# Extraspecial commutator-in-center subgroup is central factor

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## Statement

### 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 $G$.

## Definitions used

### Central factor

Further information: 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. 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.