Algebra proof involving subgroups and $[G:K]=q$ a prime number Let $K$ be a subgroup of a group $G$ with $[G:K]=q$ a prime number. 
a) If $[G:K]=q$ is a prime number, show that $\{g \in G : \alpha_g(K)=K \}$ is either $K$ or $G$. (Here, $\alpha_g$ denotes conjugation by $g$.)
b) Show that $K$ is either a normal subgroup of $G$ or that there are $q$ different conjugate copies of the subgroup $K$ in $G$.
For a) I know I must use Lagrange's Theorem somehow. Also, possibly use the fact that if $H$ is a subgroup of $G$ then $H$ is a normal subgroup of $G$ iff $\alpha(H)=H$ for all inner automorphisms $\alpha \in {\rm inn}(G)$.
For b) maybe use cosets and Lagrange's Theorem?
 A: Define $N(K) = \{ g \in G : gKg^{-1} = K \}$. This is usually called the normaliser of $K$.
(i) Show that $N(K)$ is a subgroup of $G$.
(ii) Show that $K$ is a subgroup of $N(K)$.
(iii) Use Lagrange's theorem, together with (i) and (ii) and the fact that $[G : K]$ equals the prime $q$, to prove that either $N(K) = K$ or $N(K) = G$.
Now, let $\Omega_K = \{ gKg^{-1} : g \in G \}$ be the set of subgroups conjugate to $K$. Notice that $G$ naturally acts on $\Omega_K$ by conjugation.
(iv) Let ${\rm Stab}(K)$ be the stabiliser of $K$ under this action of $G$ on $\Omega_K$. Convince yourself that ${\rm Stab}(K) = N(K)$.
(v) Let ${\rm Orb}(K)$ be the orbit of $K$ under the action of $G$ on $X$. Convince yourself that ${\rm Orb}(K) = \Omega_K$.
Now, remember that there is a relationship between the size of the stabiliser of $K$ and the size of the orbit of $K$ under the action of $G$:
$$ |G| = |Stab(K)|\times |Orb(K)|.$$
(vi) If $N(K) = K$, what is the size of $\Omega_K$? What if $N(K) = G$? What can you deduce about the possible numbers of subgroups conjugate to $K$?
