Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$G$ group , $N \lhd G$ en $K \lhd G/N$ then there is a group $N \leq K' \lhd G$ such that $K'/N = K$, i.e. $K'$ is the group with all representants of $K$ in $G/N$. Is it then true that $[G:K'] = [G/N : K]$.

Let $A' := \{gK' \mid g\in G\}$ and $A := \{(gN)K \mid gN \in G/N \}$. Then define $$ f: A' \rightarrow A: gK' \mapsto (gN)K $$

Assume $f(gK') = f(hK')$ then $(gN)K = (hN)K$ thus $(hN)^{-1}(gN) = (h^{-1}gN) \in K$ and thus $h^{-1}g \in K'$ so that $hK' = gK'$ such that $f$ is injective.

Further is $f$ surjective because let $(gN)K \in A$ then $f(gK') =(gN)K$. This proves that $|A| = |A'|$ and thus $[G:K']=[G/N:K]$.

Is this a correct proof ?

share|cite|improve this question
You don't mean $G/K'=K$, you mean $K'/N=K$. – Derek Holt Jan 8 '13 at 21:28
up vote 1 down vote accepted

Your statement is equivalent to $\frac{|G|}{|K'|}=\frac{|G|/|N|}{|G|/|K'|}=\frac{|K'|}{|N|}$, which is not true in general.

For example let $G = \mathbb{Z}_{12}$, $N=\langle 6 \rangle$, so $G/N\cong \mathbb{Z}_6$. The subgroup $K$ isomorphic to $\mathbb{Z}_2$ of $G/N$ corresponds to $K'=\langle 2 \rangle$ in $G$, but $$2=[G:K] \not= [G/N:K]=3.$$

I think the theorem you're looking for is $[G:K']=[G:K'][K':N]$. You should be able to prove this with the same sort of method you were trying, by setting up a surjection $G/N\rightarrow G/K'$.

share|cite|improve this answer
Can u tell me where I made a mistake in the proof ? – Epsilon Jan 8 '13 at 21:27
If I understand your writing correctly, the mistake occurs when you define $f(gK')=(gN)K$ but then immediately afterwards state $f(gK')=g(hK')$. How is $(gN)K=g(hK')$? – Alexander Gruber Jan 8 '13 at 21:39
In other words the error is that you've assumed that $G/K'=K$ but what you really want is for $K'/N=K$. – Alexander Gruber Jan 8 '13 at 21:40
$f(gK′)=g(hK′)$ is (was) a typo. With the last comment you are right. – Epsilon Jan 8 '13 at 21:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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