A group acting on a $p$-group Let $G$ be a finite $p$-group acting on a $p$-group $N$ with cardinality $p^n$.
If we denote $[n,g]:=n^{-1}n^g$ we can consider the subgroup
$[N,G]=\langle [n,g]:n\in N, g\in G\rangle$.
We can now define $N_1=N$ and $N_{i+1}=[N_i,G]$.
Does it always happen that $N_i\gneq N_{i+1}$ if $N_i$ is not trivial.
I am studying finite $p$-groups acting uniserially. And by the way things are written down in the book, it seems that this is true but I don't see why this should be true in general.
For example if the action is conjugation, this is $n^g=g^{-1}ng$ then is clear that $N_i\gneq N_{i+1}$ if $N_i$ is not trivial. But is it true for any action?
I fell into this question when I read the following lemma: (Note that $|N|=p^n$).
The action of $G$ on $N$ is uniserial if and only if $N\not=1$; and so $N=N_1\gneq N_2\gneq...N_n\gneq N_{n+1}$, and $|N_i:N_{i+1}|=p$ for $1\leq i\leq n.$
I hope I have been understood. Any help would be welcome.
 A: The result, as stated, is not true - take $N= \mathbb Z/p$ and $G={\rm Aut}(N)$. Then, because any element of $N\setminus\{0\}$ generates $N$, it suffices to show that $[n,g]$ can be nontrivial to show that $[N,G]= N$, in other words it suffices to show that $n^g$ can be different from $n$.
But if $p>2$, then $G$ is nontrivial and so $n^g\neq n$ for some nontrivial $n$ and some $g$.
To answer your comment : suppose $G$ is a $p$-group. I claim that then the action on $N$ is unipotent, in the sense that there are subgroups $N_i$ of $N$, stable under the action of $G$, each normal in the next, such that $N_k = N, N_0 = 1$ and such that the induced $G$-action on $N_{i+1}/N_i$ is trivial.
Indeed, working by induction on $N$, it suffices to show that $Z(N)^G$ (the subgroup fixed points inside the center of $N$) is nontrivial - this subgroup is obviously normal in $N$, so if it is nontrivial then we can consider $N/Z(N)^G$, which is strictly smaller, so we can apply the induction claim to that one. In fact this shows that we can choose the $N_i$'s to be normal in $N$.
But now $Z(N)$ is an abelian $p$-group with an action of the $p$-group $G$, so we are reduced to the abelian case, which is well-known (I can give you more details about that if you don't know it).
Now I claim that it follows that the answer is yes in this case. Indeed, it of course suffices to show that $[N,G]<N$, but now by picking $N'$ to be the biggest $N_i$ in our sequence from before, we have that each $[n,g]$ gets mapped to $1$ in $N/N'$, so $[N,G]\leq N' < N$.
