
Abstract Algebra Abstract Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 23rd, 2016, 05:13 AM  #1 
Member Joined: Oct 2013 Posts: 36 Thanks: 1  H < Z(G) and G/H nilpotent, implies G is nilpotent.
So I am doing this question from Rotman (pg 117, q5.38 ): if H is a subgroup of the center of G and G/H is nilpotent then G is nilpotent. I have looked at other proofs on other websites but they do not match what I have down so I just wanted to check if what I did was right too. I said let $G = p_1^{e_1}...p_n^{e_n}H, p_i$ are primes. Then $G/H \cong P_1 \times ... \times P_n$ where $P_i$ is a Sylow $p_i$subgroup of order $p_i^{e_i}$ since $G/H$ is nilpotent. Now since $H$ is abelian it is nilpotent too but also we can get an isomorphism from $Q_1 \times ... \times Q_n \times H$ to $G$ where $Q_i$ is the preimage of $P_i$ in $G$ by the natural map. Then since $Q_1 \times ... \times Q_n$ and $H$ are both nilpotent so is $G$ as their direct product. Does this work? Edit: damn realised this only works for finite G in any case is this still valid if G is finite? Last edited by fromage; June 23rd, 2016 at 05:22 AM. 
June 25th, 2016, 03:58 AM  #2 
Senior Member Joined: Sep 2008 Posts: 150 Thanks: 5 
The line, were you state the isomorphism of G with a Group that has H as a factor is wrong: A counterexample would be G cyclic of order 4 and H the unique subgroup of order 2. I would use the characterization of nilpotent with the Central series to prove your claim. 

Tags 
<, <, g or h, implies, nilpotent 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Nilpotent elements in rings  PeterMarshall  Abstract Algebra  2  January 20th, 2012 01:15 AM 
nilpotent  tinynerdi  Abstract Algebra  2  April 7th, 2010 11:46 PM 
Can you help me?... nilpotent group...  vananh  Abstract Algebra  0  April 9th, 2009 01:31 AM 
Nilpotent matrices  angelz429  Linear Algebra  2  May 7th, 2008 07:39 AM 
locallynilpotent group  sastra81  Abstract Algebra  2  February 24th, 2007 02:32 AM 