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.

You have $\sigma = (1234)$ and $\tau=(14)(23)$. It is easy to see that $ord(\sigma) = 4$ and $ord(\tau)=2$. Moreover, you have $\tau\sigma\tau = \sigma^{-1} = (1432)$.
So, all elements of this group are: 1, $\sigma$, $\sigma^2$, $\sigma^3$, $\tau$, $\tau\sigma$, $\tau\sigma^2$, $\tau\sigma^3$ (that equal $\sigma\tau$)
