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?

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 $\sigma_2^3=\epsilon$.
