Centralizer-commutator product decompoition for abelian groups

Statement
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:

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

Other averaging lemmas

 * Maschke's averaging lemma for modules: This is the most general form, stating that if a group acts on a module over a ring in which its order is invertible, and there is an invariant direct summand of the module, the direct summand has an invariant complement.
 * Maschke's averaging lemma for Abelian groups: A somewhat weaker reformulation of the above.
 * Maschke's averaging lemma: A variant on the lemma for modules, where the base ring is a field.

Applications

 * Centralizer-commutator product decomposition for finite nilpotent groups
 * Centralizer-commutator product decomposition for finite groups

Proof
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.

Define:

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

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 $$\pi$$:
 * 2) $$C_A(G)$$ 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 $$\pi$$.
 * 3) 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 $$C_A(G)$$.
 * 4) Computation of the kernel of $$\pi$$:
 * 5) $$[A,G]$$ 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 $$\pi$$.
 * 6) 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 $$[A,G]$$.