202101141547 Homework 1 (Q5)

Let $G=\mathbb{Z}_2$ .

(a) Describe all the elements in $\mathrm{Perm}(G)$ and the maps $l_a$ for $a\in \mathbb{Z}_2$ . Is $l$ surjective? (Use the notation $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ to denote the map sending from $0$ to $1$ and from $1$ to $0$ .)

(b) Repeat part (a) for the group $G=\mathbb{Z}_3$ .

Setup.

$\textrm{Perm}(X)$ is the set of all bijective maps from $X$ to itself. And the group $(\textrm{Perm}(X),\circ )$ is the group where $\circ$ is the composition of maps.

$l:G\rightarrow \textrm{Perm}(G),\enspace a\mapsto l_a\qquad \forall\, a\in G$

is the group embedding of $G$ into $\textrm{Perm}(G)$ , where

$l_a:G\rightarrow G,\enspace g\mapsto ag\qquad g\in G$

is the left translation of $G$ .

(a)

Thus, the elements of $\textrm{Perm}(\mathbb{Z}_2)$ are altogether the following bijections:

$\begin{aligned} & \\ [0] & \mapsto [0] \\ [1] & \mapsto [1] \end{aligned}$

and

$\begin{aligned} & \\ [0] & \mapsto [1] \\ [1] & \mapsto [0] \end{aligned}$ ,

or equivalently, in matrices,

$\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$

The left translation $l_0$ is

$l_0:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,\enspace g\mapsto 0g\qquad g\in\{ 0,1\}= \mathbb{Z}_2$

But $[0]+[0]=[0]\quad (\mathrm{mod}\,2)$ and $[0]+[1]=[1]\quad (\mathrm{mod}\,2)$ ,
hence in matrix $l_0$ is $\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$ .

The left translation $l_1$ is

$l_1:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,\enspace g\mapsto 1g\qquad g=\{ 0,1\}\in \mathbb{Z}_2$

But $[1]+[0]=[1]\quad (\mathrm{mod}\,2)$ and $[1]+[1]=[0]\quad (\mathrm{mod}\,2)$ ,
hence in matrix $l_1$ is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ .

It is hence checked that the range of the left translations $l_0$ , $l_1$ is inside (or, more precisely, equal to) the codomain $\textrm{Perm}(\mathbb{Z}_2)$

Then, I can say $l:\mathbb{Z}_2\rightarrow\mathrm{Perm}(\mathbb{Z}_2)$ , the mappings of which are explicitly the following two:

$0\mapsto \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$ and $1\mapsto \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ ;

by the definition $l(a): a\mapsto l_a\qquad \forall\, a\in\{ [0],[1]\}=\mathbb{Z}_2$ .

Recall that a function is said to be surjective if the range of function constitutes its whole codomain. Thus, $l$ is surjective.

(b)

Let $G=\mathbb{Z}_3(=\{ [0],[1],[2] \})$ .

All the one-to-one mappings are listed as follows:

$\textrm{Perm}(\mathbb{Z}_3)=\Bigg\{ \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix},\begin{pmatrix} 0 & 0 \\ 1 & 2 \\ 2 & 1 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 2 & 2 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 1 \\ 2 & 0 \\ \end{pmatrix}\Bigg\}$ .

First, the left translation $l_0$ is

$l_0:\mathbb{Z}_3\rightarrow\mathbb{Z}_3,\enspace g\mapsto 0g\qquad g\in\{ 0,1,2\}=\mathbb{Z}_3$

Since $[0]+[0]=[0]\quad (\mathrm{mod}\,3)$ , $[0]+[1]=[1]\quad (\mathrm{mod}\,3)$ , and $[0]+[2]=[2]\quad (\mathrm{mod}\,3)$ , I may write

$l_0=\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix}$ .

Secondly, the left translation $l_1$ is

$l_1:\mathbb{Z}_3\rightarrow\mathbb{Z}_3,\enspace g\mapsto 1g\qquad g\in\{ 0,1,2\}=\mathbb{Z}_3$

Since $[1]+[0]=[1]\quad (\mathrm{mod}\,3)$ , $[1]+[1]=[2]\quad (\mathrm{mod}\,3)$ , and $[1]+[2]=[0]\quad (\mathrm{mod}\,3)$ , I may write

$l_1=\begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \\ \end{pmatrix}$ .

Thirdly and lastly, the left translation $l_2$ is

$l_2:\mathbb{Z}_3\rightarrow\mathbb{Z}_3,\enspace g\mapsto 2g\qquad g\in\{ 0,1,2\}=\mathbb{Z}_3$

Since $[2]+[0]=[2]\quad (\mathrm{mod}\,3)$ , $[2]+[1]=[0]\quad (\mathrm{mod}\,3)$ , and $[2]+[2]=[1]\quad (\mathrm{mod}\,3)$ , I may write

$l_2=\begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}$ .

To sum up what is up till now, $l_a=\Bigg\{ \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}\Bigg\}$

Then, I can say $l:\mathbb{Z}_3\rightarrow\mathrm{Perm}(\mathbb{Z}_3)$ , the mappings of which are explicitly the following three:

$0\mapsto \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix}$ , $1\mapsto \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0\\ \end{pmatrix}$ , and $2\mapsto\begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}$ .

by the definition $l(a): a\mapsto l_a\qquad \forall\, a\in\{ [0],[1],[2]\}=\mathbb{Z}_2$ .

$l$ in the case of $\mathbb{Z}_3$ is $\textrm{\scriptsize NOT}$ surjective because in the codomain $\mathbb{Z}_3$ , the following three elements

$\begin{pmatrix} 0 & 0 \\ 1 & 2 \\ 2 & 1 \\ \end{pmatrix}$ , $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 2 & 2 \\ \end{pmatrix}$ , and $\begin{pmatrix} 0 & 2 \\ 1 & 1 \\ 2 & 0 \\ \end{pmatrix}$

are not inside the range of $l$ .

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