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I was trying to answer the following question recently: What is the proportion of elements of order $p$ in the symmetric group $S_n$ , where $p$ is some prime number ?

I managed to work out that in general, the number of elements of order $p$ in $S_n$ is:

$$ {n \choose p} (p-1)! +\displaystyle \sum_{k=2}^{[\frac{n}{p}]} \frac{n!}{k!p^k},$$

where $[x]$ is the greatest integer less than or equal to $x$.

Using this, as well as the Taylor series at $x=0$ for $\large e^{\frac{1}{p}}$, I was able to determine that the proportion of elements of order $p$ in $S_n$ as $n \to \infty$ is:

$\displaystyle \sum_{k=2}^{\infty} \frac{1}{k!p^k} = \large e^{\frac{1}{p}} - \small \frac{p+1}{p}$

I was wondering what significance, if any, this has - it feels somewhat similar to the result that states that the limit of the ratio of the number of derangements of $n$ elements to $n!$ is $\frac{1}{e}$, but admittedly I don't fully understand the significance of this result either.

Any help would be much appreciated.

Thanks!

Shaun
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porridgemathematics
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1 Answers1

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There is a well-known formula for the generating function of the proportion of elements of order $1$ or $p$ in $S_n$, i.e., the proportion of $g$ in $S_n$ that satisfy $g^p = 1$. Letting $c_n$ be the number of such solutions, with the convention that $c_0 = 1$, the generating function is $$ \sum_{n\geq 0} \frac{c_n}{n!}x^n = e^{x+x^p/p}. $$ In particular, since this series converges at $x=1$, the proportion $c_n/n!$ tends to $0$, so you are making an error when you say the proportion has a positive limit as $n\rightarrow \infty$.

Counting the number of elements of order $1$ or $p$ in $S_n$ is the same as counting homomorphisms from $\mathbf Z/(p)$ to $S_n$. More generally, for any finite group $G$ we have $$ \sum_{n\geq 0} \frac{\#{\rm Hom}(G,S_n)}{n!}x^n = e^{\sum_{H\subset G} x^{[G:H]}/[G:H]}, $$ where we make the convention that $S_0$ is trivial and the sum in the exponent on the right runs over all subgroups $H$ of $G$. Taking $G= \mathbf Z/(p)$ recovers the first formula. This formula goes back to Wohlfahrt. If you google "wohlfahrt group formula" you'll find references to related work.

KCd
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  • My group theory is very very rudimentary, so I'm guessing you are probably right when you say I've committed error. I think I've managed to work out an expression for $c_n$ in my post. In which case $\frac{c_n}{n!}$ (If my expression for $c_n$ is correct) definitely seems to converge to a positive real number, as $\frac{n \choose p}{n!}$ tends to zero, and the sequence $\frac{S_n}{n!}$ where $S_n$ is $\displaystyle \sum_{k=2}^{[\frac{n}{p}]} \frac{n!}{k!p^k}$ is just $\displaystyle \sum_{k=2}^{[\frac{n}{p}]} \frac{1}{k!p^k}$ which certainly has a positive limit as $n \to \infty$ – porridgemathematics Mar 20 '16 at 13:26
  • Do you mind explaining why my expression for $c_n$ is incorrect? – porridgemathematics Mar 20 '16 at 13:27
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    You are not checking far enough. For $p=2$ your formula breaks down starting at $n=6$, for $p=3$ it breaks down starting at $n=8$ (you only say you checked up to $n=6$), and for $p=5$ it breaks down starting at $n=12$. You'll have to figure out yourself why your formula is wrong. Presenting a formula that doesn't work and then asking others why it doesn't work when you never explained where your formula came from is not reasonable. – KCd Mar 20 '16 at 13:56
  • By the way, in your first comment, using $S_n$ for a number when you're talking about something involving the symmetric group is pretty bad notation. – KCd Mar 20 '16 at 13:58
  • Noted, I'll try and fill in how I derived the formula so that the error I commit becomes easier to spot – porridgemathematics Mar 20 '16 at 14:00
  • I advise you to first try to determine the error yourself before asking others to do that, e.g., take $p = 2$ and figure out why your formula is wrong when $n = 6$. – KCd Mar 20 '16 at 14:05
  • Yes, I've found the error, I made a careless mistake – porridgemathematics Mar 20 '16 at 14:11
  • Hey, you deleted one of your comments. You shouldn't do that when other comments are made in response to them, since it causes the discussion to read awkwardly. I wrote that you checked $p=3$ only up to $n=6$, but the earlier comment where you said you did that is now missing, so it appears now that I'm omniscient or something. – KCd Mar 20 '16 at 17:53