# Characteristic not implies powering-invariant in solvable group

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a solvable group. That is, it states that in a solvable group, every subgroup satisfying the first subgroup property (i.e., characteristic subgroup) need not satisfy the second subgroup property (i.e., powering-invariant subgroup)
View all subgroup property non-implications | View all subgroup property implications

## Statement

It is possible to construct a solvable group $G$ and a characteristic subgroup $H$ of $G$ such that $H$ is not a powering-invariant subgroup of $G$.

In fact, we can choose $G$ to be powered for any set of primes and $H$ to be powered over any choice of subset of the set of primes (and not over the others).

## Proof

### Example of a rationally powered group and a characteristic subgroup that is not powered for any prime

Further information: GAPlus(1,R)

Suppose $G$ is the group $GA^+(1,\R)$, given explicitly as the group of linear maps with positive coefficient from $\R$ to itself under composition. Explicitly, the elements of $G$ are maps of the form:

$x \mapsto ax + b, \qquad a,b \in \R, a > 0$

We can think of $G$ as the semidirect product $\R \rtimes (\R^\ast)^+$, i.e., the semidirect product of the reals under addition by the action of the positive reals under multiplication with their natural action by multiplication.

Suppose $H$ is the subgroup of $G$ given by the semidirect product $\R \rtimes (\mathbb{Q}^\ast)^+$. Explicitly, $H$ is the subgroup comprising the maps of the form:

$x \mapsto ax + b, a \in \mathbb{Q}, b \in \R, a > 0$

Then, we note that:

• $G$ is rationally powered: See GAPlus(1,R) is rationally powered
• $H$ is not powered for any prime: For instance, the map $x \mapsto 2x$ is in $H$. For any prime $p$, its unique $p^{th}$ root in $G$ is $x \mapsto 2^{1/p}x$. But since $2^{1/p} \notin \mathbb{Q}$, this element is not in $H$.
• $H$ is characteristic in $G$: First, note that the translation subgroup $x \mapsto x + b$ is precisely the derived subgroup $[G,G]$, hence it is characteristic. This subgroup is the base of the semidirect product and is isomorphic to the additive group of $\R$, hence is rationally powered. Now, note that $H$ is precisely the subgroup of $G$ of elements such that the automorphism of $[G,G]$ induced by its action by conjugation is a rational multiplication. Note that whether an automorphism is a rational multiplication is a purely group-theoretic condition, so the above specifies $H$ uniquely in an automorphism-invariant manner.

### Example for suitably chosen sets of primes

Consider two prime sets $\pi_1$ and $\pi_2$ with $\pi_2 \subset \pi_1$ (as a proper subset). Define the following two subsets of $\R$:

• $T_1$ is the subset of the set of positive reals comprising those numbers $x$ for which there is a natural number $n$ with all prime factors in $\pi_1$ for which $x^n \in (\mathbb{Q}^\ast)^+$.
• $T_2$ is the subset of the set of positive reals comprising those numbers $x$ for which there is a natural number $n$ with all prime factors in $\pi_2$ for which $x^n \in (\mathbb{Q}^\ast)^+$.

Note that $T_2 \subset T_1$ because $\pi_2 \subset \pi_1$.

Now define $G$ and $H$ as the following groups of linear maps from $\R$ to $\R$:

$G = \{ x \mapsto ax + b, a \in T_1, b \in \R \}$

$H = \{ x \mapsto ax + b, a \in T_2, b \in \R \}$

Then, we can see that:

• $G$ is powered precisely over the primes in $\pi_1$ and over no other primes
• $H$ is powered precisely over the primes in $\pi_2$ and over no other primes
• $H$ is a characteristic subgroup of $G$