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**1. Homework Statement**

Suppose H and K are subgroups of an abelian group G (not neccessarily finite). Let the order of H and K be a and b respectively. Prove that there exists a subgroup of order L, where L = lcm(a,b).

**2. Homework Equations**

Product Formula: |HK|/|H| = |K|/|H intersect K|

lcm(a,b)*gcd(a,b)=ab

Lagrange theorem

A finite abelian group G has a subgroup of order d for every divisor d of |G|

**3. The Attempt at a Solution**

Using the product formula, i obtain: |HK||H intersect K| = ab

Then using largrange theorem, i have |H intersect K| divides |H| and |H intersect K| divides |K|. Hence, |H intersect K| divides ab. However, i am unable to proceed further. Or is there another method to solve this problem

Thank You.