Since G has a cyclic subgroup of order k, we can suppose this group is H=<b>, where b is in G and the order of b is k. Since |G|=2k, we can find some a in G, such that a^2=e or the order of a is 2. We know, [G:H]=2 and thus H is a normal subgroup of G. Thus G=<a,b> and each element in G has the form b^n or ab^n, where 0<=n<k.
To count the number of ...
The number of subgroups that are of odd order is found. The solution is detailed and well presented.