Question

The order of an element a of an abelian group G is the smallest positive int such that ma = 0, if such exists; otherwise, the order of a is said to be infinite. The order of a thus equals the order of the subgroup generated by a. (a) Show the elements of finite order in G form a subgroup of G, called its torsion subgroup. (b) Show that if G is free abelian, it has no elements of finite order. (c) Show the additive group of rationals has no elements of finite order, but is not free abelian.

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Answer for (a)\textsf{\color{#c34632}Answer for (a)} We have to show that the elements of finite order in GG form a subgroup of GG.

It is enough to show that the subgroup generated by finite order elements in GG is closed under sums and inverses.

Let T(G)T(G) be the subgroup generated by some finite order element in GG.

Let a,bT(G)a,b \in T(G) with orders m,nm,n respectively. Then we have ma=0=nbma=0=nb. Note that

mn(a+b)=n(ma)+m(nb)=0mn(a+b) = n(ma) + m(nb) = 0

It follows that the order of the element a+ba+b is less than or equal to mnmn and so it is finite. Hence a+bT(G)a+b \in T(G).

Now note that

m(a)=(ma)=0m(-a) = -(ma) = 0

It follows that aT(G)-a \in T(G).

Therefore, T(G)T(G) is a subgroup of GG.

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