Abstract algebra, order of element

Question

I have to find the order of elements and cyclic groups but I need some help as I am not sure how to do it right.

We have the elements: \begin{gather} w=\cos(2\pi/98) + i\sin(2\pi/98) \\ \alpha = w^4 \end{gather} and the cyclic groups: \begin{align} G &= \langle w\rangle\\ H &= \langle \alpha\rangle \end{align}

Could you please explain me how to do it the right way?

Answer 1

Notice $w^k = \cos (\frac {k*2\pi}{98}) + i(\frac {k*2\pi}{98})$ and $w^{98} = 1$ and that if $1 \le k < j <98$ then $w^k \ne w^j$.

That tells you that $<w> = \{1,w, w^2, ..... , w^{97}\}$ with $|w| = 98$.

And notice $(w^4)^k = \cos (\frac {2k*2\pi}{49}) + i(\frac {2k*2\pi}{49})$. And if $k > 25$ then $(w^4)^k = \cos (\frac {2k*2\pi}{49}) + i(\frac {2k*2\pi}{49})=\cos(\frac {2k*2\pi}{49} - 2\pi) + i(\frac {2k*2\pi}{49}-2\pi)$. with $(w^4)^{49} = 1$ and that if $1 \le k < j < 49$ then $(w^4)^k \ne (w^4)^j$.

So $<w^4> = \{1,w^4, w^8,..., w^{96}, w^2, w^6, .....,w^{94}\}$

Answer 2

Since $w^{98}=1$ and $w^{k}\ne1$ for $0<k<98$ (use De Moivre’s formula), we see that $w$ has order $98$ and so $$ \langle w\rangle=\{w^k:0\le k<98\} $$ Also $\alpha^{49}=(w^{4})^{49}=(w^{98})^2=1^2=1$. Hence the order of $\alpha$ is a divisor of $49$. However it can be neither $1$ nor $7$, because $\alpha=w^4\ne1$ and $\alpha^7=w^{28}\ne1$.

Hence the order is $49$ and $\langle\alpha\rangle=\{\alpha^k:0\le k<49\}$.

I exploited the fact that if $g^n=1$, with $n>0$, then the order of $g$ is a divisor of $n$.

More generally, if $g$ has order $m$, then the order of $g^n$ is $m/\!\gcd(m,n)$.