If p is prime, prove the group (Zp*,x) has exactly one element of order 2.

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SUMMARY

The group (Zp*, x) has exactly one element of order 2 when p is prime. This is established by analyzing the multiplicative group of integers modulo p, specifically the set {1, 2, ..., p-1}. The element of order 2 can be identified by solving the polynomial equation X^2 - 1 in the context of the field \mathbb{Z}_p. Since \mathbb{Z}_p^* is cyclic, it follows that the only solution to this equation that lies within the group is the element 1, confirming its unique order of 2.

PREREQUISITES
  • Understanding of group theory, specifically cyclic groups.
  • Familiarity with modular arithmetic and the structure of \mathbb{Z}_p.
  • Knowledge of polynomial equations and their roots in finite fields.
  • Basic concepts of order of elements in group theory.
NEXT STEPS
  • Study the properties of cyclic groups in group theory.
  • Learn about the structure of \mathbb{Z}_p and its multiplicative group.
  • Explore polynomial equations in finite fields, particularly X^2 - 1.
  • Investigate the implications of element orders in group theory.
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This discussion is beneficial for mathematicians, particularly those studying abstract algebra, as well as students and educators looking to deepen their understanding of group theory and finite fields.

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Hi. I need to: prove the group (Zp*,x) has exactly one element of order 2. Here, p is prime and (Zp*,x) is the set {1, 2,..., p-1} under multiplication modulo p. Any help would be much appreciated!
 
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I think the easier half is: which element has order two?

It might help if you write down the condition of what it means to have order 2
 
Something that might help you is knowing that \mathbb{Z}_p^* is cyclic. Do you know this already??
 
If you do not know the previous. Then you might want to think about the polynomial X^2-1 in \mathbb{Z}_p[X]. What do you know about its roots??
 

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