# 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) neednotsatisfy the second subgroup property (i.e., powering-invariant subgroup)

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## Contents

## Statement

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

In fact, we can choose to be powered for any set of primes and 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

## Proof

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

`Further information: GAPlus(1,R)`

Suppose is the group , given explicitly as the group of linear maps with positive coefficient from to itself under composition. Explicitly, the elements of are maps of the form:

We can think of as the semidirect product , 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 is the subgroup of given by the semidirect product . Explicitly, is the subgroup comprising the maps of the form:

Then, we note that:

- is rationally powered: See GAPlus(1,R) is rationally powered
- is not powered for any prime: For instance, the map is in . For any prime , its unique root in is . But since , this element is not in .
- is characteristic in : First, note that the translation subgroup is precisely the derived subgroup , hence it is characteristic. This subgroup is the base of the semidirect product and is isomorphic to the additive group of , hence is rationally powered. Now, note that is precisely the subgroup of of elements such that the automorphism of 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 uniquely in an automorphism-invariant manner.

### Example for suitably chosen sets of primes

Consider two prime sets and with (as a proper subset). Define the following two subsets of :

- is the subset of the set of positive reals comprising those numbers for which there is a natural number with all prime factors in for which .
- is the subset of the set of positive reals comprising those numbers for which there is a natural number with all prime factors in for which .

Note that because .

Now define and as the following groups of linear maps from to :

Then, we can see that:

- is powered precisely over the primes in and over no other primes
- is powered precisely over the primes in and over no other primes
- is a characteristic subgroup of