# Changes

## Extraspecial commutator-in-center subgroup is central factor

, 14:09, 24 September 2008
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===Statement with symbols===
Let $G$ be a finite $p$-group, i.e. a [[group of prime power order]]. Suppose $H$ is a subgroup of $G$ satisfying the following two conditions:
# $H$ is an [[extraspecial group]]
==Facts used==
* # [[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 finite $p$-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. So, pick a $g \in G$.
First, observe that since $[G,H] \le Z(H)$, conjugation by $g$ induces the identity map on the quotient $H/Z(H)$. Since Thus, conjugation by $Hg$ is extraspecial, viewed as an automorphism of $Z(H) = H'$, so conjugation by commutes with all the inner automorphisms of $GH$ induces . In particular, the identity map on the Abelianization automorphism induced by conjugation by $H/H'g$ is in the centralizer of $\operatorname{Inn}(H)$. Hence, it is an [[IA-automorphism]] of in $\operatorname{Aut}(H)$. Now, using the above By fact(1), that for an extraspecial group, IA-automorphisms are this forces the same as inner automorphisms, we conclude that conjugation by $g$ induces an inner automorphism of to actually be in $\operatorname{Inn}(H)$.
==References==
===Textbook references===
* {{booklink-proved|Gorenstein}}, Page 195, Lemma 4.6, Section 5.4