Is H a Subgroup of GL(n,R)?

[cmj1988]

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?

 

[aPhilosopher]

you can take it as a given that $$g^{-1} \in GL(n, \textbf{R})$$ 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?

 

[HallsofIvy]

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?

 

[jambaugh]

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.

 

[wofsy]

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*.