Question

Let \(G\) be a finite Abelian group. If the subgroups \(H\) and \(K\) are of index 3 each in \(G\), then what is the index of the subgroup \(H\cap K\) in \(G\)?

• A. 9
• B. 6
• C. 5
• D. 3

Hint:

Embed G divided by (H intersect K) into (G/H) times (G/K), a group of order 9, and use H not equal K.

Answer 1

The Correct Option is A

Step 1: Set up the natural map into a product of quotients.
Since \(G\) is abelian, every subgroup is normal, so \(G/H\) and \(G/K\) are groups of order 3 each. Define \(\varphi:G\to (G/H)\times(G/K)\) by \(\varphi(g)=(gH,gK)\). This is a homomorphism with kernel exactly \(H\cap K\).

Step 2: Bound the index of \(H\cap K\).
By the first isomorphism theorem, \(G/(H\cap K)\) is isomorphic to the image of \(\varphi\), a subgroup of \((G/H)\times(G/K)\cong \mathbb{Z}_3\times\mathbb{Z}_3\), of order 9. So \([G:H\cap K]\) divides 9. Also, since \(H\cap K\subseteq H\), \([G:H\cap K]=[G:H]\cdot[H:H\cap K]\) is a multiple of \([G:H]=3\). So \([G:H\cap K]\in\{3,9\}\).

Step 3: Decide between 3 and 9 using \(H\ne K\).
The image of \(\varphi\) is a subgroup of \(\mathbb{Z}_3\times\mathbb{Z}_3\) of order 3 or 9. A subgroup of order 3 is cyclic, generated by some \((a,b)\ne(0,0)\), and its existence as the image would force the kernels of the two quotient maps \(G\to G/H\) and \(G\to G/K\) to coincide, i.e. \(H=K\). Since \(H\ne K\) is given, the image cannot have order 3, so it must be the full group of order 9.

Step 4: Conclude.
Hence \([G:H\cap K]=9\). \[\boxed{[G:H\cap K]=9}\]