Assignment 3

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Exercise $\PageIndex{1}$

For the following, either show that $ G$ is a group with a given operation or list the condition(s) that fail.

$G=\mathbb{R}$ with the binary operation $ a \star b= a+b-ab$ for $a,b \in \mathbb{R}.$
$G=3\mathbb{Z}=\{3m|m\in \mathbb{Z}\}$ with addition.
$G=\{4,8,12,16\}$ with multiplication $\mod(20).$

Answer

1. Let$G := \mathbb{R}$, and let $a,b,c \in G$. Let an operation $\star$) on$G$ be defined as$a \star b := a + b - a \cdot b$. \\

Is ($G, \star$) closed? ($\star$) is comprised of ($+$), ($-$), and ($\cdot$) over real numbers. These operations are closed over $\mathbb{R}$. Therefore, since$G = \mathbb{R}$, ($G, \star$) is closed.

Is$(G, \star)$ associative? Consider the following:\begin{eqnarray}a \star (b \star c) &=& a \star (b + c - bc), \\&=& a + (b + c - bc) - a(b + c - bc), \\&=& a + b + c - bc - ab - ac + abc, \\&=& (a + b - ab) + c - c(a + b - ab), \\&=& (a + b - ab) \star c, \\&=& (a \star b) \star c. \nonumber\end{eqnarray}

Therefore, ($G, \star$) is associative.

Does$G$ contain the identity $e$? Consider $0 \in G$.

\begin{eqnarray}a \star 0 &=& a + 0 - a \cdot 0, \\&=& a + 0 - 0, \\&=& a. \\ \\0 \star a &=& 0 + a - 0 \cdot a, \\&=& 0 + a - 0, \\&=& a. \nonumber\end{eqnarray}

Since$a\star 0 = 0 \star a = a$ for any $a \in G$, and $0 \in G$;$e=0$ is the identity of $(G, \star)$.

Does each element of $G$ have an inverse in $G$? Let's assume for contradiction that $a^{-1} \in G$ exists for all $a \in G$. Then we can say that:

\begin{eqnarray}a \star a^{-1} &=& 0, \\a + a^{-1} - a \cdot a^{-1} &=& 0, \\a^{-1}(1 - a) &=& -a, \\a^{-1} &=& \frac{-a}{1 - a}. \nonumber\end{eqnarray}

Now consider where $a = 1$. In this case, $1^{-1} = \frac{1}{1 - 1} = \frac{1}{0}$. Clearly $\frac{1}{0} \notin G$, so$1^{-1} \notin G$. This is a contradiction. Therefore, there exists $a \in G$ such that $a^{-1} \notin G$.

Since$(G, \star)$ is closed, has an identity, is associative, but does not contain an inverse for each element; $(G, \star)$ is not a group.\ ((G, \star)\) is a monoid.

2. Let $G := 3\mathbb{Z} = \{ 3m : m \in \mathbb{Z} \}$. Let $m,n,p \in \mathbb{Z}$. Clearly $3m,3n,3p \in 3\mathbb{Z}$ by definition of $3\mathbb{Z}$. Consider $(G, +)$.

Is $(G, +)$ closed? Consider the following:

\begin{eqnarray}3m + 3n &=& 3(m + n). \nonumber\end{eqnarray}

Since $\mathbb{Z}$ is closed under addition, $(m + n) \in \mathbb{Z}$, and $3(m+n) \in G$. Therefore, $(G, +)$ is closed.

Is $(G, +)$ associative? Consider the following:

\begin{eqnarray}3m + (3n + 3p) &=& 3( m + (n + p)), \\&=& 3((m + n) + p), \\&=& (3m + 3n) + 3p. \nonumber\end{eqnarray}

Therefore, $(G, +)$ is associative.

Does$G$ contain the identity $e$? Consider $0 \in G$.

\begin{eqnarray}3m + 0 = 3m. \\0 + 3m = 3m. \nonumber\end{eqnarray}

Clearly $3m + 0 = 0 + 3m = 3m$. Therefore, $e = 0$ is the identity of $(G, +)$.

Does$(G, +)$ contain the inverse for each element in$G$? Consider $3(-n)$ as the inverse for $3n$. Since $-n \in \mathbb{Z}$, $3(-n) \in G$.

$3n + 3(-n) = 3(n - n)=3(0)= 0=e. $ and $3(-n) + 3n = 3((-n) + n) = 3(0)= 0 = e.$

Therefore $3n + 3(-n) = 3(-n) + 3n = e$. So for any $a \in G$, there exists $a^{-1} \in G$.

Since $(G, +)$ is closed, associative, and contains the identity and an inverse for every element, $(G, +)$ is a group.

3. The Cayley table of $G$ with multiplication $\mod(20):$

$\cdot (mod 20)$	$4$	$8$	$12$	$16$
$4$	$16$	$12$	$8$	$4$
$8$	$12$	$4$	$16$	$8$
$12$	$8$	$16$	$4$	$12$
$16$	$4$	$8$	$12$	$16$

Is $ (G, \cdot(mod(20))$ closed? Yes, we can see from the Cayley table that all results of $ \cdot(mod(20)$ are in $ G$.

Does $ G$ contain the identity for $ \cdot(mod(20)$ ? Again, by the Cayley table, we can see that the identity $ e = 16$.

Is $ (G, \cdot(mod(20))$ associative? Yes. We know by the properties of $ \mathbb{Z}$ that multiplication of integers is associative. Also, by the properties of modular arithmetic that the modulo operation has no impact on associativity. Therefore, $ (G, \cdot(mod(20))$ is associative. \\

Does $ G$ contain an inverse for each element under $\cdot(mod(20)$? Yes, consider the following:

\begin{align*}4 \cdot 4 &\equiv& 16 mod(20) \implies 4^{-1} &= 4. \\8 \cdot 12 &\equiv& 16 mod(20) \implies 8^{-1} &= 12. \\12 \cdot 8 &\equiv& 16mod(20) \implies 12^{-1} &= 8. \\16 \cdot 16 &\equiv& 16mod(20) \implies 16^{-1} &= 16.\end{align*}

So clearly for each $a \in G$, there exists $a^{-1} \in G$. Since $(G, \cdot(mod(20))$ is closed, associative, and contains the identity and an inverse for every element, $(G, \cdot(mod(20))$ is a group.

Exercise $\PageIndex{2}$

Decide whether the following structures form a group or not by looking at the Cayley table.

$\{\pm 1, \pm i\}$ with multiplication, Where $i$ is the complex number such that $i^2=-1.$
$ 1 =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, $ I =\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}$, $ J =\begin{bmatrix}0 & i\\i & 0\end{bmatrix}$, $ K =\begin{bmatrix} i & 0\\0 & -i \end{bmatrix}$, where $i$ is the complex number such that $i^2=-1.$

Define $Q_8=\{ \pm 1,\pm I, \pm J, \pm K \}.$ $Q_8$ with multiplication.

Answer

1. Let $G = \{1, -1, \ i, -i\}$

We are to determine whether the set $G$ forms a group $(G, \cdot)$ under multiplication, where $ i $ is the imaginary unit with $ i^2 = -1 $. We will construct and examine the Cayley table to verify the group properties.

$clipboard_e5404ae73bf4921de2ebadc615e60c2ab.png$

It is a group. It satisfies closedness, identity, and inverse, according
to the table, and multiplication is associative on the complex numbers.

$clipboard_e774f496b1bece3db03807b68dd302e70.png$

It is a group. It satisfies closedness, identity, and inverse according to the table, multiplication is associative on $2 \times 2$ matrices over the complex numbers.

Exercise $\PageIndex{3}$

Prove or disprove: Let $H$ and $K$ be subgroups of the group $G$. Then

$H\cup K$ is a subgroup of $G$.
$H\cap K$ is a subgroup of $G$.

Answer

1. No, $H \cup K $ is a subgroup of $G $.

Answers may vary.

Counterexample:

Let $G= (Z_6, + \pmod{6}) $.

Thus $e_G=0 $.

We shall show that $H \cup K $ is not a subgroup of $G $.

Consider $Z_6 = \{0,1,2,3,4,5\} $.

Let $H=( \{0,2,4\}, +\pmod{6}) $.

Note $H \le G $ since$e_G=0 \in H $ and $\exists_1 h^{-1}, \forall h \in H $.

Let $K=( \{0,3\}, +\pmod{6}) $.

Note $K \le G $ since $e_K=0 \in H $ and $\exists_1 k^{-1}, \forall k \in K $.

Thus $H \cup K=(\{0,2,3,4\}, + \pmod{6}) $.

Let $a=2 $ and$b=3 $, where $2,3 \in H \cup K $.

However, since $2+3\equiv 5 \pmod{6} \not \in H \cup K $, $H \cup K $ is not a group.

Thus $H \cup K $ is not a subgroup of$G $.

2. Yes, $H \cap K $ is a subgroup of $G $ .

Proof:

Let $e $ be the identity of $G $ .

Then $e $ is the identity of both $H $ and $K $ .

Therefore $e \in H \cap K $ .

Let $g_1,g_2 \in H \cap K $ .

Then we shall show that $g_1g_2^{-1} \in H \cap K $ .

Since $g_1, g_2 \in H \cap K $ , $g_1, g_2 \in H $ and $g_1, g_2 \in K $ .

Since $H \le G $ and $K \le G $ , $g_1,g_2,g_1^{-1}, g_2^{-1} $ are in both $H $ and $K $ .

Hence $g_1g_2^{-1} \in H \cap K $ .

Therefore $H \cap K \le G $ .◻

Exercise $\PageIndex{4}$

1. Show that $U(20)\neq <k>$ for any k in $U(20).$ [Hence $U(20)$ is not cyclic.]

2. Prove that if $G$ is an abelian group, then for each element $a\in G$, $C(a)=\{ x \in G |xa=ax\}$ is a subgroup of $G$.

Answer

$ U(20)=\{1,3,7,9,11,13,17,19\}$ and $ |U(20)|=8$ .

We shall show that$U(20) \ne k, \; k \in U(20)$ .

$ k$	Calculations	Order of$k$
1	$ 1\equiv 1 \pmod{20}$	1
3	$ 3^1\equiv 3 \pmod{20}$ ;$3^2\equiv 9 \pmod{20}$ ;$3^3\equiv 7 \pmod{20}$ ;$3^4\equiv 1 \pmod{20}$	4
7	$ 7^1\equiv 7 \pmod{20}$ ;$7^2\equiv 9 \pmod{20}$ ;$7^3\equiv 3 \pmod{20}$ ;$7^4\equiv 1 \pmod{20}$	4
9	$ 9^1\equiv 9 \pmod{20}$ ;$9^2\equiv 1 \pmod{20}$	2
11	$ 11^1\equiv 11 \pmod{20}$ ;$11^2\equiv 1 \pmod{20}$	2
13	$ 13^1\equiv 13 \pmod{20}$ ;$13^2\equiv 9 \pmod{20}$ ;$13^3\equiv 17 \pmod{20}$ ;$13^4\equiv 1 \pmod{20}$	4
17	$ 17^1\equiv 17 \pmod{20}$ ;$17^2\equiv 9 \pmod{20}$ ;$17^3\equiv 13 \pmod{20}$ ;$17^4\equiv 1 \pmod{20}$	4
19	$ 19^1\equiv 19 \pmod{20}$ ;$19^2\equiv 1 \pmod{20}$	2

Since $|k| \ne |U(20)|, \; \forall \;k \in U(20)$ , there is no $k \in U(20)$ that generates $U(20)$ .

2. Since $G$ is abelian, for each $a\in G$, $C(a)=\{ x \in G |xa=ax\}= G$ is a subgroup of $G$.

\(\cdot (mod 20)\)	\(4\)	\(8\)	\(12\)	\(16\)
\(4\)	\(16\)	\(12\)	\(8\)	\(4\)
\(8\)	\(12\)	\(4\)	\(16\)	\(8\)
\(12\)	\(8\)	\(16\)	\(4\)	\(12\)
\(16\)	\(4\)	\(8\)	\(12\)	\(16\)

\( k\)	Calculations	Order of\(k\)
1	\( 1\equiv 1 \pmod{20}\)	1
3	\( 3^1\equiv 3 \pmod{20}\) ;\(3^2\equiv 9 \pmod{20}\) ;\(3^3\equiv 7 \pmod{20}\) ;\(3^4\equiv 1 \pmod{20}\)	4
7	\( 7^1\equiv 7 \pmod{20}\) ;\(7^2\equiv 9 \pmod{20}\) ;\(7^3\equiv 3 \pmod{20}\) ;\(7^4\equiv 1 \pmod{20}\)	4
9	\( 9^1\equiv 9 \pmod{20}\) ;\(9^2\equiv 1 \pmod{20}\)	2
11	\( 11^1\equiv 11 \pmod{20}\) ;\(11^2\equiv 1 \pmod{20}\)	2
13	\( 13^1\equiv 13 \pmod{20}\) ;\(13^2\equiv 9 \pmod{20}\) ;\(13^3\equiv 17 \pmod{20}\) ;\(13^4\equiv 1 \pmod{20}\)	4
17	\( 17^1\equiv 17 \pmod{20}\) ;\(17^2\equiv 9 \pmod{20}\) ;\(17^3\equiv 13 \pmod{20}\) ;\(17^4\equiv 1 \pmod{20}\)	4
19	\( 19^1\equiv 19 \pmod{20}\) ;\(19^2\equiv 1 \pmod{20}\)	2

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