Abstract algebra cyclic subgroups

tyrannosaurus
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Homework Statement


Suppose that G is a group with exactly eight elements of order 10. How many cyclic subgroups of order 10 does G have?

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The Attempt at a Solution


I really don't have a clue how to solve this, any help would be greatly appreciated.
 
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If you have one cyclic subgroup of order 10, how many elements of order 10 does that force G to have?
 
you would need four elements of order 10 (by the Euler phi function). does that mean there would be two cyclicsubgroups of order2?
 
tyrannosaurus said:
you would need four elements of order 10 (by the Euler phi function). does that mean there would be two cyclicsubgroups of order2?

You tell me. You definitely need more than one. Can two cyclic subgroups of order 10 share any elements of order 10?
 
there would only be 2 cyclic subgroups of order 10 because non of the subgroups can share an order 10 element because if they did share an element in common, that element would generate both groups, so the two groups would be the same. So this means that no two cyclic subgroups of order 10 can share an element of order 10.
Thanks a lot for the help, this made sooooooo much more sense now.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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