If C_r , C_s are cyclic groups such that g.c.d(r,s)=1, then show that C_r ×C


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Question Number 171314 by Mastermind last updated on 12/Jun/22

$I f C_{r}, C_{s} a r e c y c l i c g r o u p s s u c h t h a t$ $g . c . d (r, s) = 1, t h e n s h o w t h a t C_{r} \times C_{s} i s$ $a c y c l i c g r o u p .$ $M a s t e r m i n d$
	Answered by mindispower last updated on 12/Jun/22
	$C_r cyclic ⇒∃a∈C_r ...such ∀g∈C_r ∃n∈{0,....r−1} a^n =g C_s cyclic ⇒∃b∈C_s ..such ∀g′∈C_s .∃m∈{0,....s−1} b^m =g′ we use standar definition of product of/Groups card (C_r ∗C_s )≤rs (a,b)∗.....(a,b)...(rs) =((a^r )^s ,(b^s )^r )=(e,e′)Times (a^x ,b^x )=(e,e′)⇒r∣x,s∣x⇒rs∣x...gcd(r,s)=1 ⇒rs∣card (C_r ∗C_s )⇒card (C_r ∗C_s )=rs and C_r ∗C_s =<(a,b) >$
	$C_{r} c y c l i c \Rightarrow \exists a \in C_{r} . . . s u c h \forall g \in C_{r} \exists n \in {0, . . . . r - 1}$ $a^{n} = g$ $C_{s} c y c l i c \Rightarrow \exists b \in C_{s} . . s u c h \forall g^{'} \in C_{s} . \exists m \in {0, . . . . s - 1}$ $b^{m} = g^{'}$ $w e u s e s t a n d a r d e f i n i t i o n o f p r o d u c t o f / G r o u p s$ $c a r d (C_{r} * C_{s}) ⩽ r s$ $(a, b) * . . . . . (a, b) . . . (r s) = ({(a^{r})}^{s}, {(b^{s})}^{r}) = (e, e^{'}) T i m e s$ $(a^{x}, b^{x}) = (e, e^{'}) \Rightarrow r ∣ x, s ∣ x \Rightarrow r s ∣ x . . . g c d (r, s) = 1$ $\Rightarrow r s ∣ c a r d (C_{r} * C_{s}) \Rightarrow c a r d (C_{r} * C_{s}) = r s$ $a n d C_{r} * C_{s} =< (a, b) >$
	Commented by Mastermind last updated on 14/Jun/22

	$T h a n k s$
	Commented by mindispower last updated on 19/Jun/22

	$w i t h e p l e a s u r$
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