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Let $S_n$ denote the symmetric group on $n$ letters. We know that $\tau^{n!} = e$ for any element $\tau \in S_n,$ where $e$ denotes the identity element. Can we find a smaller positive integer $m$ with this property? That is, can we find a positive integer $m < n!$ such that $$\tau^m = e$$ for all $\tau \in S_n$?

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Let $\sigma$ be an $n$-cycle. Then $\sigma^m\neq 1$ if $m<n$ since $\sigma$ has order $n$. – Ian Coley Sep 14 '13 at 8:39
I think you are asking the wrong question. For example $(1, 2)^3=(1, 2)\neq e$ in $S_3$. The order of the group $S_n$ is $n!$ – Mark Bennet Sep 14 '13 at 8:42
I think I should have written $\tau^{n!} = e$. So is there a positive integer $m$ smaller than $n!$ such that $\tau^m=e$ for each element $\tau \in S_n$? – Saaqib Mahmuud Sep 14 '13 at 8:50
I would suggest to change the title to something like What is the exponent of $S_{n}$? See – Andreas Caranti Sep 14 '13 at 10:25
See for some interesting facts about these exponents. Note that the title of the sequence in the OEIS corresponds to Jyrki Lahtonen's answer and the first comment corresponds to mine. – Rob Arthan Sep 14 '13 at 11:27

The answer is yes for $n \ge 4$. When you write a permutation $\tau$ as a product of disjoint cycles, its period $p$ is the lcm of the lengths of the cycles. So if $2^k$ divides $p$, $\tau$ contains a $2^k$ cycle and so $2^k \le n$. For $n \ge 4$, $n! = 2^rs$ where $s$ is odd and $2^r > n$, so $p$ will divide $m = 2^ks$, where $k$ is the largest integer such that $2^k \le n$. Similarly you can reduce the exponents of other prime divisors of $n!$ as soon as $n$ is large enough

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Hint (do all the steps): The order of a $k$-cycle is $k$. Any element of $S_n$ is a disjoint (hence commuting) product of cycles of length $\le n$. The order of a product of disjoint cycles is the l.c.m. of the lengths of the individual cycles. The smallest exponent that works with all the permutations in $S_n$ is $$ l.c.m.\{k\mid 1<k\le n\}. $$

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I was surprised that I didn't find this question already answered here. The closest related question may be this. – Jyrki Lahtonen Sep 14 '13 at 9:37

You do not need actually the symmetric group to do such a statement (though you should here correct the $n$ to $n!$).

Consider G a nontrivial group of order $n$.

Now either this group is cyclic (it has an element of order $n$) or it is not. If it is not the answer to your question is trivially yes: there is at least one element with order $m \neq n$. Now suppose it is a cyclic group, then $\langle g \rangle=G$. Now let us consider the case of n non prime and let us say that $m$ divedes $n$: $mk=n \Leftrightarrow g^{mk}=e \Leftrightarrow (g^{m})^{k} = e$

This proves that $g^m$ has order $k < m$.

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