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Let $S_7$ denote the symmetric group of permutations of $7$ symbols. As $|S_7|=7!$ and $10$ divides $7!$, is there an element of order $10$ in this group. If yes then what is it? Also what would be the largest possible order of any element in $S_7$?

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  • $\begingroup$ $\begin{pmatrix} 1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 4 & 5 & 6 & 7 \end{pmatrix}$. $\endgroup$ – jgon Nov 28 '17 at 7:08
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Any product of two disjoint cycles of order $\;2,\, 5\;$ resp. is an element of order $\;10\;$ , and there are elements of order $\;12\;$ in $\;S_7\;$ , for example $\;(1\,2\,3)(4\,5\,6\,7)\;$ .

As any element in $\;S_n\;$ can be expressed as the product of different cycles, and a product of disjoint cycles has order the minimal common product of their lengths, now you answer your own last question.

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The order of an element in the symmetric group is the lcm of the sizes of its disjoint cycles. So $\begin{pmatrix} 1 & 2 \end{pmatrix}\begin{pmatrix} 3 & 4 & 5 & 6 & 7 \end{pmatrix}$ has order 10. And the order of an element must be less than the largest of $(7/k)^k$ as $k$ varies. Note that $(7/2)^2 = 49/4 > 12$, and in fact there is an element of order 12, $\begin{pmatrix} 1 & 2 & 3\end{pmatrix}\begin{pmatrix} 4 & 5 & 6 & 7 \end{pmatrix}$, and this has the largest possible order as one can see by graphing $\left(\frac{7}{x}\right)^x$ and observing that that it peaks just above 13, but 13 is impossible as an order, since this would require a cycle of size 13.

In general, to find a bound on the order of an element of $S_n$, we can look at the derivative of $\left(\frac{n}{x}\right)^x$, $$\frac{d}{dx}\exp(x(\ln n -\ln x))=((\ln n -\ln x)+x(-1/x))\exp(x(\ln n-\ln x)),$$ which is 0 when $\ln x= \ln n - 1$, or $x=\frac{n}{e}$. Hence the maximum value of $\left(\frac{n}{x}\right)^x$ is $e^{n/e}$, so this gives a bound on the order of elements of $S_n$.

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