Question

The number of subgroups of \( \mathbb{Z}_7 \times \mathbb{Z}_7 \) of order 7 is ............

Hint:

The number of subgroups of order \( p \) in an abelian group \( \mathbb{Z}_p \times \mathbb{Z}_p \) is the number of 1-dimensional subspaces in \( \mathbb{F}_p^2 \), which is \( p \).

Answer 1

Correct Answer: 7.9 - 8.1

Step 1: Structure of the group.
The group \( \mathbb{Z}_7 \times \mathbb{Z}_7 \) is an abelian group of order \( 7^2 = 49 \), and it is isomorphic to the additive group of the vector space \( \mathbb{F}_7^2 \), where \( \mathbb{F}_7 \) is the finite field with 7 elements.
Step 2: Subgroups of order 7.
The number of subgroups of order 7 in an abelian group is equal to the number of 1-dimensional subspaces of \( \mathbb{F}_7^2 \), which is the number of lines through the origin in \( \mathbb{F}_7^2 \). Since each nonzero vector in \( \mathbb{F}_7^2 \) generates a unique 1-dimensional subspace, and there are 6 nonzero vectors in \( \mathbb{F}_7^2 \), the number of subgroups of order 7 is 7.
Step 3: Conclusion.
Thus, the number of subgroups of \( \mathbb{Z}_7 \times \mathbb{Z}_7 \) of order 7 is \( \boxed{7} \).