Difference between revisions of "(n-1)th power map is endomorphism taking values in the center implies nth power map is endomorphism"

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(Created page with "==Statement== Suppose <math>G</math> is a group and <math>n</math> is an integer such that the map <math>x \mapsto x^{n-1}</math> is an endomorphism of <math>G</math> an...")
 
 
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==Statement==
 
==Statement==
  
Suppose <math>G</math> is a [[group]] and <math>n</math> is an integer such that the map <math>x \mapsto x^{n-1}</math> is an [[endomorphism]] of <math>G</math> and <math>x^{n-1}</math> is in the [[center]] of <math>G</math> for all <math>x</math> in <math>G</math>. Then, the map <math>x \mapsto x^n</math> is also an [[endomorphism]] of <math>G</math>.
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Suppose <math>G</math> is a [[group]] and <math>n</math> is an integer such that the [[fact about::universal power map|power map]] <math>x \mapsto x^{n-1}</math> is an [[endomorphism]] of <math>G</math> and <math>x^{n-1}</math> is in the [[center]] of <math>G</math> for all <math>x</math> in <math>G</math>. Then, the map <math>x \mapsto x^n</math> is also an [[endomorphism]] of <math>G</math>.
  
 
==Related facts==
 
==Related facts==
  
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===Applications===
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* [[Frattini-in-center odd-order p-group implies (mp plus 1)-power map is automorphism]]
 
===Converse===
 
===Converse===
  
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'''To prove''': The map <math>x \mapsto x^n</math> is an endomorphism of <math>G</math>, i.e., <math>(gh)^n = g^nh^n</math> for all <math>g,h \in G</math>.
 
'''To prove''': The map <math>x \mapsto x^n</math> is an endomorphism of <math>G</math>, i.e., <math>(gh)^n = g^nh^n</math> for all <math>g,h \in G</math>.
  
'''Proof''': We have the following for all <math>g,h
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'''Proof''': We have the following for all <math>g,h \in G</math>.
  
 
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Latest revision as of 20:09, 25 February 2011

Statement

Suppose G is a group and n is an integer such that the power map x \mapsto x^{n-1} is an endomorphism of G and x^{n-1} is in the center of G for all x in G. Then, the map x \mapsto x^n is also an endomorphism of G.

Related facts

Applications

Converse

The precise converse is not true, but a partial converse is: nth power map is surjective endomorphism implies (n-1)th power map is endomorphism taking values in the center

Proof

Given: A group G, an integer n such that the map x \mapsto x^{n-1} is an endomorphism taking values in the center.

To prove: The map x \mapsto x^n is an endomorphism of G, i.e., (gh)^n = g^nh^n for all g,h \in G.

Proof: We have the following for all g,h \in G.

Step no. Assertion/construction Given data used Previous steps used
1 (gh)^n = (gh)(gh)^{n-1}
2 (gh)^{n-1} = g^{n-1}h^{n-1} x \mapsto x^{n-1} is an endomorphism
3 (gh)^n = ghg^{n-1}h^{n-1} Step (2) plugged into Step (1)
4 (gh)^n = gg^{n-1}hh^{n-1} g^{n-1} is in the center of G, so commutes with h Step (3)
5 (gh)^n = g^nh^n Step (4) (simplified)