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 should also notice that $\sigma \tau = \tau \sigma^{3}$. – Morgan Rodgers Aug 5 '16 at 6:50

Note $\sigma$ and $\sigma^2\sigma^3$ are the same since $\sigma^4$ is the identity permutation.

Anyway, if you label the points of a unit square $1,2,3,4$ counterclockwise, then $\sigma$ is a right angle rotation and $\tau$ is a reflection. Evidently $\sigma\tau=\tau\sigma^{-1}$ (this rule defines all dihedral groups $D_{2n}$, and even the orthogonal group $\mathrm{O}(2)$).

The elements should be $\sigma^0,\sigma^1,\sigma^2,\sigma^3$ (rotations) and $\sigma^0\tau,\sigma^1\tau,\sigma^2\tau,\sigma^3\tau$ (reflections). The latter four can be relabelled $\tau\sigma^0,\tau\sigma^1,\tau\sigma^2,\tau\sigma^3$ if that floats your boat.

• +1. Ignoring the fact that $\sigma^5=1\cdot \sigma$ is part of what causes the miscount. Checking element orders is important! – Nij Aug 5 '16 at 7:02

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 $\tau\sigma= \sigma^{-1}\tau =\sigma^{n-1}\tau$

• Conventions differ about what subscript to put under $D$. – arctic tern Aug 5 '16 at 6:53
• @arctictern I have heard that there are other conventions. But most books commonly used adopt this convention: $n$ is t the number of sides of the regular polygon for which it is the group of symmetries. – P Vanchinathan Aug 5 '16 at 7:06
• @PVanchinathan Just as many books adopt the notation that $D_{n}$ is the dihedral group of order $n$. – Morgan Rodgers Aug 5 '16 at 7:06

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 $\sigma\tau$)