Does $G$ contain a normal subgroup that is either a 2-group, 3-group, 5-group or of order a multiple of 15? 
Suppose that a group $G$ of order $450$ has exactly two subgroups, $H$
  and $K$, of order $225$. Show that $G$ contains a non-trivial normal
  subgroup that is either a $2$-group, a $3$-group, a $5$-group or of order a
  multiple of $15$.

Note that $|G|=450=2\times3^2\times5^2$. Sylow I thus gives that $G$ contains a Sylow $2$-subgroup, $3$-subgroup, and $5$-subgroup. This is where I get stuck. I am fairly certain I want to consider $H \cap K$ and proceed on a case-by-case basis, eliminating each of the above options. 
What's the best way to go about finding the non-trivial subgroup in $G$?
 A: $\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$I don't quite get it. $H$ and $K$ are normal, as they have index $2$. So what is the subgroup $H K$? Its order is 
$$
\Size{H K} = \frac{\Size{H} \Size{K}}{\Size{H \cap K}}
= \Size{H} \cdot \frac{\Size{K}}{\Size{H \cap K}}.
$$
Since $H K$ is a subgroup, this number divides the order of $G$, so that
$$
\Size{H} \cdot \frac{\Size{K}}{\Size{H \cap K}} \text{ divides } \Size{G},
$$
and thus
$$
\frac{\Size{K}}{\Size{H \cap K}} \text{ divides } \frac{\Size{G}}{\Size{H}} = 2.
$$
But also
$$
\frac{\Size{K}}{\Size{H \cap K}} \text{ divides } \Size{K} = 225.
$$
Therefore
$$
\frac{\Size{K}}{\Size{H \cap K}} = 1,
$$
and thus $K = H \cap K$, that is, $K \le H$, and thus $H = K$, as $H$ and $K$ have the same order.
So such a group does not exist.

As noted in a comment, this is a particular case of a more general result that tells you that if $H$ is a normal subgroup of $G$, and $K$ is a subgroup of $G$ such that  $\gcd(\Size{K}, \Size{G : H}) = 1$, then $K \le H$. (The proof is really the same as above, where you get that $\dfrac{\Size{K}}{\Size{H \cap K}}$ divides both $\Size{K}$ and $\Size{G : H}$, and thus must be $1$.) In particular, if  $\gcd(\Size{H}, \Size{G : H}) = 1$, then $H$ is the unique subgroup of $G$ of its order.
