Does k Divide R's Order When R's Additive Group is Cyclic?

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SUMMARY

If R is a finite ring with a cyclic additive group, it can be expressed as R = = {nr : n an integer} for some element r in R. The discussion establishes that if r^2 = kr for some integer k, then the question arises whether k must divide the order of R. The finite field of 5 elements serves as an example, where 3 generates the ring, leading to the inquiry about the divisibility of 5 by 3.

PREREQUISITES
  • Understanding of finite rings and their properties
  • Knowledge of cyclic groups and their structure
  • Familiarity with the concept of order in group theory
  • Basic comprehension of finite fields, specifically the field of 5 elements
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  • Explore the properties of finite rings and cyclic groups
  • Study the relationship between elements and their orders in group theory
  • Investigate the structure of finite fields, particularly focusing on their additive groups
  • Learn about divisibility in the context of group orders and ring elements
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Mathematicians, students of abstract algebra, and anyone studying the properties of finite rings and cyclic groups will benefit from this discussion.

tjkubo
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If R is a finite ring and its additive group is cyclic, then
R = <r> = {nr : n an integer} for some r in R.
r^2 is in R so r^2 = kr for some integer k.
Does k have to divide the order of R?
 
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the finite field of 5 elements is a ring with a cyclic additive group.

3 generates this ring, and 4= 32= (3)(3), is it the case that 3 divides 5?
 
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