TITLE: Group of order 100: 1 or 3 subgroups of order 50?
QUESTION [5 upvotes]: 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.

REPLY [8 votes]: By Correspondence Theorem, the subgroups of $G$ containing $N$ are in one-to-one correspondence with the subgroups of $G/N$, with the map given by $H\mapsto H/N$. Since $|G/N|=4$, then there exist either $1$ or $3$ subgroups of $G/N$ of order $2$ (why?), which means that there are either $1$ or $3$ subgroups of $G$ containing $N$, aside from $N$ and $G$.
It remains for you to show that if $K$ is a subgroup of $G$ of order $50$, then $K$ contains $N$. (This isn't tricky.)