N-abelian implies n(n-1)-central

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Statement

Suppose G is a group and n is an integer other than 0 or 1. Suppose that G is a n-abelian group, i.e., the map x \mapsto x^n is an endomorphism of G (hence, it is a universal power endomorphism). Then, G is n(n-1)-central: the inner automorphism group of G has exponent dividing n(n-1).

Related facts

Converse

Facts about n-abelian groups

We say that a group is a n-abelian group if the n^{th} power map is an endomorphism. Here are some related facts about n-abelian groups.



Value of n (note that the condition for n is the same as the condition for 1-n) Characterization of n-abelian groups Proof Other related facts
0 all groups obvious
1 all groups obvious
2 abelian groups only 2-abelian iff abelian endomorphism sends more than three-fourths of elements to squares implies abelian
-1 abelian groups only -1-abelian iff abelian
3 3-abelian group means: 2-Engel group and derived subgroup has exponent dividing three Levi's characterization of 3-abelian groups cube map is surjective endomorphism implies abelian, cube map is endomorphism iff abelian (if order is not a multiple of 3), cube map is endomorphism implies class three
-2 same as for 3-abelian (based on n-abelian iff (1-n)-abelian)


Facts used

  1. n-abelian implies every nth power and (n-1)th power commute

Proof

Given: A group G such that the map x \mapsto x^n is an endomorphism of G. Elements g,h \in G (possibly equal).

To prove: g^{n(n-1)} commutes with h. This is sufficient because h being arbitrary, it shows that g^{n(n-1)} is central, and g being arbitrary, it shows that G/Z(G) has exponent dividing n(n-1).

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 g^{n(n-1)} commutes with h^n. Fact (1) The map x \mapsto x^n is an endomorphism of G. By the given data and Fact (1), every n^{th} power commutes with every (n-1)^{th} power. Treating g^{n(n-1)} = (g^n)^{n-1} as a (n-1)^{th} power and h^n as a n^{th} power, we obtain that g^{n(n-1)} commutes with h^n.
2 g^{n(n-1)} commutes with h^{n-1}. Fact (1) The map x \mapsto x^n is an endomorphism of G. By the given data and Fact (1), every n^{th} power commutes with every (n-1)^{th} power. Treating g^{n(n-1)} = (g^{n-1})^n as a n^{th} power and h^{n-1} as a (n-1)^{th} power, we obtain that g^{n(n-1)} commutes with h^{n-1}.
3 The centralizer in G of g^{n(n-1)} contains the elements h^n and h^{n-1}. Since the centralizer of any element is a subgroup, it also contains the quotient h^n(h^{n-1})^{-1} = h, and hence g^{n(n-1)} commutes with h. Steps (1), (2) The first part follows from Steps (1) and (2). The rest is self-explanatory.