Understanding Additive Factor Group Q/Z

[DanielThrice]

If we're considering the additive factor group Q/Z, can we show that Q/Z is the torsion subgroup of R/Z? I'm lost on these.

 

[DanielThrice]

Alright, so I showed this:


We have that if r+Z is an element of R/Z, and the order of r + Z = n, nr is an element of Z implies that r is an element of Q.

But what about this: Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unityin C. (That is, U∗ = {z ∈ C|zⁿ=1 for some n ∈ Z+}.)

 

[Deveno]

Alright, so I showed this:


We have that if r+Z is an element of R/Z, and the order of r + Z = n, nr is an element of Z implies that r is an element of Q.

But what about this: Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unityin C. (That is, U∗ = {z ∈ C|zⁿ=1 for some n ∈ Z+}.)

what can you say about the map f:Q/Z-->U* given by:

f(q + Z) = e^2πiq

 

[lavinia]

Alright, so I showed this:


We have that if r+Z is an element of R/Z, and the order of r + Z = n, nr is an element of Z implies that r is an element of Q.

But what about this: Show that Q/Z is isomorphic to the multiplicative group U∗ consisting of all roots of unityin C. (That is, U∗ = {z ∈ C|zⁿ=1 for some n ∈ Z+}.)

you are just restating the same question in different language.

 

[Outlined]

If we're considering the additive factor group Q/Z, can we show that Q/Z is the torsion subgroup of R/Z? I'm lost on these.

It is quite easy, let T be the torsion subgroup. It is easy to proove Q/Z \subseteq T.

If we have Q/Z \subset T then there is a real number r \notin Q such that r + Z has finite order. This implies r \in Q. Contradiction so Q/Z = T.