Retract implies central factor of normalizer

Verbal statement
Any subgroup of a group that is a retract is also a central factor of normalizer.

Statement with symbols
Suppose $$H$$ is a retract of a group $$G$$. In other words, $$H$$ is a subgroup of $$G$$ such that there exists a surjective homomorphism $$\alpha:G \to H$$ whose restriction to $$H$$ is the identity map on $$H$$. Then, $$H$$ is a central factor of normalizer of $$G$$: any inner automorphism of $$G$$ that restricts to an automorphism of $$H$$ must restrict to an inner automorphism of $$H$$.

Intermediate properties

 * Subset-conjugacy-closed subgroup

Proof
Given: A group $$G$$, a subgroup $$H \le G$$. A retraction $$\alpha:G \to H$$. An inner automorphism $$\varphi$$ of $$G$$ such that the restriction of $$\varphi$$ to $$H$$ is an automorphism of $$H$$.

To prove: The restriction of $$\varphi$$ to $$H$$ is an inner automorphism of $$H$$.

Proof:


 * 1) There exists $$g \in G$$ such that $$\varphi = c_g$$, where $$c_g = x \mapsto gxg^{-1}$$: This follows from the definition of inner automorphism.
 * 2) If $$x \in H, gxg^{-1} \in H$$, then $$gxg^{-1} = \alpha(g)x\alpha(g)^{-1}$$: By definition of homomorphism, $$\alpha(gxg^{-1}) = \alpha(g)\alpha(x)\alpha(g)^{-1}$$. If $$x, gxg^{-1} \in H$$, we have $$\alpha(x) = x$$ and $$\alpha(gxg^{-1}) = gxg^{-1}$$. This yields $$gxg^{-1} = \alpha(g)x\alpha(g)^{-1}$$.
 * 3) The restriction to $$H$$ of $$c_g$$, if well-defined, equals conjugation in $$H$$ by the element $$\alpha(g)$$: This follows from the previous step.

Thus, the restriction of $$\varphi$$ to $$H$$, if well-defined, is conjugation by an element inside $$H$$, and hence an inner automorphism.