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 $I\cap H=1$.

  • 3
    $\begingroup$ Is $G$ a finite group? $\endgroup$
    – Bernard
    Dec 9, 2019 at 22:36
  • $\begingroup$ Not necessarily... $\endgroup$
    – applebees
    Dec 9, 2019 at 22:37
  • $\begingroup$ 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. $\endgroup$
    – applebees
    Dec 9, 2019 at 22:39
  • $\begingroup$ I think this is a sort of converse to this question math.stackexchange.com/questions/828227/… $\endgroup$
    – applebees
    Dec 9, 2019 at 22:43
  • 1
    $\begingroup$ 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. $\endgroup$
    – Berci
    Dec 9, 2019 at 22:48

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


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