If you find any mistakes, please make a comment! Thank you.

A fact on the interaction of subgroups with subgroups of prime index

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.

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.