
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
December 3rd, 2017, 07:50 AM  #1 
Newbie Joined: Mar 2014 Posts: 16 Thanks: 2  Normal Subgroups of pgroups
Let G be a pgroup and N ≠ (1) be a normal subgroup of G. Then N and Z(G) have a nontrivial intersection. Is this an induction problem?

December 3rd, 2017, 04:04 PM  #2 
Member Joined: Jan 2016 From: Athens, OH Posts: 88 Thanks: 47 
Since you mention induction, I assume that G is finite; aside: I realized I don't know if this is true for infinite pgroups. Here's most of a solution that uses a minimal counterexample, a form of induction: Assume false and let G be a minimal counterexample; i.e. for any pgroup of order less than the order of G the statement is true. Now the center of any finite pgroup is nontrivial; easy proof via the class equation. So let $Z=Z(G)$ and $Z_2/Z=Z(G/Z)$. By assumption $(NZ\cap Z_2)/Z=(N\cap Z_2)Z/Z$ is not trivial; i.e. $N\cap Z_2$ is nontrivial. Let $1\neq z\in N\cap Z_2$. Then for any $g\in G$, $[g,z]=g^{1}z^{1}gzZ=Z$ in $G/Z$. That is $[g,z]\in Z\cap N=<1>$ or $1\neq z\in N\cap Z$, contradiction. 
December 4th, 2017, 02:48 AM  #3 
Newbie Joined: Mar 2014 Posts: 16 Thanks: 2 
It is assumed to be a finite group  left that out. I thought the Isomorphism THM would be used  but couldn't get the key move.
Last edited by dpsmith; December 4th, 2017 at 02:54 AM. 
December 15th, 2017, 11:33 AM  #4 
Newbie Joined: Mar 2014 Posts: 16 Thanks: 2  Found an Easier Way
Actually, you do not need the Isomorphism THM machinery to do this. We have that N is a normal subgroup of G. So if a is in N, every conjugate of a by an element of N is in N. Hence, constructing a conjugacy class equation for N makes sense. Since N is bigger than (1), you now do the same proof as that which shows that a nontrivial pgroup has a nontrivial center. (That is the case when N = G.) 
December 16th, 2017, 08:15 AM  #5 
Member Joined: Jan 2016 From: Athens, OH Posts: 88 Thanks: 47 
Well done. If you want to seriously study groups, factor groups should become second nature for you.

December 21st, 2017, 05:00 PM  #6 
Newbie Joined: Mar 2014 Posts: 16 Thanks: 2 
They are. Abstract Algebra was my favorite topic in grad school, which was decades ago. I now am getting into math history and the interesting (and sometimes crazy) people who got involved.


Tags 
normal, pgroups, subgroups 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Primary groups and pure subgroups.  stf92  Abstract Algebra  1  February 14th, 2017 06:33 PM 
Subgroups of an arbitrary groups with square commutation.  beesee  Abstract Algebra  6  January 31st, 2013 05:24 PM 
A question groups and subgroups...  Artus  Abstract Algebra  2  January 21st, 2013 07:27 AM 
Subgroups of Abelian Groups  gaussrelatz  Abstract Algebra  9  July 9th, 2012 09:09 AM 
About minimal normal groups and subnormal groups  Sheila496  Abstract Algebra  0  October 20th, 2011 09:45 AM 