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This might be obvious but it is the only thing in the proof that I can't realize. The proof is here: and my question is why the image of $N$ precisely is $NZ(G)/Z(G)$ as it says in 1.

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By the Isomorphism Theorems, if $K$ is a normal subgroup of $G$, there is a one-to-one, inclusion preserving, normality preserving correspondence between the subgroups of $G/K$ and the subgroups of $G$ that contain $K$; the correspondence is given by mapping a subgroup $M$ of $G$ that contains $K$ to the subgroup $M/K$; and mapping a subgroup $Q$ of $G/K$ to $\{g\in G\mid gK\in Q\}$.

If $H$ is an arbitrary subgroup of $G$ (which may or may not contain $K$), the image of $H$ in $G/K$ is a subgroup of $G/K$. The image corresponds to the smallest subgroup of $G$ that contains $H$ and contains $K$, and this is precisely $HK$. Thus, the image of $H$ in $G/K$ must equal the image of $HK$, which is $HK/K$.

Note that this holds regardless of whether $G$ is quasisimple, $H$ is normal, or $K$ is the center of $G$ or not. It holds for arbitrary $G$, normal $K$, and subgroup $H$.

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I prefer this answer to Geoff's, because I think it was important to point out that this is a general thing. – Tara B May 18 '12 at 8:14

The elements of the group $G/Z(G)$ are precisely the cosets of th form $xZ(G)$ for $ x \in G.$ The image of the subgroup $N$ in $G/Z(G)$ is the set of cosets $nZ(G)$ with $n \in N.$ This is the same collection as the elements of the factor group $NZ(G)/Z(G).$

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Are you confused about this possibly implying that $NZ(G)/Z(G)$ gives distinct cosets for each element of $N$? This is true if and only if $N\cap Z(G)$ is trivial:

Suppose this is true and choose $n\in N, m\in G-Z(G)$ such that the cosets $[m], [n]\in G/Z(G)$ are equal. This means $m^{-1}n=g$ for some $g\in Z(G)$. Then $mn=m^2g$ and $nm=mgm=m^2g$ because $g\in Z(G)$. Then this says that $n\in Z(G)$. Hence the coset was $[n]$ was trivial to begin with. Then $N\subseteq\ Z(G)$, a contradiction. Hence each coset generated by an element of $N$ is distinct.

If $N\cap Z(G)$ is not trivial, then it's clear that the elements of $N$ can't generate distinct cosets.

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