Extraspecial commutator-in-center subgroup is central factor

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. $[G,H]\leq Z(H)$ (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 $[G,H]\leq Z(H)$ 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 $[G,H]\leq Z(H)$, $H/Z(H)$ 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.

References

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 195, Lemma 4.6, Section 5.4