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I know there are no elements of $S_4$ with order 12 from a list of the elements of $S_4$ but how can I prove it without listing all the elements with their orders?

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    $\begingroup$ It depends how much you know about the group. There's a way to compute the order of an element in terms of its cycle decomposition. $\endgroup$ – Matt Samuel Mar 13 '16 at 19:53
  • $\begingroup$ I don't know anything about cycle decomposition. $\endgroup$ – Jammie Dodger Mar 13 '16 at 19:58
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Every permutation can be written as a product of disjoint cycles. The order is the least common multiplier of the cycle-lengths.

If we have $4$ elements, order $12$ would require a $3$-cycle, but then there is no place for another cycle.

The possible orders for permutations with $4$ elements are $1,2,3,4$ because in the case of two non-trivial cycles, we can only have two transpositions.

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  • $\begingroup$ Could you briefly explain what cycles are as well as what you mean by "3-cycle"? $\endgroup$ – Jammie Dodger Mar 13 '16 at 20:06
  • $\begingroup$ The permutation sending $1234$ to $2314$ has a $3$-cycle, namely $1-2-3$ because it sends $1$ to $2$, $2$ to $3$ and $3$ to $1$. We have to apply this permutation three times to get the original permutation, therefore the order of this permutation is $3$. $\endgroup$ – Peter Mar 13 '16 at 20:10
  • $\begingroup$ In general , if a permutation has elements $x_1,...,x_k$, and it sends $x_1$ to $x_2$, $x_2$ to $x_3$, ... ,$x_{k-1}$ to $x_k$ and $x_k$ to $x_1$, it cotains a cycle of length $k$. The order of such a permutation must be a positive multiple of $k$. A cycle of length $2$ is called a transposition and a cycle of length $k$ a $k$-cycle. $\endgroup$ – Peter Mar 13 '16 at 20:12

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