Let G be a finite group, let p be the smallest prime divisor of #G, and let H ≤ G be a subgroup of index p. Then H is normal in G.

Question

I have been spending time attempting to understand the following proof by @user:544975:

By the orbit-stabilizer theorem, the size of every orbit of cosets divides $|H|$, and hence also $|G|$. Since there are exactly $p$ cosets of $H$ and $p$ is the smallest prime dividing $|G|$, it must be that either there is a single orbit of size $p$ or there are $p$ different orbits, all of size $1$.

The first option, however, is impossible, since for every $h\in H$, $hH=H$, meaning the action fixes the coset corresponding to the identity. Hence, there exists an orbit of size $1$, so they must all be of size $1$.

This means that for every $h \in H, g \in G$ we have: $$hg^{-1}H=g^{-1}H\implies \exists h_1\in H \space s.t.\space hg^{-1}=g^{-1}h_1\implies \\ ghg^{-1}=h_1\in H$$

I am not sure however why it is necessary we need p to be the smallest prime dividing G.

Answer 1

In general, the argument "one orbit of size $p$ or $p$ orbits of size $1$" does not work. Instead, you'd have "some orbits, where each has size dividing $|G|$ and their sum is $[G:H]$". So those "sizes dividing $|G|$" cannot be greater than $[G:H]$, but they can be various values $\le[G:H]$. However, if $[G:H]$ is the smallest divisor (apart from $1$) of $|G|$, this means that that there is the dichotomy used in the proof: "one orbit of size $p$ or $p$ orbits of size $1$"