Group of order 1001 is cyclic. (Basic Group Theory Solution)

Question

I am trying to prove that if $G$ is a group of order $1001$, then $G$ has normal subgroups of order $7$, $11$, and $13$, and thus $G$ is cyclic. I know that by Cauchy's Theorem, there exist elements $p$, $q$, and $r$ of orders $7$, $11$, and $13$ respectively, as $1001=7\cdot 11\cdot 13$, and we know that they are cyclic. However, we have no guarantee of normality. One way I can think of doing this is by perhaps showing that these are the only subgroups of order $7$, $11$, and $13$ using a counting argument, but this is tricky since the order is large. I also CANNOT use Sylow Theorems or any advanced group theory. I have been thinking about this problem for a while. Is there any way I can start? A hint would be nice.