Extraspecial commutator-in-center subgroup is central factor

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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

Extraspecial group

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)


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