Fixed-point subgroup of a subgroup of the automorphism group implies local powering-invariant

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., fixed-point subgroup of a subgroup of the automorphism group) must also satisfy the second subgroup property (i.e., local powering-invariant subgroup)
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Statement

Suppose G is a group. Suppose B is a subgroup of the automorphism group of G. Suppose H is the subgroup of G comprising precisely those elements that are fixed by every element of B. In other words, H is a fixed-point subgroup of a subgroup of the automorphism group in G.

Then, H is a local powering-invariant subgroup of G: if h \in H and n \in \mathbb{N} are such that there is a unique x \in G satisfying x^n = h, then x \in H.

Proof

Given: Group G, subgroup B of \operatorname{Aut}(G). Subgroup H of G defined as the set of fixed points under B of G. h \in H and n \in \mathbb{N} are such that there is a unique x \in G satisfying x^n = h.

To prove: x \in H. In other words, \sigma(x) = x for all \sigma \in B.

Proof: We do the proof for fixed but arbitrary \sigma \in B. We have that, since \sigma is an automorphism:

(\sigma(x))^n = \sigma(x^n)

Simplifying further:

(\sigma(x))^n = \sigma(x^n) = \sigma(h) = h

where the last step follows from the fact that h \in H and every element of H is fixed by every automorphism in B.

We thus obtain that (\sigma(x))^n = h. Since x is the unique element whose n^{th} power is h, this forces \sigma(x) = x, completing the proof.