Let $G$ be a group and $H$ a subgroup of $G$ of index $2$. Then prove that $H$ is normal in $G$.
The following proof for the above statement was given in Serge Lang's book on Algebra.
Note that $H$ is contained in it's normalizer, so the index of the normalizer in $G$ is either $1$ or $2$. If it's $1$ then we are done. Suppose it is $2$. Let $G$ operate by conjugation on the set of subgroups. The orbit of $H$ has two elements, and $G$ operates on this orbit. In this way we get a homomorphism of $G$ into the group of permutations of two elements. Since there is one conjugate of $H$ unequal to $H$, then the kernel of our homomorphism is normal, of index $2$, hence equal to $H$, which is normal, a contradiction which concludes the proof.
Now I understand most of the proof besides the part marks in bold. Like what does it contradict? The fact that the normalizer has an index $2$? Why? Any help will be appreciated.
Best Answer
We’re assuming at that point that the normalizer of $H$ has index $2$ in $G$. If $H$ is normal, however, then its normalizer is $G$, which has index $1$, not $2$; this is the contradiction.