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Question Number 69568 by Ajao yinka last updated on 25/Sep/19

Commented by mathmax by abdo last updated on 25/Sep/19

$i f n = 1 H = C i f n \neq 1 w e h a v e z = z^{n} \Leftrightarrow z^{n - 1} = 1 l e t z = r e^{i θ} s o$ $z^{n - 1} = 1 \Leftrightarrow r^{n - 1} e^{i (n - 1) θ} = e^{i 2 k π} \Rightarrow r = 1 a n d θ = \frac{2 k π}{n - 1} w i t h k \in Z$ $s o t h e e l e m e n t s o f H a r e Z_{k} = e^{\frac{i 2 k π}{n - 1}} t h e o r e m o f d i v i s i o n g i v e$ $k = q (n - 1) + r w i t h 0 ⩽ r ⩽ n - 2 \Rightarrow$ $Z_{k} = Z_{q (n - 1) + r} = e^{\frac{i 2 (q (n - 1) + r) π}{n - 1}} = e^{i 2 π q + \frac{i 2 r π}{n - 1}} = e^{\frac{i 2 r π}{n - 1}} = Z_{r} s o H i s f i n i t e$ $a n d c a r d H = n - 2$
Commented by mathmax by abdo last updated on 26/Sep/19

$f o r g i v e c a r d H = n - 1$
Commented by Ajao yinka last updated on 26/Sep/19

$s o w h a t s y o u r c o n c l u s i o n$
	Answered by mind is power last updated on 25/Sep/19
	$H={z∈C∣z^n −z=0} H is set of roots/of P(X)=X^n −X∈C_n [X] C is a field So integer domain ⇒card(h)≤Deg(P)=n So finite$
	$H = {z \in C ∣ z^{n} - z = 0}$ $H i s s e t o f r o o t s / o f P (X) = X^{n} - X \in C_{n} [X]$ $C i s a f i e l d S o i n t e g e r d o m a i n \Rightarrow c a r d (h) ⩽ D e g (P) = n$ $S o f i n i t e$
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