Question

148: The set G ={1, -1} is a group with respect to​

Answer 1

$\underline{\textsf{Given:}}$

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

$\underline{\textsf{To find:}}$

$\textsf{The binary operation which makes G is a group}$

$\underline{\textsf{Solution:}}$

$\textsf{First we form cayley's table with respect to multiplication}$

$\mathsf{\begin{array}{|c|c|c|}\cline{1-3}{\times}&1&-1\\\cline{1-3}1&1&-1\\\cline{1-3}-1&-1&1\\\cline{1-3}\end{array}}$

$\textsf{1.From the cayley's table, it is clear that all the elements in the}$

$\textsf{table are elements of G}$

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

$\textsf{2.Usual multiplication is always associative}$

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

$\textsf{3.Clearly 1 is the identity element and}\;\mathsf{1{\in}G}$

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

$\textsf{4.Inverse of 1 is 1}$

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

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

$\textsf{Hence G is a group under multiplication}$

$\underline{\textsf{Answer:}}$

$\textsf{G is a group with respect to}\;\underline{\textsf{multiplicaion}}$

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