Showing the unit group is cyclic

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SUMMARY

The unit group Up of the integers modulo a prime p, denoted as Z/Zp, is cyclic and isomorphic to Z/Z(p − 1). The unit group consists of all invertible elements in Zp, specifically the integers from 1 to p-1, as 0 is not invertible. Therefore, Up contains exactly p-1 elements, confirming its cyclic nature generated by a single element.

PREREQUISITES
  • Understanding of group theory concepts, specifically cyclic groups.
  • Familiarity with modular arithmetic and the structure of Z/Zp.
  • Knowledge of invertible elements in a modular system.
  • Basic understanding of isomorphism in algebraic structures.
NEXT STEPS
  • Study the properties of cyclic groups in group theory.
  • Learn about the structure of the multiplicative group of integers modulo n.
  • Explore the concept of generators in cyclic groups.
  • Investigate the relationship between prime numbers and the unit group in modular arithmetic.
USEFUL FOR

Students of abstract algebra, mathematicians interested in group theory, and anyone studying modular arithmetic and its applications.

Locoism
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Homework Statement


Let p be a positive prime and let Up be the unit group of Z/Zp. Show that Up is
cyclic and thus Up \cong Z/Z(p − 1).


The Attempt at a Solution


What do they mean by the unit group? Is that just the identity? Is it the group [p]? I'm lost without starting the question...
 
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The unit group is just the group of all invertible elements (with multiplication as operation). So x\in U_p if and only if x is invertible in \mathbb{Z}_p. You have to show that it is cyclic (generated by 1 element).
 
OK thank you!
 
But now that I think of it, wouldn't Up be {e,...,p}, the entire equivalence class of Zp, since every element can be multiplied by another (its respective inverse) to get the identity?
 
Locoism said:
But now that I think of it, wouldn't Up be {e,...,p}, the entire equivalence class of Zp, since every element can be multiplied by another (its respective inverse) to get the identity?

0 doesn't have an inverse...
 
oh right, so then U would have p-1 elements. Got it!
 

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