Centralizer-commutator product decompoition for abelian groups

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Suppose A is an Abelian group and G \le \operatorname{Aut}(A) is a finite group such that the map a \mapsto |G|a is a bijective map on A (when A is finite, this is equivalent to requiring that the orders of A and G be relatively prime). Define:

  • [A,G] is the subgroup generated by elements of A of the form g(a) - a, where a \in A, g \in G.
  • C_A(G) is the subgroup comprising those a \in A such that g(a) = a.

Note that these correspond to the usual notions of commutator of two subgroups and centralizer if we look inside the semidirect product of A by G.

Then, A is the internal direct product of the subgroups [A,G] and C_A(G). In other words, we have:

A = [A,G] \times C_A(G).

Related facts

Other averaging lemmas



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

To prove: A = [A,G] \times C_A(G)

Proof: We define a map \pi: A \to A, prove that \pi is a projection, and show that the kernel and image of \pi are [A,G] and C_A(G) respectively.


\pi(a) = \frac{1}{|G|} \sum_{g \in G} g(a).

Note that this expression makes sense because the multiplication by |G| map is invertible in A by the condition of relatively prime orders.

We observe that:

\pi \circ g = g \circ \pi = \pi \ \forall \ g \in G.

This is because \pi \circ g and g \circ \pi yield the same summation as \pi, in a different order, and the group is Abelian.

This further yields:

\pi^2 = \pi.

Thus, \pi is an idempotent endomorphism of A. In other words, the image of \pi equals its fixed-point space.

  1. Computation of the image of \pi:
    1. C_A(G) is in the image: If a \in C_A(G), then g(a) = a for all g \in G, so \pi(a) = a. Thus, \pi(a) = a, so C_A(G) is in the fixed-point space of \pi.
    2. The image is in C_A(G): If a is fixed under \pi, then a = \pi(a) = g(\pi(a)) = g(a), so a is fixed by all g \in G. Thus, the fixed-point space of \pi, which is also the image of \pi, is precisely C_A(G).
  2. Computation of the kernel of \pi:
    1. [A,G] is in the kernel: Let's now consider the kernel of \pi. Suppose a = g(b) - b for some b \in A. Then \pi(a) = \pi(g(b)) - \pi(b) = \pi(b) - \pi(b) = 0. Thus, any element in [A,G] is in the kernel of \pi.
    2. The kernel is in [A,G]: If \pi(a) = 0, then a = - \frac{1}{|G|} \sum_{g \in G} g(a) - a. This yields that a \in [A,G]. Thus, the kernel of \pi is precisely [A,G].