Cube map is surjective endomorphism implies abelian: Difference between revisions

Latest revision as of 23:37, 4 June 2012

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, or more generally a surjective endomorphism, 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, or more generally, a surjective endomorphism. Then, $G$ is an abelian group.

Related facts

Similar facts for cube map

Similar facts for other values

Weaker facts for other values

Opposite facts

The statement breaks down if we remove the assumption of surjectivity:

Frattini-in-center odd-order p-group implies p-power map is endomorphism: In particular, for $p=3$ , we can obtain non-abelian groups of order $p^{3}=27$ , such as prime-cube order group:U(3,3) and semidirect product of Z9 and Z3, where the cube map is an endomorphism. In the former case, the cube map is the trivial endomorphism. In the latter, it is a nontrivial endomorphism.

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 surjective endomorphism 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 a surjective endomorphism of $G$ .

To prove: $G$ is abelian.

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	$g^{2}h^{3}=h^{3}g^{2}$ for all $g,h\in G$ , i.e., every square commutes with every cube	Fact (1)	Cube map is an endomorphism	--	[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 surjective	Step (1)	[SHOW MORE] Since the cube map is surjective, 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 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.

Hands-off proof (using facts (2) and (3))

Given: A group $G$ such that the map sending $x$ to $x^{3}$ is a surjective endomorphism 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 can in fact prove 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 1: / Line 1: @@
 {{elementary nonbasic fact}}
+[[difficulty level::3| ]]
 ==Statement==
 ===Verbal statement===
-If the [[cube map]] on a group is an [[automorphism]], then the group is an [[Abelian group]].
+If the [[fact about::cube map]] on a group is an [[automorphism]], or more generally a [[surjective endomorphism]], then the group is an [[proves property satisfaction of::abelian group]].
 ===Statement with symbols===
-Let <math>G</math> be a group such that the map <math>\sigma:G \to G</math> defined by <math>\sigma(x) = x^3</math> is an automorphism. Then, <math>G</math> is an Abelian group.
+Let <math>G</math> be a group such that the map <math>\sigma:G \to G</math> defined by <math>\sigma(x) = x^3</math> is an automorphism, or more generally, a surjective endomorphism. Then, <math>G</math> is an abelian group.
 ==Related facts==
-===Applications===
+===Similar facts for cube map===
 * [[Cube map is endomorphism iff abelian (if order is not a multiple of 3)]]
+* [[Cube map is endomorphism implies class four for 2-divisible group]]
-===Stronger facts for other values===
+===Similar facts for other values===
 * [[Inverse map is automorphism iff abelian]]
 * [[Square map is endomorphism iff abelian]]
@@ Line 23: / Line 23: @@
 * [[nth power map is endomorphism iff abelian (if order is relatively prime to n(n-1))]]
+* [[nth power map is automorphism implies (n-1)th power map is endomorphism taking values in the center]]
+* [[nth power map is endomorphism implies every nth power and (n-1)th power commute]]
+* [[(n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism]]
+===Opposite facts===
+The statement breaks down if we remove the assumption of surjectivity:
+[[Frattini-in-center odd-order p-group implies p-power map is endomorphism]]: In particular, for <math>p = 3</math>, we can obtain non-abelian groups of order <math>p^3 = 27</math>, such as [[prime-cube order group:U(3,3)]] and [[semidirect product of Z9 and Z3]], where the cube map is an endomorphism. In the former case, the cube map is the trivial endomorphism. In the latter, it is a nontrivial endomorphism.
 ==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 surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center]]
+# [[uses::Square map is endomorphism iff abelian]]
 ==Proof==
-===Abelian implies cube map is endomorphism===
+===Hands-on proof using fact (1)===
-This is a direct consequence of fact (1).
-===Cube map is endomorphism implies abelian===
+{{tabular proof format}}
-'''Given''': A group <math>G</math> such that the map sending <math>x</math> to <math>x^3</math> is an automorphism of <math>G</math>.
+'''Given''': A group <math>G</math> such that the map sending <math>x</math> to <math>x^3</math> is a surjective endomorphism of <math>G</math>.
 '''To prove''': <math>G</math> is abelian.
-'''Proof''':
+'''Proof''': We denote by <math>c_g</math> the map <math>x \mapsto gxg^{-1}</math>.
 {| class="sortable" border="1"
-! Step no. !! Assertion !! Given data used !! Facts used !! Previous steps used !! Explanation
+! Step no. !! Assertion !! Facts used !! Given data 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 || Fact (1) ||Cube map is an endomorphism ||  -- || <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 surjective  || Step (1) || <toggledisplay>Since the cube map is surjective, 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 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>
 |}
+===Hands-off proof (using facts (2) and (3))===
+'''Given''': A group <math>G</math> such that the map sending <math>x</math> to <math>x^3</math> is a surjective endomorphism of <math>G</math>.
+'''To prove''': <math>G</math> is 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===
-In the case of the square map, we had in fact proved something much stronger:
+In the case of the square map, we can in fact prove something much stronger:
 <math>(xy)^2 = x^2y^2 \iff xy = yx</math>