# How to prove $GL_n(\mathbb{C})$ has no subgroup with finite index? [closed]

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

• 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 $N>1)$. Commented Oct 16, 2016 at 7:47
• @AriyanJavanpeykar I'm a little confused. May I ask what is [$GL_n(Z) : kernel$]? Just N or related to n? Commented Oct 17, 2016 at 7:26
• 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 $N$. Commented Oct 17, 2016 at 11:00
• Let $G$ be the subgroup of elements with $|det(g)| \leq 1$, what is the degree of $G$? Commented Oct 17, 2016 at 11:04

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

• 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.
– zyy
Commented Jun 3, 2022 at 19:59