Elements of Subgroups in Additive Group Q/Z: G(2) and G(P)

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SUMMARY

The subgroup G(2) of the additive group Q/Z consists of elements whose order is a power of 2, specifically multiples of 2 in this context. Similarly, G(P) for any positive prime P includes elements whose order is a power of P. The discussion clarifies that elements of Q/Z do not have infinite order; rather, they have finite order, which can be understood through the group operation and the definition of order in this additive group. Understanding these concepts is crucial for grasping the structure of subgroups within Q/Z.

PREREQUISITES
  • Understanding of additive groups, specifically Q/Z.
  • Knowledge of group theory concepts, including order of elements.
  • Familiarity with subgroup definitions and properties.
  • Basic mathematical notation and operations involving sets and cosets.
NEXT STEPS
  • Study the properties of additive groups, focusing on Q/Z and its subgroups.
  • Learn about the concept of element order in group theory.
  • Explore the structure of G(n) for various integers n, particularly primes.
  • Investigate the implications of finite versus infinite order in groups.
USEFUL FOR

Mathematicians, particularly those specializing in abstract algebra, students studying group theory, and anyone interested in the properties of additive groups and their substructures.

Thorn
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The question:

If G is the additive group Q/Z, what are the elements of the subgroup G(2)? Of G(P) for any positive prime P?

Where G(n)={a e G| |a| = n^(k) for some k is greater than or equal to 0}...That is the set of all a in G, s.t. the order of a is some power of n. (But since it is the additive group, I suppose it would just a be a multiple of n)

How do I even begin with this? Aren't the elements of Q/Z sets? The collections of right cosets? and don't they have infinite order?...
 
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Thorn said:
(But since it is the additive group, I suppose it would just a be a multiple of n)
I don't see why you would suppose that. The additive notation has no bearing on this matter.

Aren't the elements of Q/Z sets? The collections of right cosets?
Sure.

and don't they have infinite order?...
No. Take an element r+Z in Q/Z. What does it mean for n(r+Z) to be the zero element of Q/Z? If you can answer this correctly, then you can easily deduce that all the elements of Q/Z have finite order.

How do I even begin with this?
My suggestion is to actually think about what Q/Z is, as a group. Make sure you understand how the group operation works, and what order means in this setting.
 

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