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Nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center

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Statement

The following statements are equivalent for a group G and an integer n. Suppose the n^{th} power map g \mapsto g^n is a surjective endomorphism (such as an automorphism) of G.

Then, the (n-1)^{th} power map g \mapsto g^{n-1} is an endomorphism of G and g^{n-1} is in the center of G for all g \in G.

Related facts

Facts used

Proof

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Given: A group G and an integer n such that \sigma = g \mapsto g^n is a surjective endomorphism of G.

To prove: \tau = g \mapsto g^{n-1} is an endomorphism of G and g^{n-1}x = xg^{n-1} for all g,x \in G.

Proof

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 g^{n-1}h^n = h^ng^{n-1} for all g,h \in G. Fact (1) n^{th} power map is endomorphism Given+Fact direct
2 g^{n-1}x = xg^{n-1} for all g,x \in G n^{th} power map is surjective Step (1) [SHOW MORE]
3 For any g,x \in G, (gx)^{n-1} = g^{n-1}x^{n-1}, so the (n-1)^{th} power map is an endomorphism n^{th} power map is endomorphism Step (2) [SHOW MORE]

Steps (2) and (3) complete the proof.