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

Weaker facts for other values

nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))

Facts used

Abelian implies universal power map is endomorphism: In an Abelian group, the $n^{th}$ power map is an endomorphism for all $n$ .
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

Abelian implies cube map is endomorphism

This is a direct consequence of fact (1).

Cube map is endomorphism implies abelian

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

To prove: $G$ is abelian.

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

Step no.	Assertion	Given data used	Facts used	Previous steps used	Explanation
1	$g^{2}h^{3}=h^{3}g^{2}$ for all $g,h\in G$ , i.e., every square commutes with every cube	Cube map is an automorphism, and hence an endomorphism	Fact (2)	--	Consider $g,h\in G$ . Then, by fact (2), $c_{g}$ is an automorphism, so we have: $\!c_{g}(h^{3})=c_{g}(h)^{3}$ , giving $gh^{3}g^{-1}=(ghg^{-1})^{3}$ . On the other hand, since the cube map is an endomorphism, we have $(ghg^{-1})^{3}=g^{3}h^{3}g^{-3}$ . Combining these, we get $gh^{3}g^{-1}=g^{3}h^{3}g^{-3}$ . Canceling the left-most $g$ and the right-most $g^{-1}$ and rearranging yields that $g^{2}h^{3}=h^{3}g^{2}$ .
2	$g^{2}x=xg^{2}$ for all $g,x\in G$	Cube map is an automorphism, hence is bijective	--	Step (1)	Since the cube map is bijective, every element of $G$ is a cube. Combining this with step (1) yields that $g^{2}x=xg^{2}$ for every $g,x\in G$ .
3	$g^{2}x^{2}=xgxg$ for all $g,x\in G$	Cube map is an automorphism, hence an endomorphism	--	--	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	We get $gx=xg$ for all $g,x\in G$ .			Steps (2), (3)	Using step (2), we can rewrite $g^{2}x^{2}$ as $xg^{2}x$ . Combining with step (3) yields that $xgxg=xg^{2}x$ . 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:

$(xy)^{2}=x^{2}y^{2}\iff xy=yx$

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 $xy\neq yx$ . 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