How do I find the order of a bijection? Question: List all bijections (permutations) from $\{1, 2, 3\}$ onto $\{1, 2, 3\}$. Find their order and sign.
I understand there will be n! permutations, namely:
$
\begin{Bmatrix}
    1 & 2 & 3 \\
    1 & 2 & 3 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
    1 & 2 & 3 \\
    1 & 3 & 2 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
    1 & 2 & 3 \\
    2 & 1 & 3 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1 \\
\end{Bmatrix}
$,$
\begin{Bmatrix}
    1 & 2 & 3 \\
    3 & 1 & 2 \\
\end{Bmatrix}
$,
$
\begin{Bmatrix}
    1 & 2 & 3 \\
    3 & 2 & 1 \\
\end{Bmatrix}
$
I understand that order of a permutation $\sigma$ is the smallest possible integer $k$ such that $\sigma^k = \epsilon$, where $\epsilon$ is the identity permutation.
But I am confused by the definition of "identity permutation". If my identity is:
$$
\begin{Bmatrix}
    1 & 2 & 3 \\
    1 & 2 & 3 \\
\end{Bmatrix}
$$
then order $= 0$ and sign = $(-1)^k = (-1)^0 = 1$.
And for: $
\begin{Bmatrix}
    1 & 2 & 3 \\
    1 & 3 & 2 \\
\end{Bmatrix}
$, order $= 1$, sign $= -1$. And for $
\begin{Bmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1 \\
\end{Bmatrix}
$ order $= 2$, sign $= 1$.
But if my first bijection from $\{1, 2, 3\}$ onto $\{1, 2, 3\}$ is:
$$
\begin{Bmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1 \\
\end{Bmatrix}
$$
then order $= 0$ and sign $= 1$. So depending on which projection I chose as an identity, the order differs. Can someone clarify?
 A: The identity is not something you choose; the identity means the permutation that fixes each element, so the first one you considered is the identity permutation. Secondly, the order of the identity permutation is $1$, because $\epsilon^1=\epsilon$ (the order must be positive). The order of $
\sigma_1=\begin{Bmatrix}
    1 & 2 & 3 \\
    1 & 3 & 2 \\
\end{Bmatrix}
$ is $2$ because $\sigma_1$ is not the identity permutation, but if you compose it with itself, you get the identity, so $\sigma_1^2=\epsilon$. The order of $\sigma_2=\begin{Bmatrix}
    1 & 2 & 3 \\
    2 & 3 & 1 \\
\end{Bmatrix}$ is $3$ because $\sigma_2\neq \epsilon$ and $\sigma_2^2\neq \epsilon$, but $\sigma_2^3=\epsilon$.
A: You obtain a direct answer to all questions decomposing permutations into a product of disjoint cycles. Concerning permutations of $\{1,2,3\}$, a permutation can be:


*

*a single $3$-cycle: $(1\,2\,3)$  or $(1\,3\,2)$

*a transposition: $(1\,2)$, $(1\,3)$ or $(2\,3)$

*the empty cycle $(\,)$ (corresponding to the identity)


Now, a cycle of length $k$ has order $k$ and signature $(-1)^{k-1}$.
