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#2 (permalink) |
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I JUST got here.
Join Date: Apr 2008
Posts: 2
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you forgot x^9 as well. This means two of the elements are the identity of the group. x^3 is the identity because that means both x^3 and x^9 are the identity (e^3=e). Thus, there are three elements in the other group: x^13, x^26, x^39=e
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#3 (permalink) | |
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I JUST got here.
Join Date: Oct 2008
Posts: 5
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what do u mean by {x^3,x^5,x^9} has 2 elements. It has 3 elements rights? and why should two of them be identities? |
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#4 (permalink) | |
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I JUST got here.
Join Date: Oct 2008
Posts: 5
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what do u mean by {x^3,x^5,x^9} has 2 elements. It has 3 elements rights? and why should two of them be identities? ![]() |
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#6 (permalink) | |
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I JUST got here.
Join Date: Oct 2008
Posts: 5
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why are u taking products of power and the element? it is supposed to be x^3= 5^3 = 125 right? why are u taking the product 5*3?? Thanks a lot for your time! ![]() |
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#8 (permalink) | |
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I JUST got here.
Join Date: Oct 2008
Posts: 5
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thus the group x^(13n) = {1,x,x^2,x^3} ie 4 elements but as this is not in the options we can take x^3=x^9 ..this is just a made up solution though. I think u r correct..but im just wondering how is it an additive group? Also can u clear one doubt of mine.. What do u mean by order of a SUBGROUP? The order of the group = number of elements in the group or equivalently we can say a^n=1 then n=order of the group. Does the same defn hold for order of a subgroup? is it the number of elements in the subgroup? Thanks a lot lime! have u given your subject gre already? ![]() |
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#9 (permalink) | ||
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Glutton!
![]() ![]() ![]() Join Date: Nov 2007
Location: Russia
Posts: 949
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I see you're thinking too much. Instead of checking the answer by specific example, you're trying to find general solution.
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#10 (permalink) |
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I JUST got here.
Join Date: Dec 2008
Posts: 5
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The group is given as a multiplicative group of order 15 (with 15 elements).
So let's keep it multiplicative. Every element x has order (least power = to identity e) which divides 15. So order of x is 1, 3, 5, 15. If order is 1, 5 or 15 then {x^3,x^5,x^{9}} = {e}, {x^3, e, x^4} or {x^3, x^5, x^9} all distinct in last two cases. So, since the set has two distinct elements, the order of x is 3. Then x^3=x^9=e and x^13=(x^12)x =ex=x. So x^{13n} = x^n. Since x has order 3, x^{13n} achieves 3 values as n goes from 1 to infinity. So {x^{13n} | n is in N} has 3 elements. |
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