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Orders of Groups 
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#1
Jan1208, 06:10 PM

P: 39

Hi i have completed the answer to this question. Just need your verification on whether it's completely correct or not:
Question: If G is a group and xEG we define the order ord(x) by: ord(x) = min{[itex]r \geq 1: x^r = 1[/itex]} If [itex]\theta[/itex]: G > H is an injective group homomorphism show that, for each xEG, ord([itex]\theta(x)[/itex]) = ord(x) My answer: Please verify If [itex]\theta(x)[/itex] = {[itex]x^r: r \epsilon Z[/itex]} then ord([itex]\theta(x)[/itex]) = ord(x). For any integer r, we have x^r = e (or 1) if and only if ord(x) divides r. In general the order of any subgroup of G divides the order of G. If H is a subgroup of G then "ord (G) / ord(H) = [G:H]" where [G:H] is an index of H in G, an integer. So order for any xEG divides order of the group. So ord([itex]\theta(x)[/itex]) = ord(x) any suggestions or changes please? thnx :) 


#2
Jan1208, 08:32 PM

Math
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Thanks
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P: 39,348




#3
Jan1308, 07:29 AM

P: 39

i believe we have to show 2 things:
i) [itex](\theta(x))^a = e'[/itex] ii) [itex]0 < b < a \implies (\theta(x))^b \neq e'[/itex]. ok so basically: If ord(x)=a then [itex]\left[ {\phi (x)} \right]^a = \left[ {\phi (x^a )} \right] = \phi (e) = e'[/itex]. Now suppose that [itex]ord\left[ {\phi (x)} \right] = b < a[/itex]. Then [itex]\left[ {\phi (x)} \right]^b = e' = \left[ {\phi (x)} \right]^a [/itex] [itex]\phi (x^b ) = \phi (x^a )[/itex] [itex]x^b = x^a[/itex] (injective) [itex]x^{a  b} = e[/itex] there seems to be a contradiction where if [itex]x^{a} = x^{b}[/itex], then [itex]\left[ {\phi (x)} \right]^b = e'[/itex] which is not what statement (ii) says. am i correct in this assumption? any ideas on how to deal with this? 


#4
Jan1608, 09:12 PM

P: 49

Orders of Groups
When you have [itex]x^{ab} = e[/itex], then ab is positive since you assumed b<a. But this contradicts the definition of order. Done.



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