Show that every group of order 100 has a subgroup of order 50. Show also that the number of subgroups of order 50 is either 1 or 3.
For the first part I did the following:
As $|G|=100=2^2 5^2$ we can deduce from Sylow's Theorems that $G$ has a normal subgroup $N$ of order 25 (normal because there is only one). There is also a subgroup $H$ of order 2 (as 2 divides 100). Since $N$ is normal there holds $NH=HN$. So $NH$ is in fact a subgroup, which has order 50.
But I am puzzled as to why there are either 1 or 3 such subgroups.