Example 1.0.2.

1.

The set of even integers with addition \(\hspace {0.5cm}2\mathbb {Z} = \{0\, ,\, \pm 2\, , \, \pm 4\, , \cdots \cdots \}\,\, (2\mathbb {Z}\, ,\, +)\)

(i)

\(2r + (2s + 2t) = (2r + 2s) + 2t.\hspace {0.3cm}\) holds

(ii)

\(a+ 0 = 0 + a = a\hspace {0.3cm}\) holds \(\, 0\in 2\mathbb {Z}=\) (even numbers)

(iii)

\(a + (-a) = (-a) + a = a\hspace {0.3cm}\) holds

\(\therefore \,\, \) Even integers with addition form a group.

2.

\(\{0\}\) with addition is a group. Since \(\, \, \big (\{0\}\, , \, +\big )\)

(i)

\(0 + (0 + 0) = (0 + 0) + 0 = 0\)

(ii)

\( 0 + 0 = 0 + 0 = 0\)

(iii)

\(0 + 0 = 0\)

3.

Positive rational numbers with multiplication. \(\hspace {0.2cm}\big (\mathbb {Q}\, , \, \times \big )\). \[\mathbb {Q}^+ = \big \{r/s\, \, , \, r\, , \, s\in \mathbb {Z}\, , \, s\neq 0\, , \, r\, , s>0\hspace {0.2cm}\text {or}\hspace {0.2cm} r\, , \, s<0\big \}\]

(i)

\(\Big (\dfrac {a}{b}\times \dfrac {c}{d}\Big )\times \dfrac {e}{f} =\dfrac {a}{b}\times \Big (\dfrac {c}{d}\times \dfrac {e}{f}\Big )\hspace {0.3cm}\) holds \(\hspace {0.3cm}\dfrac {a}{b}\, , \, \dfrac {c}{d}\, , \, \dfrac {e}{f}\in \mathbb {Q}\)

(ii)

\(\exists \, 1/1 = 1 \in \mathbb {Q}^+ \, \ni \, \dfrac {a}{b}\times 1 = 1 \times \dfrac {a}{b} = \dfrac {a}{b}\)

(iii)

\(\forall \, \dfrac {a}{b}\,\in \mathbb {Q}^+\, , \, \dfrac {a}{b}\times \dfrac {b}{a} = \dfrac {b}{a}\times \dfrac {a}{b} = 1\)

4.

Consider the set \(\, A = \{a\, , \, b\, , \, c\}\) with two operations \(\bullet \) and \(*\) described by Cayley lables (multiplication tables)

\(\bullet \)	\(a\)	\(b\)	\(c\)
\(a\)	\(a\)	\(b\)	\(c\)
\(b\)	\(b\)	\(c\)	\(a\)
\(c\)	\(c\)	\(a\)	\(b\)

     

\(*\)	\(a\)	\(b\)	\(c\)
\(a\)	\(c\)	\(a\)	\(b\)
\(b\)	\(a\)	\(b\)	\(c\)
\(c\)	\(b\)	\(c\)	\(a\)

\((a*b)*c = a*(b*c)\hspace {0.4cm}\)
To check for associativity \(\, (a*b)*c = a*(b*c)\)

(a)

(i)

\((a\cdot b)\cdot c = a\cdot (b\cdot c)\)

(ii)

\(\big (A\, , \, \cdot \big )\,\) has an identity \(a\).

(iii)

\(a\) has inverse \(a\).

\(b\) has inverse \(c\).

\(c\) has inverse \(b\).

(b)

(i)

\((a*b)*c = a*(b*c)\).

(ii)

\(\big (A\, , \, *\big )\) has identity \(b\).

(iii)

\(a\) has inverse \(c\).

\(b\) has inverse \(b\).

\(c\) has inverse \(a\)

5.

Set of invertible \(n\times n\) matrices over \(\mathbb {R}\), with operation given by multiplication. \(G(n,\mathbb {R})\,\) or \(\, G_nL({\mathbb {R}})\).

(i)

for all \(\,A\, , \, B\,, \, C\,\) in \(\, GL(n,\mathbb {R})\hspace {0.3cm} (AB)C = A(BC)\hspace {0.3cm}\) (prove!!!)

(ii)

\(\forall \, A\in G(n,\mathbb {R})\, , \, A\cdot I = I\cdot A = A\)

(iii)

\(\forall \, A\in GL(n,\mathbb {R})\,\, \exists \, A^{-1}\ni A\cdot A^{-1} = A^{-1}A = I\)

Proof. \((AB)C = A(BC) \, \implies \begin {bmatrix} \begin {pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end {pmatrix}\begin {pmatrix} b_1 & b_2\\ b_3 & b_4\\ \end {pmatrix} \end {bmatrix}\begin {pmatrix} c_1 & c_2\\ c_3 & c_4\\ \end {pmatrix}\) □

\begin {align*} & = \begin {pmatrix} a_1b_1 + a_2b_3 & a_1b_2 + a_2b_4\\ a_3b_1 + a_4b_3 & a_3b_2 + a_4b_4\\ \end {pmatrix}\begin {pmatrix} c_1 & c_2\\ c_3 & c_4\\ \end {pmatrix}\\\\ & = \begin {pmatrix} a_1b_1c_1 + a_2b_3c_1 + a_1b_2c_3 + a_2b_4c_3 & a_1b_1c_2 + a_2b_3c_2 + a_1b_2c_4 + a_2b_4c_4\\ a_3b_1c_1 + a_4b_3c_1 + a_3b_2c_3 + a_4b_4c_3 & a_3b_1c_2 + a_4b_3c_2 + a_3b_2c_2 + a_4b_4c_4\\ \end {pmatrix}\\\\ & = \begin {pmatrix} a_1(b_1c_1 + b_2c_3) + a_2(b_3c_1 + b_4c_3) & a_1(b_1c_2 + b_2c_4) + a_2(b_3c_2 + b_4c_4)\\ a_3(b_1c_1 + b_2c_3) + a_4(b_3c_1 + b_4c_3) & a_3(b_1c_2 + b_2c_4) + a_4(b_3c_2 + b_4c_4)\\ \end {pmatrix}\\\\ & = \begin {pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end {pmatrix}\begin {pmatrix} b_1c_1 + b_2c_3 & b_1c_2 + b_4c_3\\ b_3c_1 + b_4c_3 & b_2c_2 + b_4c_4\\ \end {pmatrix}\\\\ & = \begin {pmatrix} a_1 & a_2\\ a_3 & a_4\\ \end {pmatrix} \begin {bmatrix} \begin {pmatrix} b_1 & b_2\\ b_3 & b_4\\ \end {pmatrix}\begin {pmatrix} c_1 & c_2\\ c_3 & c_4\\ \end {pmatrix} \end {bmatrix} = A(BC)\\ \end {align*}

6.

Consider a non-empty set. Now the set \(M(S)\) of invertible mappings on \(S\) with operations as composition of mappings is a group. \begin {align*} & e\\ S = \{a\, , \, b\, , \, c\}\hspace {0.5cm}\implies \hspace {0.5cm} a & \longrightarrow a\\ b & \longrightarrow b\\ \{e\} \hspace {5cm} c & \longrightarrow c\\ \end {align*}