Steps in the proof of the smallest prime dividing a group

This question comes from here. It is not a duplicate but asking for explanation of the last part of the proof.

I am asked to prove that if 5 is the smallest prime dividing the order of a finite group $G$, then any subgroup of index 5 in $G$ is normal.

This is a proof I have found but don't understand the last steps. Let $H$ be a subgroup of index $5$ where $5$ is the smallest prime that divides $|G|$. Then $G$ acts on the set of left cosets of $H$, $\{gH\mid g\in G\}$ by left multiplication, $x\cdot(gH) = xgH$.

This action induces a homomorphism $G\to S_5$, whose kernel is contained in $H$. Let $K$ be the kernel. Then $G/K$ is isomorphic to a subgroup of $S_5$, and so has order dividing $5!$. But it must also have order dividing $|G|$, and since $5$ is the smallest prime that divides $|G|$, it follows that $|G/K|=5$.

I understand everything up until here: Since $|G/K| = [G:K]=[G:H][H:K] = 5[H:K]$, it follows that $[H:K]=1$, so $K=H$. Since $K$ is normal, $H$ was in fact normal.

I'm sorry I am quite new to group theory, could anybody explain how $[G:K]=[G:H][H:K]$? and if this =$5[H:K]$, how does it follow that $[H:K]=1$?

• To your specific question: for any subgroup $G_1\subset G$ we define $[G:G_1]=\frac {|G|}{|G_1|}$. That makes the formula clear, no? – lulu Feb 13 '17 at 17:49
• For your other question....well, if $[G:K]=5=5[H:K]$ then $[H:K]=1$, yes? – lulu Feb 13 '17 at 17:54

1)This is Lagrange formula: if $K\leq H\leq G$, $K,H,G$ groups, then $[G:H][H:K]=[G:K]$
2)This is because you've proof that $5=[G:K]=5\dot{}[H:K]$