Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

3 Answers

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 $H$.

share|improve this answer
    
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
add comment

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 $NZ(G)/Z(G).$

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

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.