Retract implies central factor of normalizer

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., retract) must also satisfy the second subgroup property (i.e., central factor of normalizer)
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Statement

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 $H$.

Definitions used

Retract

Further information: Retract

Central factor of normalizer

Further information: Central factor of normalizer

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 $\phi$ of $G$ such that the restriction of $\phi$ to $H$ is an automorphism of $H$.

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

Proof:

1. There exists $g\in G$ such that $\phi =c_g$, where $c_g=x\mapsto gx{g}^{-1}$: This follows from the definition of inner automorphism.
2. If $x\in H,gx{g}^{-1}\in H$, then $gx{g}^{-1}=\alpha (g)x\alpha (g{)}^{-1}$: By definition of homomorphism, $\alpha (gx{g}^{-1})=\alpha (g)\alpha (x)\alpha (g{)}^{-1}$. If $x,gx{g}^{-1}\in H$, we have $\alpha (x)=x$ and $\alpha (gx{g}^{-1})=gx{g}^{-1}$. This yields $gx{g}^{-1}=\alpha (g)x\alpha (g{)}^{-1}$.
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 $\phi$ to $H$, if well-defined, is conjugation by an element inside $H$, and hence an inner automorphism.