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April 2nd, 2014, 06:14 AM   #1
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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.

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April 2nd, 2014, 07:24 AM   #2
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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.
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April 2nd, 2014, 07:41 AM   #3
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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.
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April 2nd, 2014, 08:31 AM   #4
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Quote:
Originally Posted by Davv009 View Post
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.
In general there is no need to check associativity since this property will be “inherited” from the bigger group.

To verify that a subset $\displaystyle H$ is a subgroup of a bigger group, you have to check that
  1. $\displaystyle H$ is not empty. Usually that means verifying that the identity element is in it.
  2. $\displaystyle H$ is closed under the group operation. The product of two elements of $\displaystyle H$ must be in $\displaystyle H$.
  3. if $\displaystyle a\in H$ then $\displaystyle a^{-1}\in H$.

Quote:
Originally Posted by Davv009 View Post
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?
Let’s take an example. We have $\displaystyle H=\{2\times2\ \text{real matrices whose determinant is a nonzero integer}\}$. Is $\displaystyle H$ a subgroup of $\displaystyle \mathrm{GL}(2,\mathbb R)$? If it is, then the inverse of every member in $\displaystyle H$ must be in $\displaystyle H$ too. Let

$$A\ =\ \begin{pmatrix}1 & 0\\0 & 2\end{pmatrix}$$

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.

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Originally Posted by Davv009 View Post
c) so you are saying that because the inverse (-2/3) is irrational it is not in the ration group?
The inverse is $\displaystyle \frac23$ and it is rational. I am saying that $\displaystyle \frac23$ is not of the form $\displaystyle \frac m{2^n}$ therefore not in $\displaystyle H=\left\{\frac m{2^n}:m\in\mathbb Z\setminus\{0\},\,n\in\mathbb Z^+\cup\{0\}\right\}$.

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Originally Posted by Davv009 View Post
so in general for both of these if the inverse is not in the supposed number set it does not work?
Yes, that’s basically it.
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Last edited by Olinguito; April 2nd, 2014 at 08:33 AM.
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