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

## 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

### 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)

## 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.