TITLE: Subgroup of $C^*$ (nonzero complex) with finite index.
QUESTION [6 upvotes]: True or false:
Let $C^*$ be the set of all nonzero complex numbers and $H$ be a subgroup of $C^*$(with respect to multiplication) be such that $[C^*:H]$ is finite then $H=C^*$.
I'm guessing it true as I am thinking that if suppose there is such a proper subgroup $H$ for which the number of coset will be finite then I'm guessing that there is a gap between $C^*$ and $H$ and that gap cannot be filled up by finite union.  But I am unable to give a concrete prove.
Thanks in advance.

REPLY [8 votes]: Suppose $H$ has finite index in $\Bbb C^\times$, let $m=[\Bbb C^\times:H]$. Then for any nonzero complex number, $z^m\in H$. Now given $w\in\Bbb C$ we can always solve $z^m-w=0$, so $w\in H$. 
Alternatively, a finite divisible abelian group is trivial. Now $\Bbb C^\times /H$ is finite, and it is divisible, so it must be trivial.