Finding index 2 subgroups of the affine general linear group I need to show that AGL(1,$\mathbb{R}$) has an index 2 subgroup but that AGL(1,$\mathbb{C}$) does not. I also need to find for which finite fields, $\mathbb{F}$, AGL(1,$\mathbb{F}$) has an index 2 subgroup. 
To show AGL(1,$\mathbb{C}$) does not have an index 2 subgroup, I think that I need to show that every proper subgroup has infinite index.
 A: Hint: An index 2 subgroup has abelian quotient, so contains the derived subgroup. Calculate the commutator of two general elements to see there is a nice answer.

 The derived subgroup contains the commutator of $x\mapsto x+1$ and $x\mapsto (1+a) x$, which is $x \mapsto x-a$. Thus the derived subgroup contains all translations*. Hence for a field $K$, $\operatorname{AGL}(1,K)$ has an index 2 subgroup if and only if its maximal abelian quotient, $K^\times$, has an index 2 subgroup. $\mathbb{R}^\times$ has one; $\mathbb{C}^\times$ does not; a finite $\mathbb{F}^\times$ has one iff $\mathbb{F}$ has odd characteristic. The field $\mathbb{F}$ of size 2 is an exception*, since $\operatorname{AGL}(1,\mathbb{F})\cong \mathbb{F}_+$ has a subgroup of index 2, but $\mathbb{F}^\times\cong 1$ does not.

Tricky detail:

 *As long as the field is larger than size 2; in char 2 but bigger than $\mathbb{Z}/2\mathbb{Z}$ one can use that the commutator $x\mapsto ax+1$ and $x\mapsto cx+1$ is $x \mapsto x+a+c$ to get the translations. If $|\mathbb{F}|=2$, then we get $\operatorname{AGL}(1,\mathbb{F})$ is cyclic of order 2, so is a special case.

