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 $H\leq G\leq K$ 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 $H\leq G$, $\beta (h)=h$, and so we get $\beta (\alpha (h))=h$. Thus, $\beta \circ \alpha$ is a retraction, and thus $H$ is a retract of $K$.