Characteristic not implies powering-invariant in solvable group

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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

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 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