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$G$ 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 ?

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You don't mean $G/K'=K$, you mean $K'/N=K$. –  Derek Holt Jan 8 '13 at 21:28

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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 $G/N\rightarrow G/K'$.

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Can u tell me where I made a mistake in the proof ? –  André 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. –  André Jan 8 '13 at 21:43

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