Extraspecial commutator-in-center subgroup is central factor

Statement with symbols
Let $$G$$ be a group. Suppose $$H$$ is a subgroup of $$G$$ satisfying the following two conditions:


 * 1) $$H$$ is an extraspecial group
 * 2) $$[G,H] \le Z(H)$$ (i.e., $$H$$ is a commutator-in-center subgroup of $$G$$)

Then $$HC_G(H) = G$$, i.e., $$H$$ is a central factor of $$G$$.

Central factor
A subgroup $$H$$ of a group $$G$$ is termed a central factor of $$G$$ if $$H$$ is normal in $$G$$, and the following holds: consider the induced map $$G \to \operatorname{Aut}(H)$$, by conjugation by $$G$$. Then, the image of $$G$$ under this map is precisely $$\operatorname{Inn}(H)$$.

Equivalently, every inner automorphism of $$G$$ restricts to an inner automorphism of $$H$$.

Facts used

 * 1) uses::Extraspecial implies inner automorphism group is self-centralizing in automorphism group (Note: An equivalent formulation of this is IA equals inner in extraspecial)

Proof
Given: A group $$G$$, a subgroup $$H$$ such that $$[G,H] \le Z(H)$$ and $$H$$ is extraspecial.

To prove: $$H$$ is a central factor of $$G$$

Proof: We use the definition of central factor in terms of inner automorphisms. In other words, we strive to show that conjugation by any element of $$G$$ is equivalent to an inner automorphism as far as $$H$$ is concerned.

First, observe that since $$[G,H] \le Z(H)$$, $$H/Z(H)$$ is in the center of $$G/Z(H)$$. Thus, $$\operatorname{Inn}(H)$$ is in the center of the subgroup $$K$$ of $$\operatorname{Aut}(H)$$ obtained by the action of $$G$$ on $$H$$ by conjugation. In particular, $$K$$ is in the centralizer of $$\operatorname{Inn}(H)$$ in $$\operatorname{Aut}(H)$$. By fact (1), we see this forces that $$K = \operatorname{Inn}(H)$$, completing the proof.

Textbook references

 * , Page 195, Lemma 4.6, Section 5.4