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.

Let $N\neq 1$ be a normal subgroup of G, where G has order $p^n$, $p$ prime, $n\geq 1$. I need to show that $Z(G)\cap N\neq \{1\}$.

I have tried to do it this way: Suppose $|N|=p^m$, $m<n$. Since $N\cap Z(G)$ is normal in $Z(G)$, in both cases ($N\cap Z(G)=\{1\}$ and $N\cap Z(G)\neq\{ 1\}$), I used the Second Group Isomorphisms Theorem to get:

$G/N\cong Z(G)/N\cap Z(G)$

so $|G/N| = |Z(G)/N\cap Z(G)|$

Since N is normal in G, $|G/N|=p^{n-m}$, with $n-m>0$. So $|Z(G)/N|\cap Z(G)|=p^{n-m}\Rightarrow [Z(G):N\cap Z(G) ]=p^{n-m}$

and this means that $N\cap Z(G)\neq \{1\}$.

share|improve this question
1  
This is the same as this question. –  Mikko Korhonen May 4 '12 at 7:07
add comment

2 Answers

up vote 7 down vote accepted

Since $N$ is normal in $G$, then $G$ acts on $N$ by conjugation. So,

$$N = \sum \operatorname{Orbit}_{G}(x),$$ ($x\in N $) $ =$ 1 + $\sum \textrm{Orbit}_{G}(x)$, $x\in N-\{1\}$. Now, since the order of $\textrm{Orbit}_{G}(x)$ is a non negative power of $p$ and $p$||N|, then we see one orbit has order of 1. So, there is an non trivial element $s\in N$ that $\operatorname{Orbit}_{G}(s)$=${s}$. Therefore for all $g\in G$, $s^g=s$ and this means that $N \cap Z(G)$ is not 1.

share|improve this answer
    
Nicely presented! :+) –  amWhy Mar 5 '13 at 0:18
add comment

Since $N$ is normal, it is an union of disjoint conjugacy classes. It contains the conjugacy class $\{1\}$, and thus not all of the other conjugacy classes can have order divisible by $p$. This means that besides the conjugacy class $\{1\}$, the subgroup $N$ contains also some other conjugacy class $\{z\}$ of size $1$. Then $z$ is a nontrivial central element in $N$.

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.