Question about Symmetric groups 
Given permuations $$
\sigma=
\begin{pmatrix}
  \text{1 2 3 4} \\
   \text{2 3 4 1}\\
\end{pmatrix}
\qquad\tau=
\begin{pmatrix}
  \text{1 2 3 4} \\
   \text{4 3 2 1}\\
\end{pmatrix},
$$
  show that the subgroup $D_8:=\langle\tau,\sigma\rangle$ of $\operatorname{Sym}(4)$ has order 8 and write down its elements

With the identity element, I got 9. The elements I got are $\sigma,\sigma^2\sigma^3,\tau,\tau\sigma,\sigma\tau,\tau\sigma^2,\tau\sigma^3,1$
It's important to note that $\tau\sigma^2=\sigma^2\tau$ and $\tau\sigma=\sigma^3\tau$, so I just chose one representative.

First notational issue: It should be $D_4$, and is a group of order 8.
There is a repetition, because: $\sigma\tau=\tau\sigma^{-1}=\tau\sigma^3$.
In all dihedral groups $D_n$ assuming that $\tau$ is any reflection and $\sigma$ is the generator of  the cyclic subgroup of order $n$ we have $\tau\sigma= \sigma^{-1}\tau =\sigma^{n-1}\tau$
