(New page: Category:MA453Spring2009Walther Prove or disprove that U(20) and U(24) are isomorphic. By listing out the elements of U(20) = {1,3,7,9,11,13,17,19} and then their corresponding ord...)
 
Line 6: Line 6:
  
 
-Karen Morley
 
-Karen Morley
 +
 +
[[Category:MA453Spring2009Walther]]
 +
 +
That's an interesting way to explain that.  It looked at it a different way. In U(20), the elements whose square is 1 are 1,9,11,19. In U(24), the elements whose square is 1 are 1,5,7,11,13,17,19,23. It follows that no isomorphism can exist because it would have to math 1-1 a set of 8 elements to one with 4. I like your explanation better though.  It shows it in a simpler, easier to understand way. 
 +
 +
-Linley Johnson
 +
 +
--[[User:Johns121|Johns121]] 16:33, 10 February 2009 (UTC)

Revision as of 12:33, 10 February 2009


Prove or disprove that U(20) and U(24) are isomorphic.

By listing out the elements of U(20) = {1,3,7,9,11,13,17,19} and then their corresponding orders which are 1,4,4,2,2,4,4,2 and then the elements of U(24) = {1,5,7,11,13,17,19,23} and their corresponding orders which are 1,2,2,2,2,2,2,2 we can see that since the largest order of any element in these groups does not agree, then U(20) and U(24) cannot be isomorphic.

-Karen Morley

That's an interesting way to explain that. It looked at it a different way. In U(20), the elements whose square is 1 are 1,9,11,19. In U(24), the elements whose square is 1 are 1,5,7,11,13,17,19,23. It follows that no isomorphism can exist because it would have to math 1-1 a set of 8 elements to one with 4. I like your explanation better though. It shows it in a simpler, easier to understand way.

-Linley Johnson

--Johns121 16:33, 10 February 2009 (UTC)

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett