Cube map is surjective endomorphism implies abelian: Difference between revisions

Revision as of 18:03, 25 February 2011

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

Group acts as automorphisms by conjugation: For any $g\in G$ , the map $c_{g}=x\mapsto gxg^{-1}$ is an automorphism of $G$ .
nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center
Square map is endomorphism iff abelian

Proof

Hands-on proof using fact (1)

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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 (1)	--	[SHOW MORE] Consider $g,h\in G$ . Then, by fact (1), $c_{g}$ 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)	[SHOW MORE] 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	--	--	[SHOW MORE] 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)	[SHOW MORE] 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.

Cube map is automorphism implies abelian (using facts (3) and (4))

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

To prove: $G$ is abelian.

Proof: By fact (2), we conclude that the square map must be an endomorphism of $G$ . By fact (3), we conclude that therefore $G$ must be abelian.

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

@@ Line 26: / Line 26: @@
 ==Facts used==
-# [[uses::Abelian implies universal power map is endomorphism]]: In an Abelian group, the <math>n^{th}</math> power map is an endomorphism for all <math>n</math>.
 # [[uses::Group acts as automorphisms by conjugation]]: For any <math>g \in G</math>, the map <math>c_g = x \mapsto gxg^{-1}</math> is an automorphism of <math>G</math>.
 # [[uses::nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center]]
@@ Line 32: / Line 31: @@
 ==Proof==
-===Cube map is automorphism implies abelian (hands-on proof using fact (2))===
+===Hands-on proof using fact (1)===
 {{tabular proof format}}
@@ Line 45: / Line 44: @@
 ! Step no. !! Assertion !! Given data used !! Facts used !! Previous steps used !! Explanation
 |-
-| 1 || <math>g^2h^3 = h^3g^2</math> for all <math>g,h \in G</math>, i.e., every square commutes with every cube || Cube map is an automorphism, and hence an endomorphism || Fact (2) || -- || Consider <math>g,h \in G</math>. Then, by fact (2), <math>c_g</math> is an automorphism, so we have: <math>\! c_g(h^3) = c_g(h)^3</math>, giving <math>gh^3g^{-1} = (ghg^{-1})^3</math>. <br>On the other hand, since the cube map is an endomorphism, we have <math>(ghg^{-1})^3 = g^3h^3g^{-3}</math>. Combining these, we get <math>gh^3g^{-1} = g^3h^3g^{-3}</math>. Canceling the left-most <math>g</math> and the right-most <math>g^{-1}</math> and rearranging yields that <math>g^2h^3 = h^3g^2</math>.
+| 1 || <math>g^2h^3 = h^3g^2</math> for all <math>g,h \in G</math>, i.e., every square commutes with every cube || Cube map is an automorphism, and hence an endomorphism || Fact (1) || -- || <toggledisplay>Consider <math>g,h \in G</math>. Then, by fact (1), <math>c_g</math> is an automorphism, so we have: <math>\! c_g(h^3) = c_g(h)^3</math>, giving <math>gh^3g^{-1} = (ghg^{-1})^3</math>. <br>On the other hand, since the cube map is an endomorphism, we have <math>(ghg^{-1})^3 = g^3h^3g^{-3}</math>. Combining these, we get <math>gh^3g^{-1} = g^3h^3g^{-3}</math>. Canceling the left-most <math>g</math> and the right-most <math>g^{-1}</math> and rearranging yields that <math>g^2h^3 = h^3g^2</math>.</toggledisplay>
 |-
-| 2 || <math>g^2x = xg^2</math> for all <math>g,x \in G</math> || Cube map is an automorphism, hence is bijective || -- || Step (1) || Since the cube map is bijective, every element of <math>G</math> is a cube. Combining this with step (1) yields that <math>g^2x = xg^2</math> for every <math>g,x \in G</math>.
+| 2 || <math>g^2x = xg^2</math> for all <math>g,x \in G</math> || Cube map is an automorphism, hence is bijective || -- || Step (1) || <toggledisplay>Since the cube map is bijective, every element of <math>G</math> is a cube. Combining this with step (1) yields that <math>g^2x = xg^2</math> for every <math>g,x \in G</math>.</toggledisplay>
 |-
-| 3 || <math>g^2x^2 = xgxg</math> for all <math>g,x \in G</math> || Cube map is an automorphism, hence an endomorphism || -- || -- || Since the cube map is an endomorphism, we get <math>(gx)^3 = g^3x^3</math>, so expanding and canceling the left-most and right-most terms yields <math>xgxg = g^2x^2</math>.
+| 3 || <math>g^2x^2 = xgxg</math> for all <math>g,x \in G</math> || Cube map is an automorphism, hence an endomorphism || -- || -- || <toggledisplay>Since the cube map is an endomorphism, we get <math>(gx)^3 = g^3x^3</math>, so expanding and canceling the left-most and right-most terms yields <math>xgxg = g^2x^2</math>.</toggledisplay>
 |-
-| 4 || We get <math>gx = xg</math> for all <math>g,x \in G</math>. || || || Steps (2), (3) || Using step (2), we can rewrite <math>g^2x^2</math> as <math>xg^2x</math>. Combining with step (3) yields that <math>xgxg = xg^2x</math>. Canceling <math>xg</math> from the left, we get <math>gx = xg</math>, which is the goal.
+| 4 || We get <math>gx = xg</math> for all <math>g,x \in G</math>. || || || Steps (2), (3) || <toggledisplay>Using step (2), we can rewrite <math>g^2x^2</math> as <math>xg^2x</math>. Combining with step (3) yields that <math>xgxg = xg^2x</math>. Canceling <math>xg</math> from the left, we get <math>gx = xg</math>, which is the goal.</toggledisplay>
 |}
@@ Line 60: / Line 59: @@
 '''To prove''': <math>G</math> is abelian.
-'''Proof''': By fact (3), we conclude that the square map must be an endomorphism of <math>G</math>. By fact (4), we conclude that therefore <math>G</math> must be abelian.
+'''Proof''': By fact (2), we conclude that the square map must be an endomorphism of <math>G</math>. By fact (3), we conclude that therefore <math>G</math> must be abelian.
 ===Difference from the corresponding statement for the square map===