Subgroup of $G=\langle(123),(456),(23)(56)\rangle$ 
Let $G=\langle(123),(456),(23)(56)\rangle$ be a subgroup of order $18$ of $S_6$.
Clearly $H=\langle(123),(456)\rangle$ is a subgroup isomorphic to $C_3 \times C_3$.

SOLVED: Not sure how $H$ is a subgroup clearly. Could someone please explain how this is so obvious?

Apparently $H$ has subgroups, $A_1=\langle(123)\rangle,A_2=\langle(456)\rangle,A_3=\langle(123)(456)\rangle,A_4=\langle(123)(465)\rangle$.

How is $A_4$ a subgroup?

...these are all normal subgroups

Again how is it obvious that the $A_i$'s are normal in $G$?
 A: Your $H$ is the subgroup generated by $(123)$ and $(456)$, in the sense that your notation $\langle \cdot \rangle$ explicitly means the subgroup generated by $\cdot$. So it's a subgroup, very clearly.
Since they're disjoint 3-cycles, you also get that it's isomorphic to $C_3 \times C_3$. This is also how you know, for instance, that $A_4$ is a subgroup. It's the subgroup generated by $(123)(465)$. To see that it's a subgroup of $H$, you just need to verity that $(123)(465) \in H$ (which it is - but if you haven't, you should verify it!).
There are very many reasons why the $A_i$ are normal. One way to see is to explicitly check. This set of computations might give you a better understanding of what's going on, so perhaps it's even a good idea.
A different way is to note that $H$ is abelian, and any subgroup of an abelian group is abelian.
Yet another way is to note that these are subgroups of minimal prime index in $H$, and subgroups of minimal prime index are normal. To be fair, I do not think you should know this statement yet. But somehow it's good to get an idea of some of the general statements out there, to see how everything connects.
