Fermat's Little Theorem: Let G be a finite group of order n, and let H be a subgroup of G. If H is cyclic of order k, then the order of the group G is divis...
Fermat's Little Theorem:
Let G be a finite group of order n, and let H be a subgroup of G. If H is cyclic of order k, then the order of the group G is divisible by k.
Proof:
Assume that G is a finite group of order n. Since G is cyclic, it can be expressed as a direct product of cyclic subgroups of order k, where k divides n.
Now, let H be a cyclic subgroup of order k. Since H is a subgroup of G, it is also cyclic. Therefore, the order of H is divisible by k.
But the order of H is also equal to the order of G/H. Therefore, we have n = k.
Implications:
Fermat's Little Theorem has several important implications:
It implies that the order of any group of order n can be expressed as a multiple of a few prime numbers.
It provides a concrete way to determine the order of a group of order n that is cyclic.
It gives a necessary and sufficient condition for a group to be cyclic.
Examples:
If G = Z_12 and H = {0, 3, 6}, then G is cyclic of order 12, and H is a cyclic subgroup of order 3. Therefore, the order of G is divisible by 3, which is a multiple of 12.
If G = Z_120 and H = {0, 12, 24}, then G is cyclic of order 120, and H is a cyclic subgroup of order 6. Therefore, the order of G is divisible by 6, which is not a multiple of 120