Given a group $G$, we denote by $T(G)$ the subgroup generated by all (maximal) normal abelian subgroups of $G$.


Let define the series $(T\_i(G))$ by $T\_0(G)=1$ and $T\_{i+1}(G)/T\_i(G)=T(G/T\_i(G)$, and $n(G)$ to be the smallest integer such that $T\_n(G)=G$.


1) Can one produce finite p-groups $G$ with arbitrary large n(G)?


2) If $G$ is a finite p-group, what can one say in general about the index of $T(G)$ in $G$.


3) I found that the nilpotency class of $T(G)$ (assuming just that $G$ is finite), can not exceed $s$, where $s$ is the cardinal of the set of conjugacy class sizes in $G$. In fact if $1 \lt n\_1 \lt ... \lt n\_s$ are the conjugacy class sizes of $G$, we define $H\_i$ to be the subgroup generated by the elements with conj class sizes $\leq n\_i$, we use a result of Mann and Isaacs to show that $T(G)$ is a stability group of the series $H\_i$ and the claim follows from a well known result of Kaloujnin.


Is this result trivial ? 



Normal abelian subgroups in p-groups
The answer to question 1 is yes. We will demonstrate a way to add one to $n(G)$, so $n(G)$ can be any natural number.


Given a group $G$, and an abelian group $A$, take a faithful action of $G$ on $A$, and consider the semidirect product $A \rtimes G$ .


We need the following condition: If $g$ is a nontrivial element of $g$, then there exists $a \in A$ such that $g^2(a)-2g(a)+a\neq 0$. Such an action on an abelian group exists for every $G$. For instance, this is satisfied by the regular representation of $G$ over $\mathbb F\_p$ for $p$ odd and the regular representation of $G$ over $\mathbb Z/4$ when $p=2$.


Then we will show that $T(A \rtimes G)=A$, so $n(A \rtimes G) = n(G)+1$. $A$ is abelian and normal, so contained in $T(G)$, so we must only check that each abelian normal subgroup is contained in $A$. Suppose instead that $H$ is an abelian normal subgroup containing an element $x$ which projects to a nontrivial element $g$ of $G$. Let $a$ be an element of $H$, then


$$g(a)-a=(xax^{-1}) a^{-1} =x (ax^{-1}a^{-1}) \in H$$


by normality.


Since $H$ is abelian, $g(a)-a$ commutes with $x$, so $g (g(a)-a) - g(a)-a =0$, so $g^2(a)-2g(a) +a=0$, but this cannot hold for all $a$, so there is a contradiction.


