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Problem: Show that the general linear group $G = GL(n, \mathbb{C})$ has no proper subgroup of finite index.

I wrote down a proof, but not quite sure if it is right, especially about the part about constructing an infinitude of elements in the quotient group. Can anyone please verify it for me?

Proof: Assume such a sugroup exists, say $H \leq G$ of index $m$. Let $G$ act on the coset space $\{gH|\, g \in G\}$, we get a homomorphism $\rho: G \to S_m$, and $\{1\} \neq \ker \rho \leq H$ is normal in $G$, which has finite index since $G/\ker\rho \cong S_m$.

Denote $\ker \rho$ by $K$. Consider the quotient group $G/K$. If $G/K = \{K\}$, then $H = G$, contradicting $H$ being a proper subgroup. Hence, there exists $g_0K \in G/K$ with $g_0 \notin K$. We can take a square root for any element of $GL(n, \mathbb{C})$, so can we do for $g$. Denote one of its square roots by $g_1$. If $g_1K = K$, $g_1 \in K$ and thus $g_0 = g_1^2 \in K$, a contradiction. Also $g_1K \neq g_0K$ for the same reason. Thus we obtain three distinct elements $K, g_0K, g_1K \in G/K$.

Continue by taking $g_2$ to be a square root for $g_1$, and then $g_3$ for $g_2,\,\dots\,$ Inductively, for $g_{i-1}$ we let $g_i$ be one of its square roots. We prove next $g_iK \neq g_jK, j=0,\dots,i-1$. Otherwise, assume $g_iK = g_{i-l}K$ for some $0 \lt l \leq i$. Then $g_i^{-1}g_{i-l} = g_i^{-1}g_i^{2^l} = g_i^{2^l-1} \in K$. Since $\gcd(2^l-1, 2^i) = 1$, we know $g_0 = g_i^{2^i}$ is some power of $g_i^{2^l-1}$, hence lying in $K$, a contradiction. Thus, for each step we obtain one more distinct element of $G/K$, and continue we construct an infinitude of elements in $G/K$, contradicting $G/K$ being finite.

Shaun
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zyy
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    I don't think this quite works, for example if $g_0^3 \in K$, then we get $g_2K = g_0K$, with $g_0 = g_2^4$. Why should this imply that $g_0$ is a power of $g_2^3$? You could start with an element $g_0$ with $g_0^p \in K$ for some prime $p$, and then repeatedly take $p$-th roots of $g_0$. – Derek Holt Jun 04 '22 at 09:08
  • Special case of this: https://math.stackexchange.com/questions/201460/subgroups-of-finite-index-in-divisible-group – David A. Craven Jun 04 '22 at 22:02

1 Answers1

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It's easier to work within the potential quotient group. For any element from $GL_n(\mathbb C)$, it has $p$-th root for any prime number $p$. This property carries to any quotient group, hence we just have to show no finite subgroup $G$ has this property.

To illustrate why your idea fails, take the finite group $G=\langle a \rangle$ where $a$ has order $3$, we have $a$ and $a^2$ are square-roots of each other, as $(a^2)^2=a^4=a$.

The key point here is to use a $p$ that divides $|G|$. As pointed out in the comment, by Cauchy's theorem, we can find $a_1\in G$ with order $p$. Keep taking the $p$-th roots, we get a chain of elements $a_0=1, a_1, a_2, \cdots$ such that $a_n^{p}=a_{n-1}$ for all $n\ge 1$.

It's clear that $a_n\not=1$ for all $n\ge 1$. Otherwise, we would have $a_1=a_n^{p^{n-1}}=1$. Hence the order of $a_n$ must be a positive integer.

Since $a_i^{p^i}=a_0=1$, the order of $a_i$ must be a divisor of $p^i$, hence also a power of $p$. Now if $a_m=a_n$ for $m>n$, then $a_n=a_m^{p^{m-n}}=a_n^{p^{m-n}}$, $a_n^{p^{m-n}-1}=1$ which is impossible since $\gcd(p, p^{m-n}-1)=1$, $p^{m-n}-1$ must be coprime with the order of $a_n$. Put in another way, as both $a_n^{p^n}=1$ and $a_n^{p^{m-n}-1}=1$ hold, we must have $a_n=(a_n^{p^n})^x(a_n^{p^{m-n}-1})^y=1$ for some $x, y$ such that $xp^n+y(p^{m-n}-1)=1$.

Just a user
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