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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
  • $\begingroup$ Another hint: what is the prime factorization of $10$? $\endgroup$ – kimchi lover 7 hours ago
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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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10 divides 7!, is there an element of order 10 in this group?

Although in $S_7$ this maybe true (10 divides 7! and the group has an element of order 10), if you look at $S_5$, and 10 divides 5!, there isn't any element of order 10.

This is because although the order of the element must divide the order of the group, just because 10 is a divisor of the order of the group that doesn't mean the group must have an element of order 10.

This is only true for prime numbers. From Cauchy's Theorem:

If p, a prime, divides the order of the group G, then G has an element of order p.

In the case of $S_5$.

(1) $\#S_5=120=2^3 \times3\times 5 $, so by Cauchy's theorem we know it has elements of order 2, 3 and 5

(2) $S_5$ has all the permutations from 1 to 5

From (2) we know it won't have a single cycle of 10 numbers, so the only other way we can have an element of order 10 is if it's a product of two disjointed cycles of order 2 and 5.

However, since all elements of order 2 are translations $(i k)$ of two numbers from 1 to 5 and all elements of order 5 have to have all the numbers from 1 to 5, the cycles are never disjointed, so there aren't any elements of order 10.

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