Prove for any $K \leq G$, either $K \leq N$ or $KN=G$ and $[K:K \cap N]=p$ where N is normal in G and $[G:N]=p$ The full question: suppose N is normal in G and $[G:N]=p$ is a prime. Prove that for any $K \leq G$, either (1) $K \leq N$ or (2)$KN=G$ and $[K:K \cap N]=p$.
Since $N,K \leq G$, it is of course possible that $K \leq N$ could happen.
but for (2), we can show $KN \subseteq G$  but how can we show $KN=G$?
And what information does "$[G:N]=p$" give us?
 A: Because $N$ is normal, $KN$ is a subgroup. $[G:N]$ being prime means that any subgroup $M\subseteq G$ with $N\subseteq M$ must be either $N$ or $G$.
A: Another way to look at this and using that $G/N \cong C_p$ has only two subgroups: look at the canonical image of $K$ in $G/N$, being $KN/N$. Then $KN/N=G/N$ or $KN/N=\bar{1}=N/N$. Hence $G=KN$ or $KN=N$ and the latter is equivalent to $K \subseteq N$. Using the $2$nd isomorphism theorem: $KN/N \cong K/K \cap N$, so if $G=KN$, then $|K:K \cap N|=p$.
A: Our strategy will be to show that if (1) is false then (2) is true. Accordingly, suppose $K$ is not contained in $N$, so that $N$ is a proper subgroup  of $NK$. Consider the following equation,
$$[G:NK][NK:N]=[G:N]=p.$$
Since $p$ is prime its only factorisation is $p=1\cdot p$. We have that $[NK:N] >1$ because $N$ is a proper subgroup of $N$. So that $[NK:N]=p$ and $[G:NK]=1$; that is $G = NK$.
Finally, by the Second Isomorphism Theorem, 
$$[K:N \cap K] = [KN : N] = [G:N] = p$$
which is what we wanted.
