Question

2.6.22. Let G be a cyclic group of order n. Let m ≤ n be a positive integer. How many subgroups of order m does G have? Prove your assertion.

Answer 1

Since G is cyclic there exists an element a in G such that G=<a> , i.e. a is the generator of G.

If m does not divide n then there is no subgroup of order m as order of any subgroup of G must divide the order of G by Lagrange’s theorem for groups.

If m divides n, we claim that <a^n/m> is the only subgroup of order m.

Since |a^k| = n/gcd(n,k) for every divisor k of n, we have that |a^n/m| = n/gcd(n,n/m) = n/(n/m) = m. Hence <a^n/m> is a subgroup of order m.

Now if for some divisor k of n, <a^k> is a subgroup of order m, then m = n/gcd(n,k) = n/k,

which implies k = n/m. Hence <a^k> = <a^n/m>, which proves the uniqueness part.