Prove that the torsion subgroup T of an abelian group G is a normal subgroup of G and that G/T is torsion free?

Question

Prove that the torsion subgroup T of an abelian group G is a normal subgroup of G and that G/T is torsion free?

Answer 1

First, any subgroup of an abelian group is normal. because abelian means gT=Tg



The torsion subgroup consists of all elements of finite order.



G/T is torsion free because if we assume there is an element gT which has finite order then g^nT=T implies g^n is an element of T and so g is of finite order which implies g is an element of T which inplies that our element gT=eT



ie. the only element of G/T which is of finite order is the identity.

Answer 2

Torsion Subgroup

Answer 3

Consider T to be the set of torsion (finite order) elements of G.



Then take t and s in T. Notice that if ord(t)=k and ord(s)=j then:



      (st)^(kj) = t^(kj) s^(kj) = (t^k)^j (s^j)^k = e^j e^k = e

      (t⁻¹)^k = (t^k)⁻¹ = e⁻¹ = e



meaning:



      ord(st) ≤ kj

      ord(t⁻¹) ≤ k



so st and t⁻¹ are in T. Thus T is a subgroup. Since G is abelian, T is also normal.











Furthermore, assume gT is of finite order in G/T. Then (gT)^m = eT for some m.



This means in particular that:



      g^m T = (gT)^m = eT



which implies that g^m is in T, so g^m is of finite order. This implies that (g^m)^n = e for some value of n. Thus ord(g) ≤ mn, implying that g is in T, and thus gT = eT. This allows you to conclude that the only element gT of finite order in G/T is eT, the identity coset. Thus G/T is torsion free.



QED