Idempotent Endomorphisms and Direct Products

April 11, 2026 · Luciano Muratore

1. Statement of the Exercise

In the study of non-trivial abelian groups $G$, three specific structural conditions are found to be equivalent. These conditions bridge the gap between abstract mappings and the internal decomposition of groups.

The Three Equivalent Conditions:

1. Idempotent Endomorphism: There exists a non-trivial homomorphism $p:G\rightarrow G$ such that $p\circ p=p$.
2. Retraction and Section: There exists a non-trivial group $H$ and homomorphisms $G \xrightarrow{f} H \xrightarrow{g} G$ such that $f\circ g=Id_{H}$.
3. Direct Product Decomposition: There exist non-trivial groups $G_1$ and $G_2$ such that $G\cong G_{1}\times G_{2}$.

2. Big Picture Intuition

The relationship between these conditions is the group-theoretic version of projection operators in linear algebra.

• Projections: Idempotent maps ($p^2 = p$) behave exactly like projections.
• Decompositions: In linear algebra, a projection $P$ on a vector space $V$ implies $V=ker(P)\oplus Im(P)$.
• Correspondence: Projections onto a subspace correspond directly to direct product decompositions of the group.

3. Proving the Equivalence

(1) $\Rightarrow$ (3): From Idempotent to Product

If we assume a non-trivial idempotent $p$, we can define two subgroups: $A=Im(p)$ and $B=ker(p)$.

• For any $a \in A$, the map acts as the identity: $p(a)=a$.
• Every element $x \in G$ can be uniquely decomposed as $x = p(x) + (x - p(x))$, where $p(x) \in A$ and $(x - p(x)) \in B$.
• Since $A \cap B = \{0\}$, it follows that $G \cong A \times B$.

(3) $\Rightarrow$ (2): From Product to Maps

If $G \cong G_1 \times G_2$, we can construct the necessary homomorphisms using the natural structure of products:

• Projection ($f$): Define $f(x,y)=x$, mapping $G_1 \times G_2 \rightarrow G_1$.
• Inclusion ($g$): Define $g(x)=(x,0)$, mapping $G_1 \rightarrow G_1 \times G_2$.
• The composition $(f \circ g)(x) = x$, satisfying $f \circ g = Id_{G_1}$.

(2) $\Rightarrow$ (1): From Maps to Idempotent

Given $f \circ g = Id_H$, we define the endomorphism $p = g \circ f$.

• Verification: $p \circ p = (g \circ f) \circ (g \circ f) = g \circ (f \circ g) \circ f = g \circ Id_H \circ f = g \circ f = p$.
• A contradiction argument ensures $p$ is neither the zero map nor the identity map, fulfilling the “non-trivial” requirement.

4. Concrete Example: $\mathbb{Z}_6$

Consider the cyclic group $\mathbb{Z}_6$, which we know is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3$.

• The Map: Define $p([x]_6) = [3x]_6$.
• Idempotency: $p(p(x)) = p(3x) = 9x \equiv 3x \pmod 6$.
• Decomposition:
  • $Im(p) = \{0, 3\} \cong \mathbb{Z}_2$.
  • $ker(p) = \{0, 2, 4\} \cong \mathbb{Z}_3$.

5. Final Takeaway

Non-trivial idempotent endomorphisms of an abelian group are precisely the projections that arise from non-trivial direct product decompositions. This fundamental concept links multiple fields of mathematics, including:

• Group and Module Theory.
• Linear Algebra (Projection matrices like $P = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$).
• Functional Analysis.