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Hi: Let $G$ be a group and $G'$ it's commutator subgroup. Then $G > G' > 1$ is a series of normal subgroups of $G$. Suppose $G$ is complete. Then, if I'm not wrong, $Aut(G)$ is the stabilizer of that series. But then, by a theorem, $Aut(G)$ is nilpotent of class <= 1, that is $Aut(G)' = 1$ and $Aut(G)$ is abelian. Now, as $G$ is complete, $Aut(G) \cong G$. But then $G$ is abelian! What am I doing wrong?

EDIT: This is from Rotman's An Introduction to the Theory of Groups, 4th edition, p.165: Definition. Let $G= G_0 > G_1 > ... > G_r= 1$ be a series of normal subgroups a group $G$. An automorphism $\alpha \in Aut(G)$ stabilizes this series if $\alpha(G_i x) = G_i x$ for all $i$ and all $x \in G_{i-1}$. The stabilizer $A$ of this series is the subgroup $A =$ {$\alpha \in Aut(G): \alpha$ stabilizes the series} $\leq Aut(G)$.

Theorem 7.19. The stabilizer A of a series of normal subgroups $G = G_0 > G_1 > ... > G_r = 1$ is a nilpotent group of class $\leq r - 1$.

stf92
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    Complete groups are centerless by definition. Abelian groups aren't, ever. What do you mean by $\mathrm{Aut}(G)$ is the stabilizer of that series? You're just blabbing words out here. – Pedro Jun 03 '16 at 20:46
  • Also, please state the theorem that you are referring to. – Brent Kerby Jun 03 '16 at 20:52
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    Your comment was more useful without the last line. OP has a sincere question and is aware that s/he is failing to understand something. Criticism is unlikely to help. – jdc Jun 03 '16 at 21:21
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    @jdc Criticism is what helps, really. It's not a bad thing. – Pedro Jun 03 '16 at 21:35
  • I guess everyone's different, but I think the first part was constructive criticism and already makes the useful points without the harshness of "blabbing." – jdc Jun 03 '16 at 22:11
  • I added a definition and a theorem in the post. – stf92 Jun 04 '16 at 21:26

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I added a definition of stabilizer and a a theorem, all from Rotman's An Introduction to the Theory of Groups, 4th edition.

stf92
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