Commutator of finite nilpotent group with coprime automorphism group equals second commutator

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Let G be a Finite nilpotent group (?) and A be a subgroup of \operatorname{Aut}(G) whose order is relatively prime to the order of G. Then, we have:

[[G,A],A] = [G,A],


[G,A] = \langle g\sigma(g)^{-1} \mid g \in G, \sigma \in A \}.

Related facts

Facts used

  1. Centralizer-commutator product decomposition for finite nilpotent groups: This states that under the same conditions (G a finite nilpotent group and A a subgroup of \operatorname{Aut}(G) of order coprime to the order of G), we have G = [G,A]C_G(A), and further, if K is an A-invariant subgroup of G such that G = KC_G(A), then [G,A] \le K.
  2. Nilpotence is subgroup-closed
  3. Lagrange's theorem
  4. Order of quotient group divides order of group


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

To prove: [[G,A],A] = [G,A].

Proof: Let G_1 = [G,A], G_2 = [G_1,A], G_3 = [G_2,A].

  1. G_3 \le G_2 \le G_1 \le G: Since G_1 \le G, we have [G_1,A] \le [G,A], so G_2 \le G_1. Thus, [G_2,A]\le [G_1,A], so G_3 \le G_2. Thus, G_3 \le G_2 \le G_1 \le G.
  2. G = G_1C_G(A): This follows directly from fact (1).
  3. G_1 = G_2C_{G_1}(A): This follows by applying fact (1), replacing G by G_1, and observing that:
    • Since G is nilpotent, fact (2) tells us that G_1 is nilpotent.
    • Since G has order relatively prime to that of A, the order of G_1 is also relatively prime to A (using fact (3)). Further, since G_1 is A-invariant, there is a homomorphism A \to \operatorname{Aut}(G_1), with image A_1 of order dividing the order of A, and so that the action of A on G_1 factors through this homomorphism. The order of A_1 divides the order of A by fact (4), so we get the conditions for fact (1), yielding G_1 = [G_1,A_1]C_{G_1}(A_1). Finally, since the action of A factors via the homomorphism, we get G_1 = [G_1,A]C_{G_1}(A) = G_2C_{G_1}(A).
  4. G = G_2C_G(A): By the two preceding steps, G = G_2C_{G_1}(A)C_G(A). But C_{G_1}(A) \le C_G(A), so we get G = G_2C_G(A).
  5. G_2 is A-invariant: Since [G_2,A] = G_3 \le G_2, we in particular have that G_2 is Ainvariant.
  6. G_1 \le G_2: By the preceding two steps, G_2 is A-invariant and G = G_2C_G(A), so by fact (1), we have G_1 \le G_2.
  7. G_1 = G_2: This follows from the preceding step, combined with step (1).


Textbook references

  • Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 181, Theorem 3.6, Section 5.3 (p'-automorphisms of p-groups), More info