Why is this proof incorrect? (Group Theory) If $G$ is a finite non-trivial group and $H \leq G$ has index $p$ prime, then $H \lhd G$. This statement is actually false. However, I cannot find the error in what I wrote. Let $G$ act on the subgroups of $G$ by conjugation. Then $\text{orb}_G(H) = \\{gHg^{-1} : g \in G\\}$. The stabilizer of $H$ with respect to our $G$-action is the set $\\{g \in G : gHg^{-1} = H\\} = N_G(H)$, the largest normal subgroup of $G$ containing $H$. By the Orbit-Stabilizer Theorem, $|\text{orb}_G(H)| = (G : N_G(H))$. Since $$p = (G : H) = (G : N_G(H))(N_G(H) : H) = |\text{orb}_G(H)|(N_G(H) : H),$$ we have that $|\text{orb}_G(H)|$ divides $p$. If $|\text{orb}_G(H)| = p$, then $(N_G(H) : H) = 1$, and $H = N_G(H)$ is normal. If $|\text{orb}_G(H)| = 1$, then $H$ is invariant under conjugation and hence normal.

Your mistake is when you conclude from $H = N_G(H)$, that $H$ is normal **in the whole group** $G$.

Remember that the normalizer $N_G(H)$ is simply the set of all elements that normalize $H$, and this happens to be a subgroup of $G$. Unless $N_G(H) = G$, your group $H$ is not normal in $G$.