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Finite index not implies local powering-invariant

Statement

It is possible to have a group G and a subgroup of finite index H of G such that H is not a local powering-invariant subgroup of G, i.e., there exists an element h \in H and a natural number n such that there is a unique solution u \in G to u^n = h, but u \notin H.

In fact, we can choose H to satisfy any of these additional constraints:

Proof

Example with subgroup of index two

Let G be the infinite dihedral group:

G := \langle a,x \mid x^2 = e, xax = a^{-1} \rangle.

Here, e denotes the identity element.

Let H be the subgroup \langle a^2,x \rangle.

Then the following are true:

  • H is a subgroup of index two in G.
  • H is not local powering-invariant in G. For the element h = a^2 and n = 2, the only solution to u^2 = h for u \in G is u = a, and a \notin H.

Example with characteristic subgroup of finite index that uses derived subgroup

Consider the infinite dihedral group, given by the presentation:

G := \langle a,x \mid xax^{-1} = a^{-1}, x^2 = e \rangle

where e denotes the identity of G. We find that:

[G,G] = \langle a^2 \rangle

is an infinite cyclic group.

Now consider the element h = a^2. Let n = 2. We note that all elements outside \langle a \rangle have order two, hence any element u with u^2 = h must be inside \langle a \rangle. The only possibility is thus u = a, which is outside H. Thus, the element h = a^2 has a unique square root in G, but this is not in H, completing the proof.