**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.3**

Let $G$ be a group, $N \leq G$ a normal subgroup of prime index $p$, and $K \leq G$ a subgroup. Prove that either $K \leq N$ or $G = NK$ and $[K:K \cap N] = p$.

Solution: Suppose $K \setminus N \neq \emptyset$; say $k \in K \setminus N$. Now $G/N \cong \mathbb{Z}/(p)$ is cyclic, and moreover is generated by any nonidentity- in particular by $\overline{k}$.

Now $KN \leq G$ since $N$ is normal. Let $g \in G$. We have $gN = k^aN$ for some integer a. In particular, $g = k^an$ for some $n \in N$, hence $g \in KN$. We have $[K:K \cap N] = p$ by the Second Isomorphism Theorem.