Introduction:
In this problem, we are given a finite abelian group G and two distinct subgroups h1 and h2 of G, both having index 3. We need to find the index of the intersection of h1 and h2 in G.

Solution:

Step 1: Understand the problem
- We are given a finite abelian group G.
- We have two distinct subgroups h1 and h2 of G.
- Both h1 and h2 have index 3 in G.
- We need to find the index of the intersection of h1 and h2 in G.

Step 2: Recall the definition of index of a subgroup
- The index of a subgroup H in a group G is the number of distinct left cosets of H in G.
- It is denoted as [G:H].

Step 3: Understand the concept of cosets
- Cosets are a way to partition a group into disjoint sets.
- Let H be a subgroup of G. The left coset of H containing an element g in G is the set gH = {gh : h ∈ H}.
- The right coset of H containing an element g in G is the set Hg = {hg : h ∈ H}.
- Left and right cosets are not necessarily the same, but they have the same number of elements.

Step 4: Use Lagrange's Theorem
- Lagrange's theorem states that for a finite group G and a subgroup H of G, the order of H divides the order of G.
- In our case, since G is a finite abelian group and h1 and h2 are subgroups of G, the order of h1 and h2 divides the order of G.
- Let |G| = n, |h1| = m1, and |h2| = m2. Then, n is divisible by both m1 and m2.

Step 5: Apply the concept of indices
- Since h1 and h2 are distinct subgroups of G and both have index 3, each of them has 3 distinct left cosets in G.
- Therefore, the number of left cosets of h1 is 3 and the number of left cosets of h2 is also 3.
- The intersection of h1 and h2 is a subgroup of both h1 and h2.
- Thus, the number of left cosets of the intersection of h1 and h2 is less than or equal to the number of left cosets of h1 and h2.

Step 6: Determine the index of the intersection of h1 and h2 in G
- Let the intersection of h1 and h2 be denoted as h1 ∩ h2.
- The number of left cosets of h1 ∩ h2 in G is denoted as [G:h1 ∩ h2].
- Since the number of left cosets of h1 and h2 is 3 each, [G:h1] = 3 and [G:h2] = 3.
- By Lagrange's theorem, |G| is divisible by both |h1 ∩ h2| and |h1| = m1.
- Therefore, |h1 ∩ h2