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 $[H:K]=1$?

  • $\begingroup$ 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? $\endgroup$ – lulu Feb 13 '17 at 17:49
  • $\begingroup$ For your other question....well, if $[G:K]=5=5[H:K]$ then $[H:K]=1$, yes? $\endgroup$ – 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 $5=[G:K]=5\dot{}[H:K]$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.