# Normal abelian subgroups in p-groups

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 ?

• I changed the symbols < to \lt ; usage of the former caused problems as they have a different meaning on this site. (There are severral such issues; as a general alternative workaround one can include the math withing dollar symbols that does not render properly withing additional backticks ` both at the beginning and the end a backtick, or include the relevant paragraph in p tags, that is <p> at the start and </p> at the end) – user9072 Jun 17 '13 at 16:25
• what are examples of $p$-groups $G$ with $n(G)>1$? (i.e, $T(G)\neq G$). – YCor Jun 17 '13 at 20:45
• @Yves Cornulier : There exist p-groups of a maximal class with a maximal subgroup that is abelian (for instance the dihedral groups $D_{2^n}$), if G is such a group with A maximal abelian, then T(G)=A or G has class 2. So if the order of G is large enough, n(G)>1. – Yassine Guerboussa Jun 18 '13 at 14:10
• what do you call class? is it the nilpotency class, or something else? – YCor Jun 19 '13 at 8:37
• Anyway, OK I'm convinced that the dihedral group of order $2^n\ge 16$ satisfies $n(G)=2$. Note that it can be embedded in a group of upper triangular matrices over $\mathbf{Z}/2\mathbf{Z}$ with 1 on the diagonal, and the latter satisfies $n(G)=1$. In particular, $n$ can, unlike the nilpotency or solvability length, increase when passing to subgroups. – YCor Jun 19 '13 at 8:45

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.

• Very good answer Will Sawin. Thank you and best wishes. – Yassine Guerboussa Aug 5 '13 at 17:38

The answer to Qustion 2 is that the index of $T(G)$ in $G$ can be unboundedly large. The reason is that if $G = X \times Y$, the direct product, then $T(G) = T(X) \times T(Y)$. Thus, for example, if $G$ is the direct product of $n$ copies of a dihedral group of order $16$, then $|G:T(G)| = 2^n$. To prove this lemma on direct products, it suffices to show that if $U$ is maximal normal abelian in $G$, then $D = A \times B$, where $A$ and $B$ are maximal normal abelian in $X$ and $Y$, respectively.

Here is the argument. Let $P = XD$ and note that $P/X$ is abelian. Let $B = P \cap Y$, so $B$ is abelian and $P = XB$. Now $B$ is a direct factor of $P$ and hence is central in $P$. It follows that $B \subseteq D$, and thus $D = A \times B$, where $A = D \cap X$. Since $D$ is maximal normal abelian in $G$, both $A$ and $B$ are normal, and we see that they must be maximal abelian normal in $X$ and $Y$, respectively.

The Sylow subgroup $\Sigma_n$ of the Symmetric group $S_{p^n}$ has only one maximal normal abelian subgroup, say $B$, for $p>2$ and $n>1$. As $\Sigma_n/B$ is isomorphic to
$\Sigma_{n-1}$, the group $\Sigma_n$ yields a positive answers on Questions 1 and 2.

Y.B.