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.

if we have normal subgroups $N \lhd G$ and $G_i \lhd G_{i+1}$ it says the reason that $NG_i \lhd NG_{i+1}$ is that $N$ and $G_{i+1}$ both normalize $NG_i$ inside $G$.

What does it mean for a group to normalize a group inside a group? And/or can you give me some idea about why $NG_i \lhd NG_{i+1}$?

Is that just saying that $NG_i$ is normal in $N$ and $NG_i$ is normal in $G_{i+1}$ so you can combine it to get $NG_i$ normal in $NG_{i+1}$? but how do we prove both these facts.

source 3.3 http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/subgpseries1.pdf

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

Well, we need to prove that for every $x \in NG_{i+1}$ and $y \in NG_i$ we have $xy x^{-1} \in NG_i$. First suppose that $x$ is in $N \cup G_{i+1}$ and $y$ in $N \cup G_i$.

If $y \in N$ it is clear because $N$ is normal in $G$. If $y \in G_i$ and $x \in G_{i+1}$ then it is again clear because $G_i$ is normal in $G_{i+1}$. The last case is $y \in G_i$ and $x \in N$. But then $xyx^{-1}$ is an element of $NG_i$.

Let me now prove that restricting attention to these elements is enough. If $xyx^{-1} \in NG_i$ and $zyz^{-1} \in NG_i$ then clearly $(xz)y(xz)^{-1} = x(zyz^{-1})x^{-1} \in NG_i$. Similarly, if $xyx^{-1} \in NG_i$ and $xzx^{-1} \in NG_i$ then $x(yz)x^{-1} = (xyx^{-1})(xzx^{-1}) \in NG_i$. Because arbitrary elements of $NG_i$ and $NG_{i+1}$ can be (by definition of product of subgroups) reduced to the product of elements belonging to the union of the subgroups, we are done by the previous paragraph.

share|improve this answer
    
thanks very kind of you to help me. I understand it –  user58512 Jan 24 '13 at 15:28
add comment

Your Answer

 
discard

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