# Is it true that $\lim_{n \to \infty} {(P(\forall i,j\leq n \text{ } [X_i, X_j] = e))}^{\frac{1}{n}} = P(X_1 \in Z(G))$?

Suppose $$G$$ is a group. $$\{X_n\}_{n = 1}^{\infty}$$ is a sequence of i.i.d. random elements of $$G$$ satisfying the condition that

$$\forall H \leq G, \qquad P(X_1 \in H) = \begin{cases} \frac{1}{[G:H]} & \quad \text{if [G:H] is finite}\\ 0 & \quad \text{if [G:H] is infinite} \end{cases}$$

Is it true that

$$\lim_{n \to \infty} P(\forall i,j\leq n, \ [X_i, X_j] = e)^{\frac{1}{n}} = P(X_1 \in Z(G)) \ ?$$

What have I tried so far?

If we accept an additional supposition, that the events $$\{\forall i \leq p, \ X_i \in C_G(X_p) \}$$ and $$\{\forall i \leq q, \ X_i \in C_G(X_q) \}$$ are independent for any natural $$p \neq q$$. Then we can see, that

$$P(\forall i,j\leq n, \ [X_i, X_j] = e) = \prod_{i = 1}^{n} P(\{\forall j \leq i, \ X_j \in C_G(X_i) \}).$$

Now, let’s see, that on one hand

\begin{align*} &P(\{\forall j \leq i, \ X_j \in C_G(X_i) \}) \\ &= P(X_i \in Z(G)) + (1 - P(X_i \in Z(G))P(X_1 \in C_G(X_i))^{i - 1} \\ &\leq P(X_1 \in Z(G)) + (1 - P(X_1 \in Z(G))\left(\frac{1}{2}\right)^{i - 1} \\ &= \frac{1}{2^{i - 1}} + \left(1 - \frac{1}{2^{i - 1}}\right)P(X_1 \in Z(G)) \\ &= P(X_1 \in Z(G))\left(1 - \frac{1}{2^{i - 1}} + \frac{1}{2^{i - 1}P(X_1 \in Z(G))}\right) \end{align*}

and on the other hand

\begin{align*} &P(\{\forall j \leq i, \ X_j \in C_G(X_i) \}) \\ &= P(X_i \in Z(G)) + (1 - P(X_i \in Z(G))P(X_1 \in C_G(X_i))^{i - 1} \\ &\geq P(X_1 \in Z(G)) + (1 - P(X_1 \in Z(G))P(X_1 \in Z(G))^{i - 1} \\ &= P(X_1 \in Z(G))^{i - 1} + \left(1 - P(X_1 \in Z(G))^{i - 1}\right)P(X_1 \in Z(G)) \\ &= P(X_1 \in Z(G))\left(1 - P(X_1 \in Z(G))^{i - 1} + P(X_1 \in Z(G))^{i - 2}\right) \end{align*}

So, we have

\begin{align*} &P(X_1 \in Z(G)) \\ &= \lim_{i \to \infty} P(X_1 \in Z(G))\left(1 - P(X_1 \in Z(G))^{i - 1} + P(X_1 \in Z(G))^{i - 2}\right) \\ &= \lim_{n \to \infty} \left( \prod_{i = 1}^n P(X_1 \in Z(G)) \left(1 - P(X_1 \in Z(G))^{i - 1} + P(X_1 \in Z(G))^{i - 2} \right) \right)^{\frac{1}{n}} \\ &\leq \lim_{n \to \infty} P(\forall i,j\leq n, \ [X_i, X_j] = e)^{\frac{1}{n}} \\ &\leq \lim_{n \to \infty} \left( \prod_{i = 1}^n P(X_1 \in Z(G)) P(X_1 \in Z(G)) \left(1 - \frac{1}{2^{i - 1}} + \frac{1}{2^{i - 1}P(X_1 \in Z(G))} \right) \right)^{\frac{1}{n}} \\ &= \lim_{i \to \infty} P(X_1 \in Z(G))\left(1 - \frac{1}{2^{i - 1}} + \frac{1}{2^{i - 1}P(X_1 \in Z(G))} \right) \\ &= P(X_1 \in Z(G)). \end{align*}

However, I do not know how to prove that the events in our supposition are always independent (or is there a counterexample?). And neither do I know, how to prove the main statement of the question without using the aforementioned supposition.

• The events are not independent, they can't be. – zhoraster Feb 19 '19 at 8:38

I believe $$S_3$$ is a counterexample. Let $$H< S_3$$ be the unique subgroup of index $$2$$. Then $$H$$ is abelian, and for every $$n$$ we have $$P(\forall i,j:[X_i,X_j]=e)^{1/n} \geq P(\forall i: X_i\in H)^{1/n}=\frac{1}{2}.$$ This means the limit on the left (if it exists) is at least $$1/2$$. On the other hand, $$Z(S_3)=\{e\}$$, so $$P(X_1\in Z(S_3))=P(X_1=e)=1/6$$.
For general $$G$$, a similar argument should show that $$\liminf_{n\to\infty}P(\forall i,j\leq n:[X_i,X_j]=e) \geq \max_{\substack{H\leq G\\H\text{ abelian}}}\frac{1}{[G:H]}.$$
For finite $$G$$, the bound given in Julian Rosen's answer is the exact limit, i.e. I claim that $$\mathrm P(\forall i,j\le n,\ [X_i,X_j] = e)^{1/n}\to \max_{\text{abelian }H \le G}\frac{1}{[G:H]} = \max_{\text{abelian }H \le G}\mathrm P(X_1 \in H). \tag{\ast}$$
Indeed, denoting by $$\mathcal A$$ the collection of abelian subgroups of $$G$$, $$\limsup_{n\to \infty} \mathrm P(\forall i,j\le n,\ [X_i,X_j] = e)^{1/n} \le \limsup_{n\to \infty}\biggl(\sum_{H\in \mathcal A} \mathrm P(\forall i\le n,\ X_i\in H)\biggr)^{1/n} =\\ = \limsup_{n\to \infty}\biggl(\sum_{H\in \mathcal A} \mathrm P(X_1\in H)^n\biggr)^{1/n} = \max_{H\in \mathcal A} \mathrm P(X_1\in H).$$
Since by Julian Rosen's answer, $$\liminf_{n\to \infty} \mathrm P(\forall i,j\le n,\ [X_i,X_j] = e)^{1/n}\ge \max_{H\in \mathcal A}\mathrm P(X_1 \in H),$$ we arrive at $$(\ast)$$.
In order for (a modification of) the argument to work, it is enough to assume that for some $$n\ge 1$$, $$\sum_{H\in \mathcal A^*} \mathrm P(X_1\in H)^n<\infty$$, where $$\mathcal A^*$$ is the collection of all maximal abelian subgroups of $$G$$.