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Extraspecial commutator-in-center subgroup is central factor

28 bytes added, 14:09, 24 September 2008
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===Statement with symbols===
Let <math>G</math> be a finite <math>p</math>-group, i.e. a [[group of prime power order]]. Suppose <math>H</math> is a subgroup of <math>G</math> satisfying the following two conditions:
# <math>H</math> 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]])
'''Given''': A finite <math>p</math>-group <math>G</math>, a subgroup <math>H</math> such that <math>[G,H] \le Z(H)</math> and <math>H</math> is extraspecial.
'''To prove''': <math>H</math> is a [[central factor]] of <math>G</math>
'''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 <math>G</math> is equivalent to an inner automorphism as far as <math>H</math> is concerned. So, pick a <math>g \in G</math>.
First, observe that since <math>[G,H] \le Z(H)</math>, conjugation by <math>g</math> induces the identity map on the quotient <math>H/Z(H)</math>. Since Thus, conjugation by <math>Hg</math> is extraspecial, viewed as an automorphism of <math>Z(H) = H'</math>, so conjugation by commutes with all the inner automorphisms of <math>GH</math> induces . In particular, the identity map on the Abelianization automorphism induced by conjugation by <math>H/H'g</math> is in the centralizer of <math>\operatorname{Inn}(H)</math>. Hence, it is an [[IA-automorphism]] of in <math>\operatorname{Aut}(H)</math>. 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 <math>g</math> induces an inner automorphism of to actually be in <math>\operatorname{Inn}(H)</math>. 
===Textbook references===
* {{booklink-proved|Gorenstein}}, Page 195, Lemma 4.6, Section 5.4
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