# Retract is transitive

This article gives the statement, and possibly proof, of a subgroup property (i.e., retract) satisfying a subgroup metaproperty (i.e., transitive subgroup property)
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## Statement

Suppose $H \le G \le K$ are groups such that $G$ is a retract of $K$ and $H$ is a retract of $G$, then $H$ is a retract of $K$.

## Definitions used

### Retract

Further information: Retract

A subgroup $H$ of a group $G$ is termed a retract of $G$ if there exists a homomorphism $\sigma:G \to H$ such that $\sigma(h) = h$ for all $h \in H$.

## Proof

Given: A group $K$, a retract $G$ of $K$, a retract $H$ of $G$.

To prove: $H$ is a retract of $K$.

Proof: Let $\alpha:K \to G$ be a retraction, i.e., $\alpha$ is a homomorphism such that $\alpha(g) = g$ for all $g \in G$. Let $\beta:G \to H$ be a retraction, i.e., $\beta$ is a homomorphism such that $\beta(h) = h$ for all $h \in H$.

Now consider the composite map $\beta \circ \alpha: K \to H$. We want to argue that this is a retraction. Consider $h \in H$. Then, by construction $\alpha(h) = h$, so $\beta(\alpha(h)) = \beta(h)$. Since $H \le G$, $\beta(h) = h$, and so we get $\beta(\alpha(h)) = h$. Thus, $\beta \circ \alpha$ is a retraction, and thus $H$ is a retract of $K$.