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# In the multiplicative group of cube root of unity, the order of $w^{2}$ is

$\begin{array}{1 1}(1)4&(2)3\\(3)2&(4)1\end{array}$

Can you answer this question?

In the cube root of unity
$\omega^3=1$
$1+\omega + \omega^2 =0$
$(\omega^2)^2 =\omega^4 =\omega^3. \omega =\omega$
$(\omega^2)^3 =\omega^6 =(\omega^3)^2 = (1)^2=1$
$0(\omega^2)=3$
Hence 2 is the correct answer.
answered May 22, 2014 by