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Subgroups and matrix homework

  • Thread starter Firepanda
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  • #1
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http://img145.imageshack.us/img145/9528/matrixex6.jpg [Broken]

I have 3 subgroup criterion.

For the first i have to show H is non empty, well here the matrix

2 1
5 3

is an element of H, so it is none empty.

The second I have to show that H is closed under the binary operation of GL2R. How do I do this? By definition ig g, h are elements of H, then gh is an element of H. I have no idea how to show this is true for it?

I can find another matrix that would be in the group of H, and I can show the product of the matrices is 1, but do I have to prove it? And how?

Thirdly I have to show the inverse of each element of H belongs to H, which is easy.
 
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Answers and Replies

  • #2
Dick
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Your set of matrices are all integer matrices with the lower left entry divisible by 5 and determinant 1. Dealing with the determinant 1 condition is pretty easy if you remember det(gh)=det(g)*det(h). Let g1=[[a1,b1],[5c1,d1]] and g2=[[a2,b2],[5c2,d2]]. To show closure multiply g1*g2 and show the lower left corner is still divisible by 5 and all the entries are integers. For inverses compute (g1)^(-1) and show it has the same properties.
 
  • #3
I think the fastest way might be to exploit that theorem that says that if it's a subset (which it is) that has the property that [itex]a,b \in H \Rightarrow ab^{-1} \in H[/itex] then H is a subgroup. I say that because just exploiting the fact that the determinant is a homomorphism, it's not hard to show that ab^{-1} is in H without doing alot of math.
 

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