Cube map is surjective endomorphism implies abelian

This article describes an easy-to-prove fact about basic notions in group theory, that is not very well-known or important in itself
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Statement

Verbal statement

If the cube map on a group is an automorphism, then the group is an Abelian group.

Statement with symbols

Let $G$ be a group such that the map $\sigma :G\to G$ defined by $\sigma (x)=x^3$ is an automorphism. Then, $G$ is an Abelian group.

Related facts

Applications

• Cube map is endomorphism iff Abelian (if order is not a multiple of 3)

Stronger facts for other values

• Inverse map is automorphism iff Abelian
• Square map is endomorphism iff Abelian

Weaker facts for other values

• nth power map is endomorphism iff Abelian (if order is relatively prime to n(n-1))
• nth power map is automorphism implies Abelian (if order is relatively prime to n-1)

Facts used

1. Abelian implies universal power map is endomorphism: In an Abelian group, the $n^{th}$ power map is an endomorphism for all $n$.
2. Group acts as automorphisms by conjugation: For any $g\in G$, the map $c_g=x\mapsto gx{g}^{-1}$ is an automorphism of $G$.

Proof

Abelian implies cube map is endomorphism

This is a direct consequence of fact (1).

Cube map is endomorphism implies Abelian

Given: A group $G$ such that the map sending $x$ to $x^3$ is an automorphism of $G$.

To prove: $G$ is Abelian.

Proof:

1. (Facts used: fact (2); Given data used: Cube map is endomorphism): $g^2{h}^3=h^3{g}^2$ for all $g,h\in G$, i.e., every square commutes with every cube: Consider $g,h\in G$. Then, by fact (2), $c_g$ is an automorphism, so we have:
   • $c_g(h^3)=c_g(h{)}^3,⟹g{h}^3{g}^{-1}=(gh{g}^{-1}{)}^3$.
   • On the other hand, since the $n^{th}$ power map is an endomorphism, we have $(gh{g}^{-1}{)}^n=g^n{h}^n{g}^{-n}$.
   • Combining these, we get $g{h}^3{g}^{-1}=g^3{h}^3{g}^{-3}$.
   • Canceling the left-most $g$ and the right-most $g^{-1}$ and rearranging yields that $g^2{h}^3=h^3{g}^2$.
2. (Given data used: Cube map is bijective) $g^{2}x=x{g}^2$ for all $g,x\in G$: Since the cube map is an automorphism, it is bijective, and so every element of $G$ is a cube. Combining this with step (1) yields that $g^{2}x=x{g}^2$ for every $g,x\in G$.
3. (Given data used: cube map is endomorphism) $g^2{x}^2=xgxg$ for all $g,x\in G$: Since the cube map is an endomorphism, we get $(gx{)}^3=g^3{x}^3$, so expanding and canceling the left-most and right-most terms yields $xgxg=g^2{x}^2$.
4. Using step (2), we can rewrite $g^2{x}^2$ as $x{g}^{2}x$. Combining with step (3) yields that $xgxg=x{g}^{2}x$. Canceling $xg$ from the left, we get $gx=xg$, which is the goal.

Difference from the corresponding statement for the square map

In the case of the square map, we had in fact proved something much stronger:

$(xy{)}^2=x^2{y}^2⟺xy=yx$

In the case of the cube map, this is no longer true. That is, it may so happen that $(xy{)}^3=x^3{y}^3$ although $xy\ne yx$. Thus, to show that $xy=yx$ we need to not only use that $(xy{)}^3=x^3{y}^3$ but also use that this identity is valid for other elements picked from $G$ (specifically, that it is valid for their cuberoots).

References

Textbook references

• Topics in Algebra by I. N. Herstein, ^{More info}, Page 48, Exercise 24

External links

Links to related riders

• Mathlinks discussion forum page for Power maps being homomorphisms implies Abelianness