Show that there are y,z such that y,z commute and their order is m and n

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SUMMARY

The discussion centers on a mathematical exercise involving a group element $x \in G$ with order $mn$, where $m$ and $n$ are coprime. It is established that there exist elements $y$ and $z$ such that $x = yz$, with $y$ having order $m$ and $z$ having order $n$, and that $y$ and $z$ commute. Key hints provided include considering the powers of $x$ and utilizing the integers $s$ and $t$ that satisfy the equation $1 = sm + tn$ due to the coprimality of $m$ and $n$.

PREREQUISITES
  • Understanding of group theory concepts, particularly group orders
  • Familiarity with the properties of coprime integers
  • Knowledge of the structure of abelian groups
  • Basic experience with mathematical proofs and hints in problem-solving
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  • Study the structure of finite abelian groups and their decompositions
  • Learn about the use of the Chinese Remainder Theorem in group theory
  • Explore the implications of the order of elements in groups
  • Investigate the relationship between group elements and their powers
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Mathematics students, particularly those studying abstract algebra, group theorists, and anyone interested in the properties of group elements and their orders.

mathmari
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Hey! :o

I got stuck at the following exercise:

If $x \in G$ has order $mn$ with $ (m,n)=1 $, show that there are $y,z$ with $ x=yz $ such that $y$,$z$ commute and they have order $m$ and $n$ respectively.

Could you give me some hints?? (Wondering)
 
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mathmari said:
Hey! :o

I got stuck at the following exercise:

If $x \in G$ has order $mn$ with $ (m,n)=1 $, show that there are $y,z$ with $ x=yz $ such that $y$,$z$ commute and they have order $m$ and $n$ respectively.

Could you give me some hints?? (Wondering)
Hint: think about powers of $x$.
 
Another hint: Since $(m,n) = 1$, there are integers $s$ and $t$ such that $1 = sm + tn$.
 

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