Maybe I'm missing something simple, but if $I$ is a subgroup of a fixed group $G$, and $H \trianglelefteq G$ is such that $I\cong G/H$, why is it true that $G=IH$? Is this some isomorphism theorem that I'm forgetting how to apply? Thanks.
EDIT: I meant to also add that we know $I\cap H=1$.
This is false (because you have no assumptions on the isomorphism $I\cong G/H$).
Here's a counterexample: Let $G=\mathbb{Z}$, $H=0$, $I=2\mathbb{Z}$. Then $I$ is abstractly isomorphic to $G/H$ but $G\neq IH$.
However, your question is true if you assume that the isomorphism $I\to G/H$ is given by the composition $I\to G\to G/H$. In this case, we know that the image of the isomorphism $I\to G/H$ is the subgroup $(IH)/H$. Thus, $G=IH$.
Your question is also true if you assume that $G$ is finite. In this case, $\lvert I\rvert=\lvert G/H\rvert=\lvert G\rvert/\lvert H\rvert$ so $\lvert IH\rvert=\frac{\lvert I\rvert\cdot\lvert H\rvert}{\lvert I\cap H\rvert}=\lvert G\rvert.$