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).(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# A General Linear Group question

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