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

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 $[K:K \cap N] = p$.

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.

• Jer  On November 14, 2010 at 8:57 pm

Can you please explain this? I really don’t understand how this proves what it’s asking to prove. Thank You.

• nbloomf  On November 15, 2010 at 9:23 am

Supposing $K \not\leq N$, we find an element $k \in K$ such that $kN$ generates $G/N \cong Z_p$. Then we find that every element of $G$ has the form $k^an$ for some integer $a$ and element $n \in N$; hence $G = KN$. The index condition follows from the Second Isomorphism Theorem, since $p = [KN:N] = [K : K \cap N]$.

• Jer  On November 15, 2010 at 1:29 pm

Why do you show G=KN and not G=NK? And also, why does p=|KN:N|? Thanks

• Jer  On November 15, 2010 at 1:32 pm

Sorry, I got the index is because |KN:N|=|G/N|=p.

• Jer  On November 15, 2010 at 1:34 pm

Sorry, I’m an idiot, KN=NK since N is normal, right?

• nbloomf  On November 15, 2010 at 3:15 pm

Correct.

• Gobi Ree  On November 28, 2011 at 2:37 am

Another approach to show $G=NK$: Suppose $K \not\leq N$, then $\langle N,K \rangle = NK \supsetneq N$ so that $NK/N$ is a nontrivial subgroup of $G/N$. So $NK/N=G/N$ and by the lattice isomorphism theorem, $G=NK$.