Suppose $\{x_k\}_{k=1}^m$ are independent random elements of $S_n$ under discrete uniform distribution. What is the probability $P_m^n$, that $\langle \{x_k\}_{k=1}^m \rangle = S_n$?

It is quite easy to determine $P_m^n$ in several specific cases: For$n \geq 3$ $P_1^n = 0$. $P_m^1 = 1$. $P_m^2 = 1 - \frac{1}{2^m}$. And for any fixed $n$, $\lim_{m \rightarrow \infty} P_m^n = 1$ However, the solution of this problem in general seems to be more complicated.

In general, one can see, that $P_m^n = 1 - Q_m^n$, where $Q_m^n$ is the probability, that $\{x_k\}_{k=1}^m$ all lie in the same maximal subgroup of $S_n$. And here I am stuck, as I do not know the structure of maximal subgroups of $S_n$.

Any help will be appreciated.

  • 1
    $\begingroup$ Trivial bound: $P_m^n\le P_m^2$ for $n\ge 2$, thanks to $A_n$. $\endgroup$ – Hagen von Eitzen Jul 1 '18 at 14:37
  • $\begingroup$ I am not sure what sort of answer you are expecting. As you say yourself here is not enough information known about the maximal subgroups of $S_n$ to be able to compute an exact value, so the best you can hope for are upper and lower bounds. I think there are some bounds known for the case $m=2$. $\endgroup$ – Derek Holt Jul 1 '18 at 16:50
  • $\begingroup$ @DerekHolt: In fact an asymptotic series is known for the case $m=2$ (see the end of my answer). $\endgroup$ – joriki Jul 2 '18 at 2:06

For any given $n$, this can be done using Möbius inversion on the lattice of subgroups of $S_n$. For an introduction to Möbius inversion, see e.g. Section $5.2$ of Martin Aigner's A Course in Enumeration. In a nutshell, it's a generalized form of the inclusion–exclusion principle.

I'll work it out for $S_3$ and $S_4$; I don't know of a systematic way of doing it for general $n$.

The subgroup structure of $S_3$ is described here, including this somewhat Enterprise-esque diagram:

subgroup structure of S_3

Let's denote the subgroups by $S_3$, $A_3$, $T_1$, $T_2$, $T_3$ and $I$ (from top to bottom and from left to right). Using the superset relation $\supseteq$ as the partial order $\le$, we can find the required values of the Möbius function using Aigner's Proposition $5.4$:


$$ \mu(a,b)=-\sum_{a\le z\lt b}\mu(a,z)\;, $$

and thus

\begin{eqnarray*} \mu(S_3,S_3)&=&1\;,\\ \mu(S_3,A_3)&=&-1\;,\\ \mu(S_3,T_i)&=&-1\;,\\ \mu(S_3,I)&=&3\;. \end{eqnarray*}

If we denote the probability for the subgroup generated by $m$ random elements to be contained in $H$ by $f_m(H)$ and the probability for that subgroup to be exactly $H$ by $g_m(H)$, then

$$ f_m(H)=\sum_{K\subseteq H}g_m(K)\;, $$

so we can use Möbius inversion from above (Aigner's Theorem $5.5$ (ii)) to obtain

$$ g_m(H)=\sum_{K\subseteq H}\mu(H,K)f_m(K) $$

and thus in particular

$$ g_m(S_3)=\sum_{K\subseteq S_3}\mu(S_3,K)f_m(K)\;. $$



this becomes


We can check this against a direct calculation for $m=2$, where we have to draw either a transposition and a $3$-cycle or two different transpositions to generate all of $S_3$, with probability

$$ 2\cdot\frac36\cdot\frac26+\frac36\cdot\frac26=\frac12\;, $$

and indeed


The subgroup structure of $S_4$ is analyzed here, including this amazing diagram:

subgroup structure of $S_4$

(Attribution for the diagram: By Watchduck (a.k.a. Tilman Piesk) - Own work, CC BY-SA 4.0, Link).

Using the diagram's labels for the types of subgroups (but with $D_4$ instead of $\text{Dih}_4$) and distinguishing the left-hand and right-hand types of $C_2^2$ and $C_2$ groups in the diagram by $L$ and $R$, respectively, we find as above:

\begin{eqnarray*} \mu(S_4,S_4)&=&1\;,\\ \mu(S_4,A_4)&=&-1\;,\\ \mu(S_4,D_4)&=&-1\;,\\ \mu(S_4,S_3)&=&-1\;,\\ \mu(S_4,C_4)&=&0\;,\\ \mu(S_4,C_2^{2L})&=&3\;,\\ \mu(S_4,C_2^{2R})&=&0\;,\\ \mu(S_4,C_3)&=&1\;,\\ \mu(S_4,C_2^L)&=&0\;,\\ \mu(S_4,C_2^R)&=&2\;,\\ \mu(S_4,C_1)&=&-12\;.\\ \end{eqnarray*}

Then Möbius inversion from above yields

$$ g_m(S_4)=1-2^{-m}-3\cdot3^{-m}-4\cdot4^{-m}+3\cdot6^{-m}+4\cdot8^{-m}+12\cdot12^{-m}-12\cdot24^{-m}\;. $$

Again we can check the case $m=2$ by hand. The following pairs of elements of $S_4$ generate all of $S_4$: each one of $6$ transpositions with one of $4$ $3$-cycles or with one of $4$ $4$-cycles, each one of $8$ $3$-cycles with any one of $6$ $4$-cycles, and each one of $6$ $4$-cycles with one of $4$ $4$-cycles, for a probability of

$$ \frac{2\cdot6\cdot4+2\cdot6\cdot4+2\cdot8\cdot6+6\cdot4}{24^2}=\frac38\;, $$

and indeed

$$ g_2(S_4)=1-2^{-2}-3\cdot3^{-2}-4\cdot4^{-2}+3\cdot6^{-2}+4\cdot8^{-2}+12\cdot12^{-2}-12\cdot24^{-2}=\frac38\;. $$

For $m\to\infty$, the leading term will always be the one corresponding to $A_n$, so the probability not to generate all of $S_n$ is asymptotic to $2^{-m}$ for all $n$.

For $m=2$, the problem has been extensively studied; see Probability that two randomly chosen permutations will generate $S_n$. The main result quoted there is the following asymptotic series from John Dixon's paper Asymptotics of generating the symmetric and alternating groups:


The numerators form OEIS A113869. This series applies to both the probability that two uniformly randomly chosen elements of $A_n$ generate $A_n$ and the probability that two uniformly randomly chosen elements of $S_n$ generate either $A_n$ or $S_n$; it follows that

$$ g_2(S_n)\sim\frac34t_n\;. $$

The last section of this paper also contains a generalization of this asymptotic series for $m\gt2$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.