Mathematics GRE

Ques : A cyclic group of order 15 has an element x such that the set { x^3, x^5, x^9 } has exactly two elements.
The number of elements in the set { x^13n : n is a +ve integer} is

A) 3
B) 5
C) 8
D) 15
E) Inf.

ans is A

Please explain your ans. Thanks

if x^3 and x^5 are the elements that are the same, which means x^3=x^5, multiply both sides by x^(-3), we have e=x^2, but here comes the problem, the three elements will all be the same if x^3=x^5, so x^3=x^9 and |x|=3.

since 13 and 3 are relatively prime, the distinct elements of { x^13n : n is a +ve integer} are x^13 x^26 and x^39.

btw, do you have any good material about abstract algebra?

If the group has order 15, then any element has order that divides 15, which immediately rules out x^3 = x^5 or x^5 = x^9.

enork wrote:If the group has order 15, then any element has order that divides 15, which immediately rules out x^3 = x^5 or x^5 = x^9.

Hi enork, do you have any good book suggestion on Abstract Algebra? I think I need to read some pages before taking the test.

Do you know where the ets is putting there emphasis about the 25% additional topics?

Thanks

The book I've got is Dummit and Foote's Abstract Algebra which I really like. There are probably plenty of other good books but that's the one I know. It covers way more stuff than I think is necessary for the test, but I don't know exactly what will show up on that.