In a group, show that ord(x) = ord(yxy^-1)

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SUMMARY

The discussion centers on proving that the order of an element x in a group G is equal to the order of the conjugate element yxy-1. Participants clarify that the expression (yxy-1)n simplifies to yxny-1, not xn. This distinction is crucial for understanding the relationship between the orders of x and its conjugate. The conversation emphasizes the importance of correctly applying group theory definitions and operations.

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  • Understanding of group theory concepts, specifically group elements and their orders.
  • Familiarity with conjugation in groups and its implications.
  • Basic knowledge of algebraic manipulation involving exponents.
  • Experience with mathematical proofs and definitions in abstract algebra.
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  • Study the definition of the order of an element in group theory.
  • Learn about conjugate elements and their properties in group theory.
  • Explore examples of groups where the order of elements is calculated, such as symmetric groups.
  • Investigate the implications of conjugation on the structure of groups, including normal subgroups.
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Students of abstract algebra, mathematicians focusing on group theory, and anyone interested in understanding the properties of group elements and their orders.

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Homework Statement



If x and y are elements of a group G show that ord(x)=ord(yxy^-1)

Homework Equations





The Attempt at a Solution



Some hints to how to do this would be great.
 
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What's (yxy^(-1))^n?
 


y^n.x^n.y^(-n) ?
 


Fairy111 said:
y^n.x^n.y^(-n) ?

You aren't going to get anywhere taking wild guesses. Try simplifying (yxy^(-1))^2=(yxy^(-1))(yxy^(-1)).
 
Last edited:


that would be x^2
so (yxy^(-1))^n would be x^n
 


Yes, that is correct. Now, what does that have to do with the definition of "order"?

Oops! Dick is correct. I forgot about the first and last y and y-1!

But, "what does that have to do with the definition of 'order'" still stands.
 
Last edited by a moderator:


Fairy111 said:
that would be x^2
so (yxy^(-1))^n would be x^n

No. (yxy^(-1))^n would be yx^ny^(-1), NOT x^n. Be careful!
 

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