Number of nth roots is a multiple of n
This article states a result of the form that one natural number divides another.
View other divisor relations |View congruence conditions
is a multiple of .
Note that since the identity element is itself in this set, the size of the set is at least .
- Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup
- At most n elements of order dividing n implies every finite subgroup is cyclic
- Number of nth roots of a subgroup is divisible by order of subgroup
The proof follows from fact (1); in fact, the given statement is a special case of fact (1).