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

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Statement

Suppose G is a group and n is an integer such that G is n-abelian: the n^{th} power map on G is an endomorphism of G. Then, for every g,h \in G we have h^ng^{n-1} = g^{n-1}h^n.

Related facts

Applications

Other related facts

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)


Analogues in other algebraic structures

Facts used

  1. Group acts as automorphisms by conjugation: For any g \in G, the map c_g:x \mapsto gxg^{-1} is an automorphism of G.

Proof

From the power map being an endomorphism

Given: A group G and a natural number n such that g \mapsto g^n is an endomorphism of G.

To prove: For every g,h \in G, h^ng^{n-1} = g^{n-1}h^n.

Proof: We denote by c_g the map x \mapsto gxg^{-1}.

Step no. Assertion Facts used Given data used Previous steps used Explanation
1 For g,h \in G, we have gh^ng^{-1} = (ghg^{-1})^n Fact (1) By fact (1), c_g is an automorphism of G, so c_g(h^n) = (c_g(h))^n. Applying the definition of c_g gives the result.
2 We have (ghg^{-1})^n = g^nh^ng^{-n} for all g,h \in G. n^{th} power map is endomorphism Since the n^{th} power map is an endomorphism, we have (abc)^n = a^nb^nc^n. Setting a = g, b = h, c = g^{-1} gives the desired result.
3 We get h^ng^{n-1} = g^{n-1}h^n for all g,h \in G. In other words, every n^{th} power commutes with every (n - 1)^{th} power. Steps (1), (2) Combining steps (1) and (2) gives that gh^ng^{-1} = g^nh^ng^{-n}. Cancel a left-most g and a right-most g^{-1} on both sides, and get h^n = g^{n-1}h^ng^{-(n-1)}. Multiply both sides on the right by g^{n-1} to get the desired result.

Thus, every n^{th} power commutes with every (n-1)^{th} power.