yachoo Posted April 8, 2008 Share Posted April 8, 2008 hello, I have another problem to explain the answer:are the two elements in the given set x^3, x^5? how to find 3 elements in set of x^13n, n is a positve integer?:rolleyes: Quote Link to comment Share on other sites More sharing options...
mraudiofreak Posted April 11, 2008 Share Posted April 11, 2008 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 Quote Link to comment Share on other sites More sharing options...
anushrimali Posted October 31, 2008 Share Posted October 31, 2008 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 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? Quote Link to comment Share on other sites More sharing options...
anushrimali Posted October 31, 2008 Share Posted October 31, 2008 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 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? :whistle: Quote Link to comment Share on other sites More sharing options...
lime Posted November 3, 2008 Share Posted November 3, 2008 Consider group Z15 under addition. Then element x=5 satisfies given conditions. Indeed: 5^3=3*5=15=0 5^5=5*5=25=10 5^9=9*5=45=0 Then x^13=65=5 Element 5 generates subgroup of order 3. Quote Link to comment Share on other sites More sharing options...
anushrimali Posted November 6, 2008 Share Posted November 6, 2008 Consider group Z15 under addition. Then element x=5 satisfies given conditions. Indeed: 5^3=3*5=15=0 5^5=5*5=25=10 5^9=9*5=45=0 Then x^13=65=5 Element 5 generates subgroup of order 3. 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! :) Quote Link to comment Share on other sites More sharing options...
lime Posted November 6, 2008 Share Posted November 6, 2008 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?? The operation in this group is addition. That means that x^3 = x+x+x x^n = x+x+...+x (n times) = nx Quote Link to comment Share on other sites More sharing options...
anushrimali Posted November 6, 2008 Share Posted November 6, 2008 The operation in this group is addition. That means that x^3 = x+x+x x^n = x+x+...+x (n times) = nx hi lime, theres so mention of it being a group with addition operation. Also if we consider it to a normal power and equate 2 elements x^5=x^9 ..we get x^4=1 ..4=order 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?:) Quote Link to comment Share on other sites More sharing options...
lime Posted November 7, 2008 Share Posted November 7, 2008 I see you're thinking too much. Instead of checking the answer by specific example, you're trying to find general solution. hi lime, theres so mention of it being a group with addition operation.Despite it doesn't say anything about addition, I brought up Z15 just as the example of group that satisfied given condition. Also...The order of the group = number of elements in the group ... Does the same defn hold for order of a subgroup?Yes. Quote Link to comment Share on other sites More sharing options...
oldmathguy Posted December 16, 2008 Share Posted December 16, 2008 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. Quote Link to comment Share on other sites More sharing options...
amintwist Posted March 25, 2009 Share Posted March 25, 2009 Excuse me, What is this "Practice Book"? Could you please give me a link to download? Quote Link to comment Share on other sites More sharing options...
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