# If $I\cong G/H$, where $I$ is a subgroup of $G$, why is $G=IH$?

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 $$I\cap H=1$$.

• Is $G$ a finite group? Dec 9, 2019 at 22:36
• Not necessarily... Dec 9, 2019 at 22:37
• This came up in Washington's cyclotomic fields book, and I feel like an idiot for not knowing why it's true... Basically, the case I'm looking at is where $G$ is a Galois group of an infinite extension, $I$ is an inertia group, and $H$ is an abelian subgroup of $G$. The author said that $I\cong G/H$, which I'm okay with, but then just says this implies $G=IH$, with no justification. Dec 9, 2019 at 22:39
• I think this is a sort of converse to this question math.stackexchange.com/questions/828227/… Dec 9, 2019 at 22:43
• I think it's not true in this generality, unless the isomorphism $I\to G/H$ is the restriction of the quotient map $G\to G/H$, or $G$ is finite. Dec 9, 2019 at 22:48

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 $$\lvert IH\rvert=\frac{\lvert I\rvert\cdot\lvert H\rvert}{\lvert I\cap H\rvert}=\lvert G\rvert.$$