# Conjugacy classes in towers of groups

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?

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

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 $$ in absolute value, we first deduce that $$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 $$m(1,0,0)$$.

• 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 $\mathbb{Z}^2$. Oct 2 at 15:41
• @DavidESpeyer I believe $m(1,a_n,0)$ should be $m(1,0,a_n)$. Also $m_n^{a_n^2+1} = m(a_n^2+1, 0, a_n (a_n^2+1))$. Oct 2 at 15:58
• 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.
– YCor
Oct 2 at 16:24
• 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 $m(1,0,0)$. 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 $$\Gamma_n$$.