Centralizer-commutator product decomposition for finite nilpotent groups

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Suppose G is a finite nilpotent group and H \le \operatorname{Aut}(G) is such that the orders of G are relatively prime. Define:

  • [G,H] as the subgroup generated by all elements of the form g\sigma(g)^{-1} where g \in G, \sigma \in H.
  • C_G(H) as the subgroup of G comprising those g \in G such that \sigma(g) = g for all \sigma \in H.

Then, we have the following:

G = [G,H]C_G(H).

Further, if K \le G is a H-invariant subgroup such that G = KC_G(H), then [G,H] \le K.

Facts used

  1. Stability group of subnormal series of finite group has no other prime factors
  2. Centralizer-commutator product decomposition for Abelian groups: This states that the result holds when G is Abelian (in fact, something stronger holds: G is an internal direct product of the two subgroups).
  3. Commutator of a group and a subgroup of its automorphism group is normal
  4. Nilpotent implies center is normality-large
  5. Nilpotence is quotient-closed


Given: A finite nilpotent group G, a subgroup H of \operatorname{Aut}(G) such that the orders of G and H are relatively prime.

To prove: [G,H]C_G(H) = G and if KC_G(H) = G, then [G,H] \le K.

Proof: We first prove that if KC_G(H) = G, then [G,H] \le K:

Pick any g \in G, \sigma \in H. We want to show that g\sigma(g)^{-1} \in K.

Since G = KC_G(H), we have g = ab, where a \in K, b \in C_G(H). Thus, we have:

g\sigma(g)^{-1} = (ab)\sigma(ab)^{-1} = ab\sigma(b)^{-1}\sigma(a)^{-1}.

Since b \in C_G(H), \sigma(b) = b, so we get:

g\sigma(g)^{-1} = a\sigma(a)^{-1}.

Since a \in K and K is H-invariant, \sigma(a) \in K, so a\sigma(a)^{-1} \in K, completing the proof.

We now prove that [G,H]C_G(H) = G.

Case that [G,H] \le Z(G)

We first consider the case that [G,H] is contained in the center of G.

  1. The map \alpha_\sigma: g \mapsto g\sigma(g)^{-1} is an endomorphism of G for any \sigma \in H: For any a,b \in G, we have ab(\sigma(ab))^{-1} = ab\sigma(b)^{-1}\sigma(a)^{-1}. Since b\sigma(b)^{-1} \in [G,H] \le Z(G), we can commute b\sigma(b)^{-1} with \sigma(a)^{-1} to get ab(\sigma(ab))^{-1} = (a\sigma(a)^{-1})(b\sigma(b)^{-1}). This proves the condition for being a homomorphism.
  2. C_G(H) contains the subgroup [G,G]: Note that C_G(H) can be described as the intersection of kernels of \alpha_\sigma for all \sigma \in H. Each \alpha_\sigma has image inside the center of G by definition, hence is a map to an Abelian group. Thus, the kernel of each \alpha_\sigma contains the commutator subgroup [G,G]. Hence, the intersection, which is C_G(H), also contains [G,G].
  3. Let A = G/[G,G] and \pi:G \to A be the quotient map. Then, H acts on A by the natural induced action, [A,H] = \pi[G,H], and C_A(H) = \pi(C_G(H)):
    1. Observe that since [G,G] is characteristic in G, it is H-invariant, so H acts on the quotient group. Note that this descent satisfies \pi(\sigma(g)) = \sigma(\pi(g)) for all g \in G.
    2. [A,H] = \pi[G,H]: This follows directly from the definition.
    3. Now, suppose L = \pi^{-1}(C_A(H)). Clearly, if g \in C_G(H), then \sigma(g) = g for all \sigma \in H. Thus, \sigma(\pi(g)) = \pi(\sigma(g)) = \pi(g), so \pi(g) \in C_A(H). Thus, \pi(C_G(H)) \le C_A(H)), so C_G(H) \le L.
    4. Consider the subnormal series \{ e \} \le [G,G] \le L for L. H acts as the identity on [G,G] since [G,G] \le C_G(H), and it acts as the identity on L/[G,G] since L/[G,G] = \pi(L) = C_A(H). Thus, H acts as stability automorphisms of this subnormal series, and its order is relatively prime to G, and hence to L. Fact (1) thus forces that H acts as the identity on L, so L \le C_G(H).
    5. Combining the previous two steps yields L = C_G(H).
  4. We have A = C_A(H) \times [A,H]: This follows from fact (2), and the observation that since the orders of G and H are relatively prime, so are the orders of A and H.
  5. G = C_G(H)[G,H]: From steps (3) and (4), we have A = \pi(C_G(H))\pi([G,H]) = \pi(C_G(H)[G,H]). Note that C_G(H)[G,H] is a subgroup, since [G,H] \le Z(G), hence is normal. Thus, C_G(H)[G,H] is a subgroup of G whose image via \pi is the whole quotient A. On the other hand, step (2) says that \operatorname{ker}\pi = [G,G] \le C_G(H), so C_G(H)[G,H] contains the kernel [G,G] and intersects every coset of it -- hence must be equal to the whole group G.

Case that [G,H] is not contained in the center Z(G)

  1. [G,H] is normal in G: This follows from fact (3).
  2. [G,H] intersects the center Z(G) nontrivially: This follows from fact (4), and the fact that [G,H] is normal.
  3. Let M = [G,H] \cap Z(G). Then, M is nontrivial, H-invariant, and normal: Z(G) is characteristic, and hence H-invariant, while we have [[G,H],H] \le [G,H], so [G,H] is H-invariant. Thus, the intersection is Hinvariant. Further, M is normal since it is the intersection of the normal subgroups [G,H] and Z(G) (or alternatively, is a subgroup of Z(G)).
  4. Let N = G/M, and \rho:G \to N be the quotient map. Then, the action of H on G descends to an action on N: This follows from step (3), where it is observed that M is normal and H-invariant.
  5. For the induced action of H on N, we have N = [N,H]C_N(H): Since N is the quotient of G by a nontrivial subgroup, it has strictly smaller order. Since G is nilpotent, so is N (fact (5)). Thus, the induction hypothesis applies to N.
  6. Let P = \rho^{-1}(C_N(H)). Then, G = [G,H]P: Clearly, we have \rho([G,H]P) = \rho([G,H])\rho(P) = \rho([G,H])C_N(H) = [N,H]C_N(H) = N. Thus, \rho([G,H]P) = N, while [G,H] also contains the kernel of \rho. Hence, [G,H]P is a subgroup (it is one because [G,H] is normal, step (1)) containing the kernel of \rho and intersecting every coset of this kernel, forcing [G,H]P = G.
  7. P = [P,H]C_P(H): Note that since \rho(P) = C_N(H), we have \rho([P,H]) = [\rho(P),H] = [C_N(H),H] which is trivial. Thus, [P,H] \le M = [G,H] \cap Z(G), and we know that M is in the center of G, hence M \le Z(P). Thus, [P,H] \le Z(P), and the previous case kicks in for P.
  8. G = [G,H][P,H]C_P(H): This follows by piecing together the previous two steps.
  9. G = [G,H]C_G(H): Since P \le G, we have [P,H] \le [G,H]. Also, C_P(H) \le C_G(H). Substituting these gives the required result.


Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 180, Theorem 3.5, Section 5.3 (p'-automorphisms of p-groups), (proved only for groups of prime power order, but the same proof technique)More info