Let $\Gamma$ be a group and $\Gamma_1\supset\Gamma_2\supset\dots$ subgroups of finite index, such that $\bigcap_{j=1}^\infty \Gamma_j=\{1\}$. Let $1\ne\gamma\in\Gamma$ and let $[\gamma]=[\gamma]_\Gamma$ denote the conjugacy class of $\gamma$ in $\Gamma$. Is it true that there exists $n\in\mathbb N$ such that $[\gamma]\cap\Gamma_n=\emptyset$? This is the case, if every $\Gamma_n$ is normal in $\Gamma$, but what about the general case?

If it is not true, are there non-trivial estimates of the number of $\Gamma_n$ conjugacy classes within $[\gamma]$? More precisely, let $c_n$ denote the cardinatlity of the set $\big([\gamma]\cap\Gamma_n\big)/\Gamma_n$, where $\Gamma_n$ acts by conjugation. Does $\frac{c_n}{[\Gamma:\Gamma_n]}$ tend to zero?

If any of this does not hold in general, are there interesting classes of groups, for which it holds?

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    $\begingroup$ @DerekHolt But those subgroups don't have finite index $\endgroup$ Oct 2 at 11:42
  • $\begingroup$ @SeanEberhard Yes sorry! $\endgroup$
    – Derek Holt
    Oct 2 at 12:49
  • $\begingroup$ Yes, right, I corrected this. $\endgroup$
    – Echo
    Oct 2 at 13:01

2 Answers 2


Here is a counterexample, with the Heisenberg group. Define by induction $a_0=1$ and $a_{n+1}=(a_n^2+1)a_n$. Clearly $a_n$ tends to infinity.

Let $\Gamma$ be the Heisenberg group, consisting of matrices $m(x,y,z)=\begin{pmatrix}1&x&z\\ 0&1&y\\0&0&1\end{pmatrix}$ with $x,y,z\in\mathbf{Z}^3$. Let $C_n$ be the $n$-th congruence subgroup, namely those $m(x,y,z)$ with $x,y,z\in n\mathbf{Z}$.

Let $\Gamma_n$ be the subgroup generated by the congruence subgroup $C_{a_n^2}$ and $m_n=m(1,0,a_n)$.

First check: $\Gamma_{n+1}\subset\Gamma_n$. Indeed $C_{a_{n+1}^2}\subset C_{a_n}\subset\Gamma_n$ because $a_n$ divides $a_{n+1}$. So it remains to check $m_{n+1}\in\Gamma_n$. Indeed, $m_n^{a_n^2+1}=m(a_n+1,0,a_n(a_n^2+1))=m(a_n,0,0)m_{n+1}$. Since $m(a_n,0,0)\in C_{a_n}$, we deduce $m_{n+1}\in\Gamma_n$.

Second check: $\bigcap_n\Gamma_n=\{m(0,0,0)\}$. Indeed, consider a nontrivial element $m(x,y,z)$. Choose $n$ such that $a_n$ is greater than $|x|,|y|,|z|$. Let us show that $m(x,y,z)\notin \Gamma_n$. Otherwise, it can be written as $m_n^km(aa_n^2,ba_n^2,ca_n^2)$, hence as $m(k+aa_n^2,ba_n^2,(ca_n+kba_n+k)a_n)$. Since the three coefficients are $<a_n$ in absolute value, we first deduce that $b=0$ so it equals $m(k+aa_n^2,0,(ca_n+k)a_n)$, then $k=-ca_n$, so it equals $m(a_n(aa_n-c),0,0)$, and finally it equals $m(0,0,0)$.

Finally, $m_n\in\Gamma_n$ is conjugate to the fixed element $m(1,0,0)$.

  • $\begingroup$ I'm missing something dumb. You write that $m(1,a_n,0)$ is conjugate to $m(1,0,0)$. But there is a quotient map $\Gamma \to \mathbb{Z}^2$ sending $m(x,y,z) \mapsto (x,y)$, and $(1,0)$ is not conjugate to $(1, a_n)$ in $\mathbb{Z}^2$. $\endgroup$ Oct 2 at 15:41
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    $\begingroup$ @DavidESpeyer I believe $m(1,a_n,0)$ should be $m(1,0,a_n)$. Also $m_n^{a_n^2+1} = m(a_n^2+1, 0, a_n (a_n^2+1))$. $\endgroup$ Oct 2 at 15:58
  • $\begingroup$ Thanks for noticing: indeed it's $m_n=m(1,0,a_n)$, that is, $\begin{pmatrix}1 & 1 & a_n\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$. Now fixed. $\endgroup$
    – YCor
    Oct 2 at 16:24
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    $\begingroup$ No problem, nice example. Now that I understand it, here is a variant that seems nice to me: Let $c$ be an irrational $p$-adic integer. Let $\Gamma_n = \{ m(x,y,z) : z \equiv c x \bmod p^n,\ y \equiv 0 \bmod p^n \}$. Then $\bigcap \Gamma_n$ is trivial but, if $c_n$ is an integer congruent to $c \bmod p^n$, then $m(1,0,c_n)$ is in $\Gamma_n$ and is $\Gamma$-conjugate to $m(1,0,0)$. $\endgroup$ Oct 2 at 16:37

YCor beat me to it, but I will post my answer anyway because it is rather different. I will construct a counterexample to the first question with $\Gamma = F_2 = F\{x,y\}$, the free group on two generators.

Lemma: Let $w \in F_2$ be a nontrivial word. Let $M_n$ be the set of surjective homomorphisms $f:F_2 \to A_n$ such that $1\notin \operatorname{Fix}(f(w))$ and $\operatorname{Fix}(f(x)) \ne \emptyset$. Then as $n\to\infty$ the proportion $|M_n|/|\operatorname{Hom}(F_2, A_n)|$ tends to the constant $1 - e^{-1}$.

Sketch: Let $f : F_2 \to A_n$ be a random homomorphism. It is well-known that $f$ is surjective with high probability. One can also check that $f(w)$ does not fix $1$ with high probability. The number of fixed points of $f(x)$ is asymptotically Poisson with mean $1$, so the probability that there are no fixed points tends to $e^{-1}$. $\square$

Now let $w_1, w_2, \dots$ be an enumeration of the nontrivial words in $F_2$. Let $n_0 = 5$. For each $i$ in turn choose $n_i > n_{i-1}$ and a surjective homomorphism $f_i : F_2 \to A_{n_i}$ such that $f_i(w_i)$ does not fix $1$ but $f_i(x)$ does have some fixed point. Let $H_i \le F_2$ be the inverse image of the stabilizer of $1$, so $w_i \notin H_i$. Let $\Gamma_n = H_1 \cap \cdots \cap H_n$. Each $H_i$ has index $n_i < \infty$, so $\Gamma_n$ has finite index. Since $w_i \notin H_i$ for each $i$, $\bigcap_{n=1}^\infty \Gamma_n = \{1\}$.

I claim that for any $n$, there is some conjugate of $x$ contained in $\Gamma_n$. Since the groups $A_{n_i}$ are distinct simple groups, the product map $f = f_1 \times \cdots \times f_n$ maps $F_2$ onto the direct product $\prod_{i=1}^n A_{n_i}$. Since $f_i(x)$ has a fixed point for each $i$, some conjugate of $f(x)$ is such that its projection to each $A_{n_i}$ fixes $1$. This implies that some conjugate of $x$ is contained in $\Gamma_n$.


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