#59 in practice book

[yachoo]

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?

[mraudiofreak]

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

[anushrimali]

Originally Posted by mraudiofreak

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?

[anushrimali]

Originally Posted by mraudiofreak

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?

[lime]

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.

[anushrimali]

Originally Posted by lime

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!

[lime]

Originally Posted by anushrimali

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

[anushrimali]

Originally Posted by lime

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?

[lime]

I see you're thinking too much. Instead of checking the answer by specific example, you're trying to find general solution.

Originally Posted by anushrimali

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.

[oldmathguy]

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.