Order of a permutation, how to calculate
I know this is a basic question however I am slightly confused what to do if the permutation contains the same element twice in cycle notation.
For example: the permutation $(1 2 3)(2 4 1)$, how would I calculate the order when $2$ maps to $3$ and $4$? Is it just the same? Does $p=3$
$\endgroup$ 21 Answer
$\begingroup$First you'll need to express $(123)(241)$ in terms of the product of disjoint cycles.
$(123)$ and $(241)$ are not disjoint cycles, as you note, since both share the elements $1, 2$.
To do so, you start from the right cycle, and compose with the left cycle. So, in the right-hand cycle, we have $1\mapsto 2$ and in the lefthand cycle, $2\mapsto 3$. Now since $3\mapsto 3$ (righthand) and $3\maps 1$ (lefthand, we have the cycle (13).
In the end, you'll find $$\phi = (123)(241)=(13)(24)$$
Second, the order of a one-cycle permutation is its length; to find the order of a product of more than one permutation cycle, as is the case here, the order of $\phi$ is the $\operatorname {lcm}$ of the lengths of the cycles.
So we have the product of two $2$-cycles, and hence the order of $\phi$ is equal to the $\operatorname {lcm}(2, 2) = 2$.
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