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Problem: Let $G$ be a group and $H$ a subgroup of index $2$. Then $H$ is normal in $G$.

Proof: Consider the group action by conjugation on the orbit, $Orb(H) = \{xHx^{-1} : x \in G\}$. So we have a homomorphism $f:G \rightarrow \pi(Orb(H))$, given by $x \mapsto \pi_x$, where $\pi_x(s) =xsx^{-1}$ for $s \in Orb(H)$.

$N_H = \{ x : xHx^{-1}=H\}$ contains $H$. So $(G:N_H) =1$ or $2$. If $1$ then we are done, otherwise if index is $2$, then $|Orb(H)| = 2$, from bijection, $g:G/N_H \rightarrow Orb(H)$ given by $xN_H \mapsto xHx^{-1}$.

We know $H \subseteq \ker f $. Also, as $(G: \ker f) = 2$, we deduce $\ker f = H$, and $H$ is normal, contradiciton.

Source:Serge Lang's Algebra, pg 28 I did not understand his proof, so this is the proof I derived, I hope one could check if my proof is correct, and if there is a neater way to write it out.

Bryan Shih
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Lang's proof is, indeed, slick. You can try to combine all this argumentation in one picture (I do not say anything that is not in your post) $$H < ker \, \pi \vartriangleleft G \xrightarrow{\pi} Sym(\text{orbit of }H)$$ Here are two inclusons with index 1 or 2 each and total index 2, and orbit has 1 or 2 elements. If inclusion of $H$ in kernel is strict, then index of kernel in $G$ is 1 and $\pi$ is trivial, but it means exactly that $H$ is normal. Let's assume that $H$ is not normal; then kernel of $\pi$ is a strict subgroup, as action is nontrivial. Then it's index is at most 2 and should be equal 2; so $H$ is unavoidably of index 1 in $ker \, \pi$, and therefore normal, contradiction.

In my opinion, this proof is, albeit original, horrendously unenlightening, because it looks like we "only used that 2 is prime", and it's not even remotely true. There're at least two much more direct ones.

  1. Group is normal iff cosets are double-sided. But coset decomposition is $G = H \bigsqcup G \setminus H$ and $H = He = eH$ is already both a left and right coset, therefore $G \setminus H$ is too. This example shows why it is not true for higher index (consider symmetric group on three element set).

  2. Group is normal iff cosets are group under multiplication. Easy verification shows that out of 16 possible binary operations (:^)) we can have the only one corresponding to cyclic group of order two.

xsnl
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