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)
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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).

Related facts

• Center is local powering-invariant, hence the center is powering-invariant
• Fixed-point subgroup of a subgroup of the automorphism group implies local powering-invariant
• Derived subgroup not is local powering-invariant
• Characteristic not implies powering-invariant in nilpotent group

Similar facts for Lie rings

• Characteristic not implies powering-invariant in solvable Lie ring

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

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

We can think of $G$ as the semidirect product $\mathbb{R}⋊({\mathbb{R}}^{\ast }{)}^{+}$, 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 $\mathbb{R}⋊({\mathbb{Q}}^{\ast }{)}^{+}$. Explicitly, $H$ is the subgroup comprising the maps of the form:

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

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 $\mathbb{R}$, 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 $\mathbb{R}$:

• $T_1$ 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 $\mathbb{R}$ to $\mathbb{R}$:

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

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

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 $G$