If $|G|=p^n,n>1$, and $1=G_n \subseteq G_1\subseteq \cdots \subseteq G_0=G$, is a composition series, with cyclic-$p$ quotients, and $P$ is set of all automorphisms of $G$, such that for every $u\in P$, $u(x)x^{-1}\in G_{i+1}$ $\forall x\in G_i$, then $P$ is a $p$ subgroup of $Aut(G)$.

What about converse: given $p$ subgroup $P$ of $Aut(G)$,(where $|G|=p^n, n>1$) does there exist a composition series $1=G_n \subseteq G_1\subseteq \cdots \subseteq G_0=G$ of $G$ with cyclic p-quotients such that for $u\in P$, $u(x)x^{-1}\in G_{i+1}$ $\forall x\in G_i$? Can $G_i$ chosen to be normal in $G$?

(This will be useful in finding some explicit automorphisms of groups of order $p^3$).

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    $\begingroup$ $G$ is normal in the semi-direct product $G\rtimes P =: H$. Define $H_0 := G$ and $H_{i+1} := [H_i, H]$ for $i\ge 0$. All $H_i$ are normal in $H$ and $H_{i+1} < H_i$ as long as $H_i \ne 1$. Refining the chain $H_0 > H_1 > H_2 >...> 1$ such that all factors are cyclic instead of abelian, you are done. But I doubt that you can choose the refinement to contain only normal subgroups. $\endgroup$ – Someone Feb 10 '11 at 12:59
  • $\begingroup$ I should have mentioned that $u(x) x^{-1}$ is simply the commutator $[u, x]$ as elements of $G \rtimes P$. $\endgroup$ – Someone Feb 10 '11 at 13:01
  • $\begingroup$ The subgroups $[H_i,H]$ are characteristic subgroups of $G$, isn't it? Then the series $1\subseteq H_r\subseteq \cdots \subseteq H_0=G$ will be a normal series (i.e. all subgroups in the series will be normal in $G$), and then we can refine the series. $\endgroup$ – Martin David Feb 10 '11 at 13:08
  • $\begingroup$ @Martin: You might call them $P$-characteristic. They don't have to be invariant under automorphisms not in $P$. Yes, the $H_i$ are a normal series which you can refine. $\endgroup$ – Someone Feb 10 '11 at 13:11
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    $\begingroup$ Actually the construction provided by Someone does provide a composition series made by groups that are normal in $G$: in fact, $H_{i+1}=[H_i,H]$ is the biggest quotient of $H_i$ on which $H$ acts trivially by conjugation, and in particular $G$ acts trivially and all the intermediate subgroups $H_{i+1} \subseteq K \subseteq H_i$ are normal in $G$. $\endgroup$ – Maurizio Monge Feb 10 '11 at 14:13

Elaborating on Someone's comment, we have that the answer to the question is the following:

Let's consider the semidirect product $H := G\rtimes P$, having $G$ as normal subgroup and where the elements $u \in P$ act on $G$ via the given action, i.e. in the semidirect product we have as $u \cdot x \cdot u^{-1} = u(x)$. Consider now the filtration defined by $$ H_0 := G,\qquad H_{i+1} = [H_i,H]\quad\text{ for }i>0. $$ The $H_i$ are normal in $H$, $H_{i+1} \subsetneq H_i$ and $H_i/H_{i+1}$ is precisely the biggest quotient of $H_i$ on which $H$ acts trivially by conjugation. Since in particular also $G$ acts trivially all the intermediate subgroups are normal in $G$. Consequently refining the chain of the $H_0 \supset H_1 \supset \dots$ so that all the factors are cyclic we obtain the requested composition series.

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