Question

Let \( F \) be a finite field and \( F^{\times} \) be the group of all nonzero elements of \( F \) under multiplication. If \( F^{\times} \) has a subgroup of order 17, then the smallest possible order of the field \( F \) is \(\underline{\hspace{1cm}}\) .

Hint:

The order of the multiplicative group of a finite field is one less than the size of the field, and it must be divisible by the order of any subgroup.

Answer 1

Correct Answer: 103

The multiplicative group \( F^{\times} \) of a finite field \( F \) is cyclic and has order \( |F| - 1 \), where \( |F| \) is the order of the field. Since \( F^{\times} \) has a subgroup of order 17, the order of \( F^{\times} \) must be a multiple of 17. Therefore, we have: \[ |F^{\times}| = |F| - 1 \geq 17. \] This implies that the smallest possible value for \( |F| \) is \( 17 + 1 = 18 \). Since the order of a finite field is always a power of a prime, the smallest possible field size is a power of 2. Thus, the smallest possible field size is \( 2^5 = 32 \), as 32 is the smallest power of 2 greater than or equal to 18. Therefore, the smallest possible order of the field \( F \) is \( \boxed{103} \).