April 2nd, 2014, 06:14 AM  #1 
Newbie Joined: Mar 2012 Posts: 2 Thanks: 0  homework help
hi yall i have a homework problem that i am working on. it is about: determining if a subset H of the given group forms a subgroup. a)the set of all matrices having an integer determinant, in GL(2,R), i.e. H={A into GL(2,R)det(A) into Z*}. b)the set of all matrices having a rational determinant, in GL(2,R), i.e. H={A into GL(2,R)det(A) into Q*}. c)the set of all nonzero rational numbers with denominators of the form 2^(n) in the multiplicative group (Q*,x) of nonzero rational numbers, i.e. H={m/2^(n)m,n into Z, m does not = 0, n less than or equal to 0}. d)the set of all functions such that lim(x>infinity)f(x) = 0 in the additive group (F,+) of all real valued functions f:R>R, i.e. H = {f into F  lim(x>infinity)f(x) = 0 }. sorry about it not being in the math form, im not sure how it works on this site yet. and i am having trouble with just trying to find groups in general so this is starting to become a real annoyance not being able to even figure this out. thanks 
April 2nd, 2014, 07:24 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
Hint for (a). If a matrix has a determinant which is an integer other than $\displaystyle 0$ and $\displaystyle \pm1$, then what is the determinant of its inverse? Can it be an integer? Hint for (c). Suppose $\displaystyle m$ is not a power of $\displaystyle 2$. Take $\displaystyle \frac32$, say. What’s its multiplicative inverse? Does that inverse belong to the set? For (b) and (d), check the axioms for a group. 
April 2nd, 2014, 07:41 AM  #3 
Newbie Joined: Mar 2012 Posts: 2 Thanks: 0 
just making note of it here to make sure im right. to determine if a subset is a subgroup of a group, the subset must have: associativity, identity, and an inverse. a) the determinant of the inverse would most likely be a fraction which is not in Z. but why would you find the determinant of the inverse because the only point you would find the inverse is to prove the subset is a subgroup, so it is necessary for the inverse to also be an integer? c) so you are saying that because the inverse (2/3) is irrational it is not in the ration group? so in general for both of these if the inverse is not in the supposed number set it does not work? (b) and (c) listed at top. 
April 2nd, 2014, 08:31 AM  #4  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
To verify that a subset $\displaystyle H$ is a subgroup of a bigger group, you have to check that
Quote:
$$A\ =\ \begin{pmatrix}1 & 0\\0 & 2\end{pmatrix}$$ Since $\displaystyle \det A=2$ (a nonzero integer), $\displaystyle A\in H$. But $$A^{1}\ =\ \begin{pmatrix}1 & 0\\0 & \frac12\end{pmatrix}$$ and $\displaystyle \det A^{1}=\frac12$ – not an integer. The inverse of $\displaystyle A\in H$ is not in $\displaystyle H$. Therefore $\displaystyle H$ is not a subgroup in this case. Quote:
Yes, that’s basically it. Last edited by Olinguito; April 2nd, 2014 at 08:33 AM.  

Tags 
groups, homework, subgroups 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
homework  kempmaer  Math Software  1  August 28th, 2010 12:41 PM 
can anyone do my homework for me please?  oahz  Complex Analysis  1  July 25th, 2010 05:03 PM 
Little Help With homework pls!  kabal  Calculus  2  May 22nd, 2010 11:50 AM 
homework help!  andi7  Algebra  2  April 21st, 2009 04:55 AM 
homework help!  andi7  Algebra  6  March 13th, 2009 10:55 AM 