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The answer to question $1$ is yes. We will demonstrate a way to add one to $n(G)$.

Given a group $G$, and an abelian group $G$ 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 $g^2(a)-2g(a)+a=0$. 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 $g$. Let $H$ be 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.

So $n(G)$ can be any natural number.

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.

The answer to question $1$ is yes. Let $p$ be odd, and let $G_k$ be the Sylow $p$-subgroup of $S_{p^k}$, that is, the wreath product of $G_{k-1} Wr_{\mathbb Z/p} \mathbb Z/p$. Alternately, we can express it as $\mathbb Z/p Wr_{[p^{k-1}] } G_{k-1}$$.

We will show that $n(G_k) = k$. To prove this, we will show that $T(G_k)$ is the kernel of the quotient map $G_k \to G_{k-1}$. Clearly the kernel, $(\mathbb Z/p)^{p^{k-1}}$, is contained in $T(G_k)$, since it is an abelian normal subgroup. We will show it is maximal.

Suppose there were an abelian normal subgroup $H$ whose projection to $G_{k-1}$ included a nontrivial permutation $\sigma$ in $G_{k-1}$. Then if $v$ is a vector in $(\mathbb Z/p)^{p^{k-1}}$, $\sigma(v)-v$ is in $H$, because it's the commutator of $v$ with an element of $H$. Since $H$ is abelian, $\sigma(\sigma(v)-v) = \sigma(v)-v$, so $\sigma^2(v) - 2 \sigma(v) +v=0$. For this to hold for all $v$, $\sigma$ must be trivial or of order $2$ and acting on a vector space of characteristic $2$. But $\sigma$ is nontrivial and $p$ is odd, so this is impossible.

The answer to question $1$ is yes. We will demonstrate a way to add one to $n(G)$.

Given a group $G$, and an abelian group $G$ 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 $g^2(a)-2g(a)+a=0$. 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 $g$. Let $H$ be 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.

So $n(G)$ can be any natural number.

The answer to question $1$ is yes. Let $p$ be odd, and let $G_k$ be the Sylow $p$-subgroup of $S_{p^k}$, that is, the wreath product of $G_{k-1} Wr_{\mathbb Z/p} \mathbb Z/p$. Alternately, we can express it as $\mathbb Z/p Wr_{[p^{k-1}] } G_{k-1}$$.

We will show that $n(G_k) = k$. To prove this, we will show that $T(G_k)$ is the kernel of the quotient map $G_k \to G_{k-1}$. Clearly the kernel, $(\mathbb Z/p)^{p^{k-1}}$, is contained in $T(G_k)$, since it is an abelian normal subgroup. We will show it is maximal.

Suppose there were an abelian normal subgroup $H$ whose projection to $G_{k-1}$ included a nontrivial permutation $\sigma$ in $G_{k-1}$. Then if $v$ is a vector in $(\mathbb Z/p)^{p^{k-1}}$, $\sigma(v)-v$ is in $H$, because it's the commutator of $v$ with an element of $H$. Since $H$ is abelian, $\sigma(\sigma(v)-v) = \sigma(v)-v$, so $\sigma^2(v) - 2 \sigma(v) +v=0$. For this to hold for all $v$, $\sigma$ must have order $2$. But $p$ is odd, so this is impossible.

The answer to question $1$ is yes. Let $p$ be odd, and let $G_k$ be the Sylow $p$-subgroup of $S_{p^k}$, that is, the wreath product of $G_{k-1} Wr_{\mathbb Z/p} \mathbb Z/p$. Alternately, we can express it as $\mathbb Z/p Wr_{[p^{k-1}] } G_{k-1}$$.

We will show that $n(G_k) = k$. To prove this, we will show that $T(G_k)$ is the kernel of the quotient map $G_k \to G_{k-1}$. Clearly the kernel, $(\mathbb Z/p)^{p^{k-1}}$, is contained in $T(G_k)$, since it is an abelian normal subgroup. We will show it is maximal.

Suppose there were an abelian normal subgroup $H$ whose projection to $G_{k-1}$ included a nontrivial permutation $\sigma$ in $G_{k-1}$. Then if $v$ is a vector in $(\mathbb Z/p)^{p^{k-1}}$, $\sigma(v)-v$ is in $H$, because it's the commutator of $v$ with an element of $H$. Since $H$ is abelian, $\sigma(\sigma(v)-v) = \sigma(v)-v$, so $\sigma^2(v) - 2 \sigma(v) +v=0$. For this to hold for all $v$, $\sigma$ must be trivial or of order $2$ and acting on a vector space of characteristic $2$. But $\sigma$ is nontrivial and $p$ is odd, so this is impossible.