$G$ has at most one subgroup of order $p$ Let $G$ a group, $H$ and $K$ two subgroups of G of finite order such that $H \cap K = \{1_G\}$. 
I already show the first exercise which says that the cardinal of $HK$ is $|H||K|$.
The second exercise ask to deduce that if $|G|=pm$ where $p$ is a prime number and $p>m$, then $G$ has at most one subgroup of order $p$. Show that if the subgroup exist, it is normal in $G$.
I think a can use the Cauchy and Sylow theorem, but I'm a bit block off on this problem. Is anyone is able to give me a hint?

No need for sophistication. If $G$ has two distinct subgroups of order $p$ then those subgroups satisfy the conditions on $H$ and $K.$
Now since the subgroup is the only one of its order, it is identical to all of its conjugates and therefore normal.
