Order of subgroup of an abelian group

[H12504106]

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).

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|

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.

 

[mighty2000]

First of all, let's define a subgroup of order L as a subgroup that contains exactly L elements.

Using the product formula, we have |HK||H ∩ K| = ab. Since G is an abelian group, we know that HK is also a subgroup of G. Therefore, |HK| divides |G|.

Now, let's consider the subgroup H ∩ K. By Lagrange's theorem, we know that |H ∩ K| divides |H| and |H ∩ K| divides |K|. This means that |H ∩ K| divides both a and b.

Since |HK| divides |G| and |H ∩ K| divides ab, we can say that |HK||H ∩ K| divides |G|ab.

Now, let's consider the subgroup of G that contains all possible products of elements from HK and H ∩ K. This subgroup has |HK||H ∩ K| elements, which divides |G|ab. This means that this subgroup has order L = lcm(a,b) by the definition of lcm.

Therefore, we have proven that there exists a subgroup of order L = lcm(a,b).