Abstract Algebra – Physicspupil

May 10, 2023May 11, 2023

202305101606 Problem 3.16.1

Let $G=\{ \mathbf{e},a,a^2,a^3\}$ be a cyclic group of order $4$ and $G'$ be the multiplicative group $\{1,-1,\mathbf{i},-\mathbf{i}\}$ of $4$ complex numbers.

i. Tabulate two isomorphisms from $G$ to $G'$ .
ii. Explain why there are no other isomorphisms from $G$ to $G'$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Background.

The group $G$ is cyclic if and only if every element of $G$ can be expressed as the power of one element of $G$ :

$\exists\, g\in G,\,\forall\, h\in G:\enspace h=g^n\textrm{ for some }n\in\mathbb{Z}$ .

if and only if it is generated by one element $g\in G$ called a generator of $G$ :

$G=\langle g\rangle$ .

_{ProofWiki on Cyclic Group}

Let $(F,+,\times )$ be a field and let $F^*:=F\backslash\{ 0\}$ be the set $F$ less its zero. The group $(F^*,\times )$ is known as the multiplicative group of $F$ .

_{ProofWiki on Multiplicative Group}

A bijective homomorphism is called an isomorphism; that is, a group isomorphism $\theta$ from $G$ to $G'$ is a bijection $\theta$ such that $\theta (ab)=\theta (a)\theta (b)$ for all $a$ and $b$ in $G$ .

_{Text on 3.2 Group Isomorphism, pg.141}

Let $\theta$ be a group isomorphism from $G$ to $G'$ . Then:

(a) $\mathbf{e}\stackrel{\theta}{\mapsto}\mathbf{e}'$ , where $\mathbf{e}$ and $\mathbf{e}'$ are the identities of $G$ and $G'$ .
(b) If $a\stackrel{\theta}{\mapsto}a'$ , then $a^n\stackrel{\theta}{\mapsto}(a')^n$ for all integers $n$ . In particular,

$a^{-1}\stackrel{\theta}{\mapsto}(a')^{-1}$ .

(c) If $a\stackrel{\theta}{\mapsto}a'$ , then $a$ and $a'$ have equal orders.

_{Text on 3.2 Group Isomorphism, pg.143}

Roughwork.

$\begin{array}{c|cccc} (G,*) & \mathbf{e} & a & a^2 & a^3 \\\hline \mathbf{e} & \mathbf{e} & a & a^2 & a^3 \\ a & a & a^2 & a^3 & \mathbf{e} \\ a^2 & a^2 & a^3 & \mathbf{e} & a \\ a^3 & a^3 & \mathbf{e} & a & a^2 \\ \end{array}$

$\begin{array}{c|cccc} (G',\times ) & 1 & -1 & \mathbf{i} & -\mathbf{i} \\\hline 1 & 1 & -1 & \mathbf{i} & -\mathbf{i} \\ -1 & -1 & 1 & -\mathbf{i} & \mathbf{i} \\ \mathbf{i} & \mathbf{i} & -\mathbf{i} & -1 & 1 \\ -\mathbf{i} &-\mathbf{i} & \mathbf{i}} & 1 & -1 \\ \end{array}$

Observe that

$G=\langle a\rangle$ ; $G'=\langle \mathbf{i}\rangle=\langle -\mathbf{i}\rangle$ .

This problem is not to be attempted.

April 13, 2023April 22, 2023

202304131118 Problem 2.1.1

For each of the following parts, tell whether it specifies $\theta$ as a permutation on $\mathbf{X_5}=\{ 1,2,3,4,5\}$ . If not, explain how Definition 1 is not satisfied.

(a) $\theta:1\mapsto 1,2\mapsto 3,3\mapsto 4,4\mapsto 5,5\mapsto 1$ .
(b) $\theta:1\mapsto 5,2\mapsto 1,3\mapsto 2,4\mapsto 3,5\mapsto 4$ .
(c) $\theta:1\mapsto 1,2\mapsto 2,3\mapsto 3,4\mapsto 4,5\mapsto 5$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Roughwork.

(a)

The arrow form

$\theta:1\mapsto 1,2\mapsto 3,3\mapsto 4,4\mapsto 5,5\mapsto 1$

may also be expressed as in function form

$\begin{aligned} \theta (1) & = 1\\ \theta (2) & = 3\\ \theta (3) & = 4\\ \theta (4) & = 5\\ \theta (5) & = 1\\ \end{aligned}$

For any finite non-empty set $S$ , $A(S)$ the set of all $\textrm{1--1}$ transformations (mappings) of $S$ onto $S$ forms a group called permutation group and any element of $A(S)$ , i.e., a mapping from $S$ onto itself, is called permutation.

_{From Wikibooks on Permutation groups}

As is seen, function $\theta$ here is $\textrm{\scriptsize{NOT}}$ injective (i.e., $\textrm{\scriptsize{NOT}}$ $\textrm{1--1}$ ), for $\theta$ sends both $1$ and $5$ to $1$ . Hence it cannot specify a permutation.

(b)

The arrow form

$\theta:1\mapsto 5,2\mapsto 1,3\mapsto 2,4\mapsto 3,5\mapsto 4$

or two-line form

$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 5 & 1 & 2 & 3 & 4 \end{pmatrix}$

or cyclic form

$(15432)$

also in diagram

is obviously a one-to-one function (/an injection). Thus function $\theta$ here specifies a permutation.

(c)

In arrow form

$\theta:1\mapsto 1,2\mapsto 2,3\mapsto 3,4\mapsto 4,5\mapsto 5$

or matrix form

$\begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 1 & 2 & 3 & 4 & 5 \end{pmatrix}$

or cyclic form

$()$

is the identity permutation.

This problem is not to be attempted.

April 13, 2023April 13, 2023

202304131021 Problem 1.2.1

Which of the following integers are in the set $6\mathbb{Z}$ of integral multiples of $6$ ?

(a) $10$ ;
(b) $-10$ ;
(c) $12$ ;
(d) $-12$ ;
(e) $4002$ ;
(f) $-4002$ ;
(g) $4003$ ;
(h) $-4003$ .

_{Extracted from A. P. Hillman. (1999). Abstract Algebra A First Undergraduate Course.}

Roughwork.

(a), (b)

$\begin{aligned} \pm 10\div 6 & = \frac{\pm 10}{6} \\ & = \pm 1\frac{2}{3}\enspace\notin \mathbb{Z} \\ \end{aligned}$

$\begin{aligned} \because \quad & 6\nmid \pm 10 \\ \therefore \quad & \mathrm{\pm 10}\notin 6\mathbb{Z} \\ \end{aligned}$

Because $6$ is not a divisor (/factor) of $\pm 10$ in $\mathbb{Z}$ , integers $10$ and $-10$ are $\textrm{\scriptsize{NOT}}$ in the set $6\mathbb{Z}$ of integral multiples of $6$ .

(c), (d)

$\begin{aligned} \pm 12\div 6 & = \frac{\pm 12}{6} \\ & = \pm 2\enspace\in \mathbb{Z} \end{aligned}$

$\begin{aligned} \because\quad & 6\mid \pm12 \\ \therefore\quad & \mathrm{\pm 12}\in 6\mathbb{Z} \\ \end{aligned}$

Since $6$ is an integral divisor of $\pm 12$ , integers $12$ and $-12$ are in the set $6\mathbb{Z}$ of integral multiples of $6$ .

(e), (f)

$\begin{aligned} \pm 4002\div 6 & = \frac{\pm 4002}{6} \\ & = \pm 667\enspace \in\mathbb{Z} \\ \end{aligned}$

$\begin{aligned} \because\quad & 6\mid \pm 4002 \\ \therefore\quad & \mathrm{\pm 4002}\in 6\mathbb{Z} \\ \end{aligned}$

As $\pm 4002$ is a multiple of $6$ in $\mathbb{Z}$ , integers $4002$ and $-4002$ are in the set $6\mathbb{Z}$ of integral multiples of $6$ .

(g), (h)

$\begin{aligned} \pm 4003\div 6 & = \frac{\pm 4003}{6} \\ & = \pm 667\frac{1}{6}\enspace \notin\mathbb{Z} \\ \end{aligned}$

$\begin{aligned} \because\quad & 6\nmid \pm 4003 \\ \therefore\quad & \mathrm{\pm 4003} \notin 6\mathbb{Z} \\ \end{aligned}$

Now that $\pm 4003$ is not an integral multiple of $6$ , integers $4003$ and $-4003$ are $\textrm{\scriptsize{NOT}}$ in the set $6\mathbb{Z}$ of integral multiples of $6$ .

This problem is not to be attempted.

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 $G$ .

September 29, 2020November 3, 2022

202009290227 Exercise 2.3.5

Prove that $10^{n+1}+10^n+1$ is divisible by $3$ for $n\in\mathbb{N}$ .

_{Extracted from T. W. Judson. (2021). Abstract Algebra Theory and Applications.}

Proof.

Let $P(n)$ be the statement:

$P(n):\qquad 3\,\Big| (10^{n+1}+10^n+1)$ for any $n\in\mathbb{N}$

Determine whether or not $P(n)$ is true when $n=1$ :

$P(1): \qquad 3\,\Big| (10^{(1)+1}+10^{(1)}+1)$

As $10^{(1)+1}+10^{(1)}+1 = 111 = 37\cdot 3$ is divisible by $3$ , $P(1)$ is true.

Suppose $P(n)$ is true for some $n\, (\geqslant 1) \in\mathbb{N}$ , try and prove the statement $P(n+1)$ :

$P(n+1): \qquad 3\,\Big| (10^{(n+1)+1}+10^{(n+1)}+1)$

$\begin{aligned} &\quad 10^{(n+1)+1}+10^{(n+1)}+1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +1 \\ = &\quad 10^{n+1}\cdot 10 + 10^n\cdot 10 +10 - 9 \\ = &\quad 10\cdot (10^{n+1}+10^n+1) - 9 \\ \end{aligned}$

As $P(n)$ is true and by the fact that three divides nine, $10\cdot (10^{n+1}+10^n+1) - 9$ is therefore divisible by $3$ . That is,

$P(n)\textrm{\enspace is true\enspace}\Rightarrow P(n+1)\textrm{\enspace is true\enspace}$

That $P(n)$ is true for $n=1$ , by the principle of mathematical induction, I have thus proven $10^{n+1}+10^n+1$ is divisible by $3$ for $n\in\mathbb{N}$ .

September 29, 2020May 8, 2023

202009290114 Exercise 2.3.1

Prove that

$1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$

for $n\in\mathbb{N}$ .

Proof.

Let $P(n)$ be the statement

$1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$ for $n\in\mathbb{N}$ .

When $n=1$ , check for the validity of $P(1)$ :

$\begin{aligned} \textrm{LHS} & = 1^2 = 1 \\ \textrm{RHS} & = \frac{(1)\big( (1)+1\big) \big( 2(1)+1\big) }{6} = 1 \\ \end{aligned}$

$\because \textrm{LHS}=\textrm{RHS}$

$\therefore P(n)$ is true for $n=1$ .

Suppose $P(n)$ holds true, let’s see if $P(n+1)$ holds too:

$P(n+1):\qquad 1^2+2^2+\cdots + n^2 + (n+1)^2 \stackrel{?}{=} \displaystyle{\frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6}}$

Beginning with the left hand side,

$\begin{aligned} \textrm{LHS} & = 1^2+2^2+\cdots + n^2 + (n+1)^2 \\ & = P(n) + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\ & = \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\ & = \frac{(n+1)\big[ (n)(2n+1) + 6(n+1) \big]}{6} \\ & = \frac{(n+1)(2n^2+n+6n+6)}{6} \\ & = \frac{(n+1)(2n^2+7n+6)}{6} \\ & = \frac{(n+1)(2n+3)(n+2)}{6} \\ \end{aligned}$

Then turn to the right hand side,

$\begin{aligned} \textrm{RHS} & = \frac{(n+1)\big( (n+1) +1 \big) \big( 2(n+1) +1 \big) }{6} \\ & = \frac{(n+1)(n+2)(2n+3)}{6} \\ \end{aligned}$

$\because \textrm{LHS} = \textrm{RHS}$

$\therefore P(n+1)$ holds when $P(n)$ holds.

As is proven $P(1)$ is true, by the principle of mathematical induction, $P(n)$ is also true for $n\geqslant 1$ , i.e.,

$\forall\, n\in\mathbb{N},\qquad 1^2+2^2+\cdots +n^2=\displaystyle{\frac{n(n+1)(2n+1)}{6}}$

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.

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 $2$ .