Show that if G is a cyclic group of order n and dn d divides

Show that if G is a cyclic group of order n and d|n (d divides n) then G has a subgroup of order d. Please be concise, and if you are doing a contradiction you must say what is contradicted. Thank you.

Solution

G is a cyclic group of order n.

d is a divisor of n=|G|.

Consider H={xG:xd=1} .Then H is a subgroup of G and H contains all elements of G that have order d (among others).

If K is a subgroup of G of order d, then K is cyclic, generated by an element of order d.

Hence, KH.

On the other hand, xH iff x=g^k with 0k<n and g^(kd)=1 , where g is a generator of G. Hence, kd=nt and so k=(n/d)t.      The restriction 0k<n implies 0t<d, and so H has exactly d elements. Therefore, K=H.