Question

show that the set {1,-1,i,-i} is an abelian finite group of order 4 under multiplication?​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{\{1,-1,i,-i\}}$

$\underline{\textbf{To prove:}}$

$\mathsf{\{1,-1,i,-i\}}\;\textsf{is an abelian group under multiplication}$

$\underline{\textbf{Solution:}}$

$\mathsf{G=\{1,-1,i,-i\}}$

$\textsf{First we form a cayley's table}$

$\left|\begin{array}{|c|cccc|}\cline{1-5}.&1&-1&i&-i\\\cline{1-5}1&1&-1&i&-i\\-1&-1&1&-i&i\\i&i&-i&-1&1\\-i&-i&i&1&-1\\\cline{1-5}\end{array}\right|$

$\textsf{(i) All the elements in the cayley's table are elements of G}$

$\implies\textsf{Closure axiom is true}$

$\textsf{(ii) Usual multiplication is always associative}$

$\implies\textsf{Associative axiom is true}$

$\textsf{(iii) Clearly, 1 is the identity element and  satisfies identity axiom}$

$\implies\textsf{Identity axiom is true}$

$\textsf{(iv)}$

$\textsf{Inverse of 1 is 1}$

$\textsf{Inverse of -1 is -1}$

$\textsf{Inverse of i is -i}$

$\textsf{Inverse of -i is i}$

$\implies\textsf{Inverse axiom is true}$

$\textsf{(v) Usual multiplication is always commutative}$

$\therefore\textbf{G is an abelian group under multiplication}$

Answer 2

Answer:

Step-by-step explanation: