Semidirect product need not preserve powering

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It is possible to have a group G, a complemented normal subgroup H, and a prime number p such that the following hold:

  • H is powered over p.
  • The quotient group G/H (which is also isomorphic to any permutable complement to H in G) is also powered over p.
  • G is not powered over p.

Note that it is not possible to construct finite examples, because in the finite case, being powered over a prime p is equivalent to p not dividing the order (see kth power map is bijective iff k is relatively prime to the order).


Below is an example where both H and G/H are rationally powered (i.e., powered over all primes), but G is not powered over any prime. There may be simpler examples.

Let W be the subgroup \exp(\mathbb{Q}) inside (\R^*)^+. Recall that the group GAPLus(1,R) is the group of affine maps from \R to \R where the multiplication is positive, i.e.:

GA^+(1,\R) = \R \rtimes (\R^*)^+

Let U be the vector space over \mathbb{Q} generated by W as a subset of \R. Note that U is a subring of \R, because its generating set is a multiplicative monoid:

G = U \rtimes W

Explicitly, it is the set of maps:

\{ x \mapsto ax + b \mid a \in W, b \in U \}

Let H be the base of the semidirect product here, so it is isomorphic to U.

  • H is powered over all primes: That's because it is isomorphic to U, a vector space over \mathbb{Q}.
  • G/H is powered over all primes: That is because it is isomorphic to W = \exp(\mathbb{Q}), which is isomorphic to \mathbb{Q}.
  • G is not powered over any prime. Consider the element of G of the form x \mapsto ex + 1 (here, e = \exp(1) is transcendental). For a prime p, the unique p^{th} root of this in GA^+(1,\R) is:

x \mapsto e^{1/p}x + \frac{1}{1 + e^{1/p} + e^{2/p} + \dots + e^{(p-1)/p}}

We would like to claim that the number \frac{1}{1 + e^{1/p} + e^{2/p} + \dots + e^{(p-1)/p}} is not an element of U, so that this p^{th} root is not in G. This can be deduced from the fact that e is trancendental.