Question

(2) Let De = {1, 0, 02, 03, 04, 0%, 7, OT, O2T, 07,041, 0 T} where o = 1 2 3 4 5 6 2 3 4 5 6 1 and T = 2 3 4 5 6 6 5 4 3 2 so that De is the dihedral group of degree 6. (i) (2 marks) Calculate to. (ii) (2 marks) Give two elements of De that have order 3. No explanation required. (iii) (4 marks) Give two subgroups of De that have order 6. No explanation required.

Answer 1

(i)

We calculate where each element is sent.

1246

$2\overset{\sigma}{\longrightarrow} 3 \overset{\tau}{\longrightarrow} 5$

34号 4

$4\overset{\sigma}{\longrightarrow} 5 \overset{\tau}{\longrightarrow} 3$
$5\overset{\sigma}{\longrightarrow} 6\overset{\tau}{\longrightarrow} 2$
$6\overset{\sigma}{\longrightarrow} 1\overset{\tau}{\longrightarrow} 1$ .

That is, $\tau \sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 5 & 4 & 3 & 2 & 1 \end{pmatrix}$ .

(ii)

Let
p= 1 2 3 4 5 6 5 6 1 2 3 4
. Then $\rho^2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 4 & 5& 6 & 1 & 2 \end{pmatrix}$ and $\rho^3 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 1 & 2 & 3 & 4 & 5 & 6 \end{pmatrix}= e$ so that $\rho$ has order three. $\rho$ can be thought as the rotation by $120^\circ$ of the hexagon.

Other element of order 3 is, for example, $\rho^{-1} = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 3 & 4 & 5 & 6 & 1 & 2\\ \end{pmatrix}$ . It is easy to see that $\rho^{-1}$ has order three since its orbit is the same of $\rho$ .

(iii) Let $\sigma$ be the rotation by $60^\circ$ . That is, $\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6\\ 2 & 3 & 4 & 5 & 6 & 1\\ \end{pmatrix}$ . It is easy to see that $\sigma$ has order six. Then, $H = \langle \sigma\rangle = \{e,\sigma,\sigma^2,\sigma^3,\sigma^4,\sigma^5\}$ has order six.

Let $\tau = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 1 & 6 & 5 & 4 & 3 & 2\end{pmatrix}$ and $\rho$ as in (ii). Then the group $\langle \rho, \tau \rangle$ has order six. Here $\rho$ is a reflection of order two.