subnormal depth of H.
  3. Consider the sequence K_i of subgroups of G defined as follows: K_0 = G, and K_{i+1} = [H,K_i] (the commutator), This sequence of subgroups eventually enters inside H. The number of steps taken is termed the subnormal depth of H.

Facts used

  1. Subnormal subgroup has a unique fastest descending subnormal series
  2. Product with commutator equals join with conjugate: If H \le G and A is a subset of G, we have  \langle H, H^A \rangle = H[A,H].

Proof

Fact (1) covers the equivalence of definitions (1) and (2). We thus need to prove the equivalence of definitions (2) and (3).

(2) implies (3)

Given: A subgroup H of G. A descending subnormal series for H in G given by G_0 = G, and G_{i+1} is the normal closure of H in G_i. We have G_n = H.

To prove: Consider the sequence K_0 = G, K_i = [H,G_i]. Then, K_n \le H.

Proof: Observe that since G_{i+1} is normal in G_i, we have:

[G_{i+1}, G_i] \le G_{i+1}.

Further, since H \le G_{i+1}, we get:

[H,G_i] \le G_{i+1}.

We now prove, by induction on i, that K_i \le G_i.

  • The base case for induction is true, since for i = 0, K_0 = G_0 = G.
  • The induction step: Suppose K_i \le G_i. Then K_{i+1} = [H,K_i] \le [H,G_i] \le G_{i+1}.

Thus, if G_n = H, we get K_n \le H.

(3) implies (2)

Given: A subgroup H of G. A descending chain of subgroups given by K_0  = G, and K_{i+1} = [H,K_i]. We have K_n \le H.

To prove: Define G_0 = G, and G_{i+1} as the normal closure of H in G_i. Then, G_n = H.

Proof: We do this by establishing a relation between K_i and G_i inductively, namely:

G_i = HK_i.

We prove this inductively. The base case is direct since both sides are equal to G for i = 0. We thus need to prove the induction step, i.e., if G_i = HK_i, we need to prove that G_{i+1} = HK_{i+1}.

By fact (2), we have:

H^{K_i} = H[K_i, H] = H[H,K_i] = HK_{i+1}.

On the other hand, we have by induction that G_i = HK_i, so:

G_{i+1} = H^{G_i} = H^{HK_i} = H^{K_i}.

Putting these together, we get:

G_{i+1} = HK_{i+1},

as desired.

Now that we have the general relation G_i = HK_i, it is clear that if K_n \le H, we have G_n = H.