Have you encountered the fundamental theorem of finitely-generated abelian groups yet?
Let H and K be subgroup of an abelian group G (not neccessarily finite). Suppose order of H and K are a and b respectively. Prove that there exists a subgroup with order L, where L = lcm(a,b).
I tried to use the Lagrange theorem but only manage to prove it for the case when g is finite. I tried to use the product formula: |HK|/|H| = |K|/|H intersect K|. Sub a and b in, i obtain: |HK||H intersect K| = ab.
Hence, it suffices to show that |H intersect K| = gcd (a,b) since lcm(a,b)*gcd(a,b) = ab. But how to show that? or is there anothe method to solve this?
Thank You.