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How to prove $GL_n(\mathbb{C})$ has no subgroup with finite index? And does $GL_n(\mathbb{Z})$ has subgroup with finite index?

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    $\begingroup$ The kernel of the natural homomorphism $GL_n(\mathbb Z)\to GL_n(\mathbb Z/N\mathbb Z)$ is a non-trivial finite index subgroup (for all $N>1)$. $\endgroup$ Commented Oct 16, 2016 at 7:47
  • $\begingroup$ @AriyanJavanpeykar I'm a little confused. May I ask what is [$GL_n(Z) : kernel$]? Just N or related to n? $\endgroup$ Commented Oct 17, 2016 at 7:26
  • $\begingroup$ Let $G$ be a group and let $H$ be a finite group. Let $G\to H$ be a morphism. Then the kernel is a finite index (normal) subgroup. Its index is the cardinality of the image of $G\to H$. Thus, in your case, as $GL_n(\mathbb Z/N\mathbb{Z})$ is a finite group, the index of the kernel is the cardinality of the image of $GL_n(\mathbb Z)\to GL_n(\mathbb Z/N\mathbb Z)$. This clearly depends on $N$. $\endgroup$ Commented Oct 17, 2016 at 11:00
  • $\begingroup$ Let $G$ be the subgroup of elements with $|det(g)| \leq 1$, what is the degree of $G$? $\endgroup$ Commented Oct 17, 2016 at 11:04

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Assume the result is false, then, by Fact $1$ , $G=GL_n(\mathbb C)$ has a proper normal subgroup $H$ of finite index, say $|G/H|=r<\infty $, then for any $M\in G$ , one has $M^r\in H$. By Fact $2$, for any $A\in G$, there exists some $B∈ G$ such that $A=B^r\in H$ and hence $H=G$, a contradiction.

Fact $1$

If a group $G$ has a proper subgroup of finite index, say $H$, then $G$ must have a proper normal subgroup of finite index.

This follows from that consider the action of $G$ on the set $\Sigma=\{gH|g \in G\}$ by left mulitiplication, one has a homomorphism $\rho:G\to S(\Sigma)\cong S_n$, then the kernel is the normal subgroup as required (note that $Ker(\rho)=\bigcap_{g\in G}gHg^{-1}\leqslant H\neq G$ and $|G/Ker(\rho)||n!$).

Fact $2$

For any $A\in GL_n(\mathbb C)$ and any positive integer $m\in \mathbb N$, there exists $B\in GL_n(\mathbb C)$ such that $A=B^m$.

This is a fact of linear algebra. Consider Jordan block and use induction on n

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  • $\begingroup$ There is a similar proof. Start from the fact $1$, consider the finite group $G/H$. There is $g_0\in G$ with $g_0H\neq H$. Denote the square root of $g_0$ by $g_1$. It is easy to see that $g_1H\neq H$ and $g_1H\neq g_0H$, because $g_1\notin H$. Continue by taking the square root $g_i$ of $g_{i-1}$. We prove $g_iH \neq g_jH,0\leq j\lt i$. Suppose $g_iH = g_jH = g_i^{2^{i-j}}H$, then $g_i^{2^{i-j}-1}\in H$. Since $({2^{i-j}-1},2^k)=1$, $g^{2^k}$ is a power of $g_jg_i^{-1}$ which also lies in $H$, a contradiction. We conclude the existence of a infinitude of elements in $G/H$, contradiction. $\endgroup$
    – zyy
    Commented Jun 3, 2022 at 19:59

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