A General Linear Group question

  • Thread starter cmj1988
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Let A be a subgroup of R* (the reals closed under multiplication). Let H = {g in GL(n,R) : det g in A} Prove that H is a subgroup of GL(n,R).

Well my problem here was just proving an inverse exists. I'm just wondering if it has an easy answer.

Claim: For every g in GL(n,R), there exists a g^{-1} in GL(n,R).

Proof:

Suppose a g and g^{-1} in GL(n,R)

det(g)det(g^{-1})=det(gg^{-1})=det(I)=1 which is a nonzero determinant. In addition to that, the multiplication of reals produces a real numbers. So it is closed under R*.

Is this correct, or is there more?
 

Answers and Replies

  • #2
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you can take it as a given that [tex] g^{-1} \in GL(n, \textbf{R})[/tex] by virtue of the fact that it's a group.

But what you have doesn't address the question at all.

What would the subgroups of R* be?
 
  • #3
HallsofIvy
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Let A be a subgroup of R* (the reals closed under multiplication). Let H = {g in GL(n,R) : det g in A} Prove that H is a subgroup of GL(n,R).

Well my problem here was just proving an inverse exists. I'm just wondering if it has an easy answer.

Claim: For every g in GL(n,R), there exists a g^{-1} in GL(n,R).

Proof:

Suppose a g and g^{-1} in GL(n,R)
No, you cannot suppose g-1 is in GL(n, R) when that is what you want to prove!

det(g)det(g^{-1})=det(gg^{-1})=det(I)=1 which is a nonzero determinant. In addition to that, the multiplication of reals produces a real numbers. So it is closed under R*.

Is this correct, or is there more?
You can say that, since det(g) is a non-zero real number, g has an inverse whose determinant is also non-zero. It is that simple, after you know that det(g) is non-zero.

Now, what part of "Let A be a subgroup of R* (the reals closed under multiplication)" tells you that 0 is not in A?
 
  • #4
jambaugh
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No, you cannot suppose g-1 is in GL(n, R) when that is what you want to prove!
No it must be in GL(n,R) since it is a group and g is in GL(n,R). The issue is proving that the inverse is in H and thus H is also a group (subgroup) which is pretty straightforward since the determinant of products is the product of determinants.
 
  • #5
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Let A be a subgroup of R* (the reals closed under multiplication). Let H = {g in GL(n,R) : det g in A} Prove that H is a subgroup of GL(n,R).

Well my problem here was just proving an inverse exists. I'm just wondering if it has an easy answer.

Claim: For every g in GL(n,R), there exists a g^{-1} in GL(n,R).

Proof:

Suppose a g and g^{-1} in GL(n,R)

det(g)det(g^{-1})=det(gg^{-1})=det(I)=1 which is a nonzero determinant. In addition to that, the multiplication of reals produces a real numbers. So it is closed under R*.

Is this correct, or is there more?

Your argument confuses me. The inverse must be in H because A is a subgroup of R*.
 

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