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Given a finite group $H$, How can one prove that the equation $Aut(X)=H$ has only finitely many solutions in the class of finite p-groups. (This would be the case if the divisibility conjecture is true : that is the order of a p-group divides the order of its automorphism group)

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  • $\begingroup$ Are you sure that your assertion is true? $\endgroup$ Jun 17, 2013 at 16:27
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    $\begingroup$ See H. K. Iyer, On solving the equation Aut(X)=G, Rocky Mountain Journal of Mathematics 9 (1979), no. 4, 653--670. The paper can be accessed here: projecteuclid.org/DPubS/Repository/1.0/… $\endgroup$
    – Primoz
    Jun 17, 2013 at 21:07
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    $\begingroup$ Your divisibility conjecture is not true. The order of $C_2\times C_2$ does not divide the order of its automorphism group $\text{GL}_2(\mathbb{F}_2)\cong S_3$. $\endgroup$
    – Alex B.
    Jun 17, 2013 at 22:34
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    $\begingroup$ @ Alex Bartel : I'm sorry, for the divisibility conjecture we have to assume that our p-group has order at least p^3. This conjecture is trivially false for p-groups of order p and p^2, for the cyclic one, the automorphism groups have order p-1. and a group of order p^2 is either cyclic, so its automorphism group has order p(p-1), or is elementary abelian, so its automorphism group is Gl(2, p) which has p (p-1)^2 elements. $\endgroup$ Jun 18, 2013 at 14:23
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    $\begingroup$ @Primoz : Thank you so much. Just I note that an affirmative answer to my question follows from THEOREM 2.10. in Iyer's paper. This result is due to K. H. Hyde : "On the order of the automorphism group of a finite group, Ph.D.Thesis, University of Utah, 1969. $\endgroup$ Jun 19, 2013 at 17:17

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