Question

Let \( G \) be an abelian group of order 35. Let \( m \) denote the number of elements of order 5 in \( G \), and let \( n \) denote the number of elements of order 7 in \( G \). Then, the value of \( m + n \) is equal to ...............

Hint:

For a finite abelian group of order \( pq \) (where \( p \) and \( q \) are primes), the group can be written as the direct product of \( \mathbb{Z}_p \) and \( \mathbb{Z}_q \), and the number of elements of each order is \( p-1 \) for \( \mathbb{Z}_p \) and \( q-1 \) for \( \mathbb{Z}_q \).

Answer 1

Step 1: Analyze the structure of the group.
By the Fundamental Theorem of Finite Abelian Groups, the group \( G \) of order 35 can be expressed as a direct product of cyclic groups. Since 35 is the product of two primes, 5 and 7, we know that: \[ |G| = 35 = 5 \times 7. \] Therefore, \( G \) is isomorphic to \( \mathbb{Z}_5 \times \mathbb{Z}_7 \), where \( \mathbb{Z}_5 \) is the cyclic group of order 5 and \( \mathbb{Z}_7 \) is the cyclic group of order 7. Step 2: Determine the number of elements of each order.
In \( \mathbb{Z}_5 \), the number of elements of order 5 is \( 4 \), as there are 4 elements in a cyclic group of order 5 that are generators. Similarly, in \( \mathbb{Z}_7 \), the number of elements of order 7 is \( 6 \), as there are 6 elements in a cyclic group of order 7 that are generators. Since \( G \) is the direct product of \( \mathbb{Z}_5 \) and \( \mathbb{Z}_7 \), the number of elements of order 5 in \( G \) is 4, and the number of elements of order 7 in \( G \) is 6. Step 3: Calculate \( m + n \).
Thus, we have: \[ m = 4, \quad n = 6, \quad m + n = 4 + 6 = 10. \] Final Answer: \[ \boxed{10}. \]