
#1
Oct805, 08:04 PM

P: 1,119

I have a problem that states
Define the Special linear group by: (Let R denote real numbers) [tex]SL(2,R) = \{ A\in GL(2,R): det(A)=1\}[/tex] Prove that SL(2,R) is a subgroup of GL(2,R). ___ Now a subset H of a group G is a subgroup if: i) [tex]1 \in H[/tex] ii) if [tex] x,y \in H[/tex], then [tex]xy \in H[/tex] iii) if [tex]if x\in H[/tex], then [tex]x^{1} \in H[/tex] I have very little knowledge of matricies and I don't even see how 1 could be in SL(2,R), other than maybe something saying that GL(2,R) has 1, so SL(2,R) must have it too, but I bet there is a more appropriate way. Also what would H be here? Would it a set containing matricies or numbers? 



#2
Oct805, 08:26 PM

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P: 1,322

In your case you should identify G with the group/set GL(2) and H with the group/set (to be proved) SL(2). G is the set of all 2x2 matrices with real entries, and H is the subset of G consisting of all 2x2 matrices that have determinant 1. When trying to prove something is in H, you've got to figure out it if its determinant is 1, right? So you're goal is always to show that the determinant is 1 when trying to prove that something is a member of H.
First, is the identity in H? H is the set of all 2x2 matrices with unit determinant, so does I fit the bill? Second, if x and y are in H, then it means det(x) = 1 and det(y) = 1, so what can you say about det(xy)? Third, if det(x) = 1, then what can you say about det(x^1)? 



#3
Oct805, 08:34 PM

P: 1,119

So 1 is not in H then.
1 is instead the identity matrix: [tex] \begin{array}{cc}1&0\\0&1\end{array}[/tex] 



#4
Oct805, 08:35 PM

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P: 1,322

Linear Groups
You are right that the number 1 is not in H since H consists of 2x2 matrices. The identity matrix is in H, however.




#5
Oct805, 08:38 PM

P: 1,119

Ok, so I then prove that the identity matrix is in H, that seems much more possible.
Well certainly the identity matrix is in G, and also the det(identity Matrix) = 1. 



#8
Oct905, 08:03 PM

P: 1,119

Ok a similar problem.
Prove that GL(2,Q) is a subgroup of GL(2,R) I have done the first two steps, but the inverse is throwing me off. Is there something that says the determinant cannot be 0? Because my book defines the following if x = [tex] \begin{array}{cc}d&c\\b&a\end{array}[/tex] then x1 = [tex]\begin{array}{cc}d/det&c/det\\b/det&a/det\end{array}[/tex] So unless there is something that says these two groups cannot have determinants of 0, then I am not sure what is going on with the problem. Thanks! 



#9
Oct905, 08:20 PM

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P: 1,322

In GL(2,Q) what does the Q mean? I don't believe I've ever seen that before.
Also, the definition of the general linear group includes the requirement that the matrices be invertible, so they must all have determinant not equal to zero. 



#10
Oct905, 08:30 PM

P: 1,119

Sorry. The Q represents rational numbers (a/b: a,b are integers, and b [tex]\neq[/tex] 0)




#11
Oct905, 08:34 PM

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P: 1,322

Right, of course. Well if you've got everything except the inverse then you're basically done. You know the inverse exists because the matrices come from [tex] GL(2,Q) \subset GL(2,R) [/tex]. Now all you have to do is prove all the elements of the inverse are rational numbers.



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