Answer to Question #141537 in Abstract Algebra for Vivek kumar

Question

|a|=(2*n+1);
aba^(-1)=b^(-1);
b^2=?

Answer 1

Given that,

$|a|=2^{n+1} \ and \ aba^{-1}=b^{-1}$

we have to find $b^2=?$

since $aba^{-1}=b^{-1} \implies (aba^{-1})^{-1}=(b^{-1})^{-1}$

$\implies ab^{-1}a^{-1}=b$

$\implies (ab^{-1}a^{-1})^2=b^2$

$\implies b^2=ab^{-1}a^{-1}ab^{-1}a^{-1}$ $=ab^{-2}a^{-1}$