A group of order $75$ and a subgroup of order $25$ Prove that a group of order $75$ can have at most one subgroup of order $25$.
My attempt
(Proof by contradiction)
Suppose $H$ and $K$ are two subgroups of order $25$.
$ | H \cap K | $ divides $25$, $ | H \cap K | $ can be $1$ or $5$.
Therefore
$|HK| = |H||K|/ | H \cap K |  = 625$ or $125$.
I am lost after this. Any help or suggestion will be much appreciated.
I am trying to solve without using Sylow's theorem.
 
 A: Hint: $HK$ is a subset (although not necessarily a subgroup) of the original group. That limits how large it can be.
A: With somewhat more sophistication, if $|G|=75$ and $H \lt G$, with $|H|=25$, then $|G:H|=3$, the smallest prime dividing the order of $G$. Hence $H \lhd G$, see here for my simple general proof (which you will not find easily in textbooks) along the same lines as your counting/reasoning above.
Now let $K \lt G$, also with $|K|=25$. Then look at the image of $K$ in the quotient group $G/H$ and use the second isomorphism theorem: $KH/H \cong K/(K \cap H)$. $|KH/H|$ divides $|G/H|$ and $|K/(K \cap H)|$ divides $|K|=|H|$. But gcd$(|H|,|G:H|)=1$, so $KH/H=\bar{1}$ (and $K/(K \cap H)=\bar{1}$), whence $K \subseteq H$ and since both subgroups have equal order, $K=H$. Hence, $H$ is unique. So what we see here in general, if $N \lhd G$, gcd$(|N|,|G:N|)=1$, then $N$ is a unique subgroup of order $|N|$. And this implies that it is characteristic (fixed by every automorphism of $G$). As an example: a normal Sylow subgroup satisfies these properties.   
