Understanding why complicated argument is necessary when looking for index 2 subgroups I have question concerning the proof of Theorem 1 in On ordered skew fields. The author uses Zorns lemma to construct a maximal subgroup $P \subset K^*$ with $-1 \notin P$ and $S \subseteq P$. We know that in $K^*/S$ every element has order $1$ or $2$.
Isn't it obvious now that $P$ is of index $2$ in $K^*$ (or as the author phrases it $P \cup (-1)P = K^*$)?
In my opinion this is a purely group theoretic result (we don't need the skew field structure for that) which follows from my theorem. I don't understand why the author uses a much more complicated argument envolving the skew field structure of $K$.
My theorem: Let $G$ be a group in which every element is of order $1$ or $2$, then each maximal subgroup $H$ is of index $2$.
My proof: Take $g \notin H$ then $gH \cup H$ is a group. Hence $gH \cup H=G$.
 A: First: "maximal subgroup not containing $-1$ and containing $S$" means a subgroup that is maximal among those that do not contain $-1$ and contain $S$, not necessarily a subgroup that is maximal, and that also fails to contain $-1$ and happens to contain $S$. So the subgroup in question need not be maximal in $K^*$, and your result would not apply.
For instance, consider the group $G$ of integers modulo $8$, $\mathbb{Z}/8\mathbb{Z}$. Among all subgroups that do not contain $2+8\mathbb{Z}$, the subgroup generated by $4+8\mathbb{Z}$ is maximal. This is a maximal subgroup not containing $4+8\mathbb{Z}$.  However, this subgroup is not a maximal subgroup of $\mathbb{Z}/8\mathbb{Z}$, since it is properly contained in $\langle 2+8\mathbb{Z}\rangle$. 
More generally, the Frattini subgroup of $G$, $\Phi(G)$, is the intersection of all maximal subgroups. If $x\in\Phi(G)$, you can look for subgroups that do not contain $x$, and you may be able to use Zorn's Lemma (or, you can always do it in a finite group) to prove that there is a subgroup $H$ that is maximal among all those that do not contain $x$. However, such a "maximal" subgroup would not be a "maximal subgroup of $G$", because all maximal subgroups of $G$ contain $\Phi(G)$ and hence contain $x$, and $H$ does not contain $x$. 
So you do not actually know that $P$ is maximal in $K^*$. You only know that it is maximal among those that contains $S$ but not $-1$. 
Second: you can't apply the result to $K^*$ because in $K^*$ you don't have the property. You have it in $K^*/S$. So you know that if you have a maximal subgroup $H$ of $K^*/S$ that contains $P/S$, then that $H$ has index $2$ in $K^*/S$. This subgroup would correspond to a maximal subgroup $M$ of $K^*$ that contains $P$, and this $M$ would have index $2$. But you do now know that this $M$ is equal to $P$. You know that either $M=P$, or else $-1\in M$ (because no proper subgroup of $K^*$ that does not contain $-1$ can properly contain $P$, by the maximality of $P$), but you do not know that it is the former and not the latter. So you still don't know that $P$ is maximal in $G$, and so you do not know whether you can apply the result to $P$. 
(Luckily, you can prove that such a maximal subgroup $H$ of $K^*/S$ will exist: a nontrivial group has no maximal subgroups if and only if it is divisible, and a group of exponent $2$ cannot be divisible; so moding out by $P/S$ gives you a nontrivial group of exponent $2$ that has a maximal subgroup $N$, which you can lift to a maximal subgroup $H$ of $K^*/S$ that contains $P/S$, which you can lift to the maximal subgroup $M$ of $K^*$ that contains $P$)
