Subgroup with Finite Index of Multiplicative Group of Field Let $F$ be an infinite field such that $F^*$ is a torsion group. We know that $F^*$ is an Abelian group. So every subgroup of $F^*$ is a normal subgroup.
My question:
Does $F^*$ have a proper subgroup with finite index?
 A: Let $F$ be the algebraic closure of a finite field. Each element of $F^*$ belongs to a finite field, so is a torsion element. On the other hand $F^*$ cannot have a subgroup of index $n>1$. For if $A$ is such a subgroup, then $x^n\in A$ for all $x\in F^*$. But if $z\in F^*\setminus A$, then $z$ has an $n$th root in $F^*$ contradicting the previous sentence.

Returning to the general case. If $F^*$ is torsion, then obviously $F$ has finite characteristic, and is algebraic over its prime field. Therefore $F$ is contained in an algebraic closure of a finite field. If $F$ itself is finite, then it obviously has finite index subgroups, but this was excluded by the OP. 

And as an example of an infinite field such that $F^*$ has a subgroup of a finite index let's try the following. Consider the nested union of extensions of $\mathbb{F}_2$ of degrees $2^n$ 
$$
F=\bigcup_{n\ge0}\mathbb{F}_{2^{2^n}}.
$$
The union can be formed inside an algebraic closure of $\mathbb{F}_2$.
For $m\ge n$ let $N^m_n:\mathbb{F}_{2^{2^m}}\to\mathbb{F}_{2^{2^n}}$
be the relative norm map.
 For $n\ge1$ define the groups
$$
A_n=\{z\in\mathbb{F}_{2^{2^n}}\mid N^n_1(z)=1\}\le \mathbb{F}_{2^{2^n}}^*.
$$
Transitivity of norm in a tower of field extensions means that
$N^n_1\circ N^m_n=N^m_1$ for all $1\le n\le m$.
Let us define
$$
A=\bigcup_{n\ge1}A_n.
$$
I claim that $A$ is a subgroup of $F^*$. It is obviously closed under inverses, as all the $A_n$ are groups as kernels of $N^n_1$. If $z\in A_n$ and $n<m$, then
$$
N^m_1(z)=N^n_1(N^m_n(z)).
$$
Here $N^m_n(z)=z^{2^{m-n}}$ is just a power of $z$, so we get that also $z\in A_m$. We have seen that $A_n\le A_m$, and the claim follows from this.
To close off this example I claim that $A$ is of index three in $F^*$. Let 
$\mathbb{F}_4^*=\{1,\omega,\omega^2=1+\omega\}$, where $\omega$ is a primitive third root of unity. Because $N^n_1(\omega)=\omega^{2^{n-1}}$ is either $\omega$ or $\omega^2$, it follows that for every element $z\in F^*$ exactly one of $z,\omega z,\omega^2 z$ belongs to the subgroup $A$. The claim follows from this.
Note that the argument from the case of an algebraically closed field does not apply for this $F$. For example, the field $F$ does not have ninth roots of unity because those reside in the field $\mathbb{F}_{64}$, and won't be included in this tower.
A: Note: This uses the very same reasoning as Jyrki's answer but with a different, perhaps slightly more groupwise,  approach:
(1) An abelian group $\,A\,$ (with multiplicative operation, to fit within our problem) is divisible if
$$\forall\,g\in G\;\wedge\;\forall n\in\Bbb N\;\exists\, x\in G\;\;s.t.\;\;g=x^n$$
(2) Any homomorphic image of a divisible group is divisible.
(3) Finite abelian groups can not be divisible
(4) Divisible groups cannot have finite-index subgroups (this is just (2)+(3))
(5) The multiplicative group of an algebraically closed group is abelian 
Now apply Jyrki's answer...
