I would appreciate some verification of my proof for a problem in Gallian's "Contemporary Abstract Algebra" (4th ed), chapter 9, problem 34:
Determine all subgroups of $\mathbb{R}^*$ (nonzero reals under multiplication) of index 2.
For some context, this chapter introduces the notion of normal subgroups of a group as well as factor groups. I'm assuming from the statement of the problem that there is some relatively simple form for such subgroups, and I think I've found it. Is there anything I'm missing?
Proposition
The only subgroup of $\mathbb{R}^* $ of index 2 is the group $S$ of all squares of elements of $\mathbb{R}$.
Proof
Let $H < \mathbb{R}^* $ be a subgroup of index 2. Then there are only two cosets of $H$ in $\mathbb{R}^* $, namely $\{ H, aH \}$, where $a$ is any element in $\mathbb{R}^* \setminus H$. By basic properties of cosets, $\left|H\right| = \left|aH\right|$ and the two sets are disjoint. Since $\mathbb{R}^* $ is an Abelian group, $H$ is an Abelian (sub)group and is thus a normal subgroup (this is an easy lemma). Hence, we know that the set of two cosets listed above forms a group (the factor group of $G$ by $H$; this implication is a major theorem in the chapter). This factor group has order 2, and since there is up to isomorphism only a single group of order 2, $\mathbb{Z}_2$, we have the Cayley table for $\mathbb{R}^* / H = \{ H, aH \}$ uniquely determined. In particular, for $a\notin H$, $$H = \left(aH\right)^2 = a^2 H $$ where the left equality comes from the Cayley table for this factor group, and the right equality is the definition of the factor group's operation. This shows that $a^2 \in H$ for all $a\notin H$; on the other hand, $a\in H \implies a^2 \in H$ since $H$ is a group. Thus, $H$ contains all squares of elements in $\mathbb{R}^* $.
[All of this so far has ignored the specific nature of our set $\mathbb{R}^* $. Next, I want to determine what constraints on these cosets are imposed by $\mathbb{R}^* $.]
Note that for any $x > 0$ we can write $x$ as the square of some positive real number. Thus, we have $$ \mathbb{R}^* = S \cup -S $$ where $S$ is the set $$ S = \left\{ x^2 | x \in \mathbb{R}_+ \right\}$$ so that $\left|S\right| = \left|-S\right|$ and both partition the set $\mathbb{R}^* $. The proof so far has shown that $S \subseteq H$, and since $H$ and $aH$ also partition $\mathbb{R}^* $, we must have
$$ 2 \left|S\right| = \left|S\right| + \left|-S\right| =\left|\mathbb{R}^* \right| = \left|H\right| + \left|aH\right| = 2 \left|H\right| $$ so that $\left|H\right| = \left|S\right|$. This, along with the result that $S\subseteq H$, allows us to conclude that $H=S$.
$\blacksquare$
Let me suggest an alternate way to complete the proof.
Since you have proven that $S \leq H \leq \Bbb R^{\ast}$, and since you also know that $[\Bbb R^{\ast}:S]=2$, then since $2=[\Bbb R^{\ast}:S]=[\Bbb R^{\ast}:H] \cdot [H:S]$, and the indices are all positive integers, you have either $[\Bbb R^{\ast}:H]=1$ (in which case $H = \Bbb R^{\ast}$) or $[H:S]=1$ (in which case $H=S$).