# 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:

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

## 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 $\pi$:
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 $[A,G]$.