Suppose that G is an abelian group that satisfies the follow

Suppose that G is an abelian group that satisfies the following properties: for all m epsilon Z, G {m} is either equal to G or is of finite order; for some m epsilon Z, {0_G,} G{m} G. Show that G(m) is finite for all non-zero m epsilon Z.

Solution

The solution uses the following two easily verifyable facts:

G{k}+G{m}=G{lcm(k,m)}

G{k}G{m}=G{gcd(k,m)}

Indeed, if d:=gcd(k,m), then sk+tm=d for some integers s,t, and so if xG{k}G{t} then dx=skx+tmx=0. Thus G{d}G{k}G{t}. The other direction is trivial, and the other equality is proven in a similar fashion.

If G is finite there is nothing to prove, so assume G is infinite. Let n be the exponent of G. If n=0 then we again have nothing to prove, so assume n>1 (n=1 is impossible, since it would imply that G is finite).

Let p₁e¹p_re^r be the prime decomposition of n. We have G=G{n}=G{p_ieⁱ}++G{p_ieⁱ}, and therefore at least one of the subgroups G{p_ieⁱ} must be infinite. Hence, by property (i) we have m=p^e for some prime p and integer e. From property (ii) we know there is some integer 0<f<e such that G{p^f} is finite, and hence e2. In particular, Since G{p^e-1} is a non-trivial subgroup of G, and so by property (i), it must be finite. We note that all of the elements in G{p^e}G{p^e-1} have order pe, so there are infinitely many such elements. Since every x G{p^e}G{p^e-1} maps into some non-zero element in G{p^e-1} when multiplied by p, we can find an infinite sequence {xi}i of distinct elements such that px_i=px_j for all i,j (via the pigeonhole principle). We examine the quotient group H:=G{p^e}G{p^e-1}. This group contains infinitely many elements of order p, and among them are the elements [x_i]. Therefore we can find two indices i,j such that [x_i][x_j]=0_H. However, [x_i][x_i]=0, which implies that [x_i][x_i][x_j] - a contradiction.