Trial and Error – Physicspupil

January 14, 2021October 24, 2022

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 $l$ .

January 14, 2021May 4, 2023

202101141538 Homework 1 (Q4)

$\phi :G_1\rightarrow G_2$ is said to be a group embedding if it is an injective group homomorphism.

For a group $G$ and $a\in G$ , we define the left translation of $G$ by $a$

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

Thus $l$ is a mapping from a value into a function of the value.

I would like to first check that $l$ is a homomorphism,

i.e., $l(a_1a_2)=l(a_1)l(a_2)$ .

$\begin{aligned} \textrm{LHS} & = l(a_1a_2)\\ & =l_{a_1a_2}(g)\qquad\qquad (g\in G)\\ & = a_1a_2g\qquad\qquad (a_1,a_2\in G)\\ & = l_{a_1}(l_{a_2}(g))\\ & =(l_{a_1}\circ l_{a_2})(g)\qquad\qquad (\circ\,\textrm{as the operation in }\textrm{Perm}(G))\\ & = (l_{a_1}l_{a_2})(g)\\ & = l(a_1)l(a_2)\\ & = \textrm{RHS}\\ \end{aligned}$

$\therefore$ $l$ is a homomorphism.

Secondly, I would like to check that $l$ is injective.

If $l(a_1)\neq l(a_2)$ , or $l_{a_1}(g)\neq l_{a_2}(g)$ , or, $a_1g\neq a_2g$ for $a_1,a_2, g\in G$ , then $a_1gg^{-1}\neq a_2gg^{-1}$ guaranteed that the inverse of $g$ always exists in $G$ . It follows that $a_1\neq a_2$ .

$\therefore$ $l$ is injective.

In conclusion, $l$ is a group embedding.

January 14, 2021May 20, 2024

202101141527 Homework 2 (Q3)

Let $G$ be a group and $e\neq a\in G$ in which $e$ is the identity of $G$ . Suppose $\mathrm{ord}(a)=n$ .

i. If $a^h=e$ , show that $n\big| h$ .

ii. Evaluate the order of $a^m$ where $m$ is an integer.

iii. Show that $|H|=n/(m,n)$ if $G=\langle a\rangle$ and $H=\langle a^m\rangle$ where $m\in\mathbb{N}$ .

iv. If G is cyclic of order $n$ , show that for any $d\big| n$ , there exists a subgroup of order $d$ in $G$ . Could $G$ have more than one subgroup of order $d$ ? (Clearly the answer is no if $d=1$ or $n$ .) Justify with a proof (if you think yes) or give a counterexample (if you think no).

Attempts.

i. By Division Algorithm, $h=qn+r$ for some $q\in\mathbb{Z}$ and $0\le r<n$ . Then one may write $a^r=a^{h-qn}=a^h(a^n)^{-q}=e$ . It can be seen that $r$ must be $0$ lest $a^r=e$ where $1\le r<n$ contradicts to the assumption that $n=\mathrm{ord}(a)$ where $n$ is the smallest positive integer such that $a^n=e$ . Thus $h=qn$ , and $n|h$ .

ii. First, $(a^m)^{\frac{n}{(n,m)}}=a^{\frac{mn}{(n,m)}}=(a^n)^{\frac{m}{(n,m)}}=e^{\frac{m}{(n,m)}}=e$ . Secondly, if $k\in\mathbb{N}$ such that $(a^m)^k=e$ , from the result of part i. previously, it follows that $\frac{n}{(n,m)}|k$ because if $a|bc$ and $(a,b)=1$ , then $a|c$ (here $a=\frac{n}{(n,m)}$ ), $b=\frac{m}{(n,m)}$ , and $c=k$ . Such a $k$ will always be greater than or equal to $\frac{n}{(n,m)}$ . Thus $\mathrm{ord}(a^m)=\frac{n}{(n,m)}$ .

iii. If $H=\langle a^m\rangle$ then $|H|=\mathrm{ord}(a^m)\stackrel{\textrm{by (ii)}}{=}\frac{n}{(n,m)}$ .

iv. Let $G=\langle a\rangle$ as it is cyclic. Let $m=n/d$ where $d|n$ is given in the problem. The order of the cyclic subgroup $\langle a^m\rangle$ of $G$ is then $\frac{n}{(n,m)}=\frac{n}{m}=d$ , in view of the result in part iii. We have constructed a subgroup of order $d$ in $\langle a\rangle$ where $d|n$ .

We wish to prove that it is unique: If $H$ is a subgroup of $G=\langle a\rangle$ and $|H|=d$ , then $H=\langle a^{\frac{n}{d}}\rangle$ . Note that $H=\langle a^m\rangle$ and $\frac{n}{(n,m)}=d$ . Consider the subgroup $\langle a^{\frac{n}{d}}\rangle$ . Because $\frac{n}{d}=(n,m)$ divides $m$ , we have $a^m\in \langle a^{\frac{n}{d}}\rangle$ and thus $H\subset \langle a^{\frac{n}{d}}\rangle$ . From the result of part iii. it follows that $|\langle a^{\frac{n}{d}}\rangle | =\frac{n}{(n,\frac{n}{d})}=d$ . Thus $H=\langle a^{\frac{n}{d}}\rangle$ is unique.

January 14, 2021October 24, 2022

202101141501 Homework 2 (Q1)

Let $H$ be a subset of the group $G$ . Show that $H$ is a subgroup of $G$ if and only if $H\neq \emptyset$ and $gh^{-1}\in H$ whenever $g,h\in H$ .

Attempts.

(only-if part) If $H$ is a subgroup of $G$ , then $H$ must contain the identity $e$ and both $x^{-1}$ and $xy$ is inside $H$ whenever $x,y\in H$ . Thus $H\neq \emptyset$ and for any $g,h\in H$ , $h^{-1}$ is inside $H$ . Thus $gh^{-1}\in H$ .

(if-part) First, because $H\neq \emptyset$ , there exists some $x$ in $H$ such that $e=xx^{-1}\in H$ (by the condition $gh^{-1}\in H$ whenever $g,h\in H$ ). This proves the existence of the identity. Secondly, having proven $e$ is in $H$ , and noted that $x^{-1}=ex^{-1}\in H$ whenever $e,x\in H$ , one proves the existence of the inverse. Thirdly, $x,y\in H$ implies $xy\in H$ because $y\in H\Rightarrow y^{-1}\in H$ and also $xy=x(y^{-1})^{-1}$ . $H$ is therefore a subgroup of $G$ .

April 24, 2020February 5, 2023

202004240713 Homework 1 (Q3)

Suppose $G$ be a finite group of even order. Prove that there exists an element $a\in G$ such that $a\neq e$ and $a^2=e$ .

Attempts.

Try negating the statement by the following:

$\forall\, a\in G$ , either $a=e$ or $a^2\neq e$ ,

then looking for contradiction to the assumption that $G$ should be a finite group of even order.

CASE 1: Would it be possible that $a=e$ ?

Given so, the inverse and the identity of $G$ would be $e$ . And the set $G$ would have contained $e$ only, i.e., the singleton set $G=\{ e\}$ . This contradicts with the assumption that its order be even.

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .

If $a^2\neq e$ , then $a\neq a^{-1}$ for all $a\in G$ . It would then become an ill-posited negation, because the identity element $e$ is one of all $a\in G$ , and both statements $e^2\neq e$ and $e\neq e^{-1}$ are abhorrent to the basic axioms of a group. So perhaps to my discretion $e$ should be precluded from $a$ ‘s, and understood as one ( $e=e^{-1}$ ) member apart from the other members.

Proceeding to observe that if $a\neq a^{-1}$ for all $a\in G$ , I may then construct a pair of $a_i$ and $a_i^{-1}$ for $i\in \{ 1,2,\dots ,n:n\in\mathbb{Z}^+\}$ , provided from the definition of a group that the inverse of each element must exist and, as it follows, that it must be unique. That is, if $a_5=a_{27}^{-1}$ , I may relabel $a_{27}$ as $a_5^{-1}$ . By such a construction I can guarantee that all elements, except the identity $e$ , are now in pairs.

One therefore, by counting in total how many elements there are in group $G$ , will get an odd number $(2n+1)$ , the $2n$ counted from the pairs, and the single $1$ the identity $e$ itself alone counts.

This contradicts with the assumption that $|G|$ be of even parity.

From the negation of statement arises contradiction, the original statement is therefore proven by contradiction.

Remark.

I could not prove otherwise than loosely to so naive myself. Not until I had survived mathematical rigor.

April 24, 2020October 24, 2022

202004240630 Homework 1 (Q2)

Call $G$ a group and $e$ the identity of $G$ . If the order of $G$ is $2$ , then $a^2=e$ for all $a\in G$ and $G$ is abelian (Prove it yourself). Show that

$G$ is abelian if $a^2=e$ for all $a\in G$ .

Give an example of such a group whose order is greater than $2$ .

Solution.

Recall that a group $G$ is said to be abelian if $xy=yx$ for all $x,y\in G$ .

For any $a,b\in G$ , there is $ab=(bb^{-1})ab(a^{-1}a)$ , because a group satisfies, first, the inverse axiom $xx^{-1}=x^{-1}x=e$ under the notation $x^{-1}$ being the inverse of $x$ , and secondly, the identity axiom $ex=xe=x$ .

Thence by (associativity axiom) $x*(y*z)=(x*y)*z$ I may change brackets and obtain as follows:

$\begin{aligned} ab & =(bb^{-1})ab(a^{-1}a) \\ & =b(b^{-1}a)(ba^{-1})a \\ & =b(b^{-1}a)(b^{-1}a)^{-1}a \end{aligned}$

Provided that $a^2=e$ for all $a\in G$ , one may deduce $a=a^{-1}$ , $b=b^{-1}$ , etc. Thus,

$ab=b(b^{-1}a)(b^{-1}a)a$

In addition $(b^{-1}a)(b^{-1}a)=(b^{-1}a)^2=e$ , hence $ab=ba$ , and $G$ abelian.

The product group $\mathbb{Z}_2\times \mathbb{Z}_2$ is an example of one abelian group whose order is greater than $2$ and the order of whose elements is $2$ .

April 24, 2020October 24, 2022

202004240606 Homework 1 (Q1)

Let $G=\{ a,b,c,d\}$ and the Cayley table of $G$ be

$\begin{aligned} * \quad | &\textrm{} \quad a &\textrm{}\quad b \quad &\textrm{}\quad c &\textrm{}\quad d \\ \hline a \quad | &\quad a &\quad b \quad &\quad c &\quad d \\ b \quad | &\quad b &\quad a \quad &\quad c &\quad d \\ c \quad | &\quad c &\quad b \quad &\quad a &\quad d \\ d \quad | &\quad d &\quad d \quad &\quad b &\quad c \end{aligned}$

Is $(G,*)$ a group?

Attempts.

I recall that a group $(G,*)$ is a set $G$ altogether with an operation $*:G\times G\rightarrow G$ , $(x,y)\mapsto x*y$ such that the following axioms be satisfied:

i. (Associativity) $(x*y)*z=x*(y*z)$ ;

ii. (Identity) There exists $e\in G$ such that $e*x=x*e=x$ for all $x\in G$ ; and

iii. (Inverse) For each $a\in G$ , there exists $b\in G$ such that $a*b=b*a=e$ .

Let me check them one by one.

i. For $x,y,z\in \{ a,b,c,d \}$ , it suffices to check $4\times 4\times 4=64$ operations. And I found that the following operation failed axiom (i):

$d*(c*b)=d*b=d$ but $(d*c)*b=b*b=a$ .

That said, $G$ under the operation $*$ is not a group.

ii. By inspection, the identity of $G$ exists, and that is $a$ because $a*x=x*a=x$ for all $x\in \{ a,b,c,d\}$ . Axiom (ii) is thus satisfied.

iii. There exists $d\in G$ , such that for all $x^i\in\{ a,b,c,d\}=G$ ,

$d*x^i=x^i*d\neq a$ .

The claim above is validated by simply checking the identity found to be $a$ in part ii. cannot be found in the entries, and by the fact that $a$ is unique.

Axioms i. and iii. having been failed, I may conclude that $G$ under the operation $*$ cannot form a group.

CASE 1: Would it be possible that ?

CASE 2: If for all elements in the group , their order cannot ever be .

CASE 1: Would it be possible that $a=e$ ?

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .