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Let $G = GL_n(\mathbb{R})$ be the general linear group. Does $G$ have any subgroups of finite index?


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Surely you mean "proper subgroup". Otherwise, just take $G$! – Arturo Magidin Jan 27 '11 at 20:00
up vote 4 down vote accepted

Here's one, though you may consider it a "trivial" example:

The map $\det\colon G\to\mathbb{R}$ maps $G$ to the nonzero reals under multiplication via the determinant function, which is a group homomorphism since $\det(AB)=\det(A)\det(B)$. The positive real numbers are a subgroup of the real numbers under multiplication, of index $2$ (map the nonzero real numbers to $C_2=\{1,x\}$, the cyclic group of order $2$, by mapping positive numbers to $1$, negative numbers to $x$. The composition $G\to C_2$ is a map from $G$ onto a finite group, hence the kernel is a proper subgroup of finite index.

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Thanks (for both you and Mariano) for the answer! Are there any less trivial examples? – the L Jan 27 '11 at 20:05

You can take the preimage under $\det:\mathrm{GL}_n(\mathbb R)\to\mathbb R^\times$ of $\mathbb R^+$.

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