contemporary Abstract Algebra?

Question

I dont really understand this question, actually I have problems with the subject, any help I'll appreciate, thanks

Q: suppose that G is a cyclic group and that 6 divides |G|. How many elements of order 6 does G have? If 8 divides |G|, how many elements of order 8 does G have? If a is one element of order 8, list the order elements of order 8

still confuse, dont get it

Answer 1

Let g be generator. Let the order be 6k

(g^k)^a = g^ka, for any a. Clearly this equals the identity when a = 6, but not for any smaller positive a. Hence, order (g^k) = 6. So the inverse of g^k, namely g^5k, also has order 6.

Next, let's look at an arbitrary element g^j with order 6 Then the order of the group, 6k, divides 6j. Hence k divides j. So the only possible elements with order 6 are g to the power of multiples of k, specifically g^nk for n = 0, 1, 2, 3, 4, 5

Well, we've already found the orders of g^1k and g^5k to be 6. You can check and see that the others are, in order, 1, 3, 2, and 3 -- i.e., not 6. So there are exactly two elements with order 6. (It is not a coincidence that 1 and 5 are relatively prime to 6, while 2, 3, 4, and of course 6 are not.)

In the second case, you'll find that the answer is 4, not 2, namely g^nk for n = 1, 3, 5, and 7. Not coincidentally, those four n are relatively prime to 8, while 2, 4, 6, and 8 are not.

Answer 2

Let |G|=6n for some n>0 and g be the generator of G. So g^(6n)=1. Assume an element g^k (k<6n) in G has order 6, then g^(k6)=1. But then k must be a multiple of n, let k=mn. Since k<6n, m<6. Furthermore, m is relatively prime to 6, or else the order of g^k=g^(mn) will not be 6. Thus, m=1 or 5 and we have exactly two elements that have order 6. The case for 8 is done similarly.