Retract is transitive

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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.