If we are looking at the multiplication group modulo 17,
because it is a prime number, every member of the group (except the identity member of course) is a generator?
or –
there is a subgroup (non-trivial) ({1,2,4,8,16,15,13,9}, X mod 17 ) for example. and from its existenc,e we can assume there is some members which arenot generators?
Plus I saw a sentence that says that: "a group whose order is a prime number is necessarily cyclic, and each element except the identity is its generator (since the order of each element except the single element equals the order of the group)."
So I am really confused!! Also I found out that only 3 is a generator of modulo 17.
Best Answer
There seems to be confusion here between the additive and multiplicative groups of $\mathbb Z/17$.
The additive group, i.e., the group whose operation is addition modulo 17, consists of all 17 congruence classes mod 17. So it is a group of prime order (namely 17), which implies that it is cyclic and every element, except for the identity 0, is a generator. (Check it: Randomly choose a non-zero congruence class, and repeatedly add; you'll get all 17 congruence classes. For example, if you start with 4, you get (in order) 4, 8, 12, 16, 3, 7, 11, 15, 2, 6, 10, 14, 1, 5, 9, 13, 0.)
The multiplicative group, i.e., the group whose operation is multiplication modulo 17, consists of the 16 non-zero congruence classes. It does not contain 0, because 0 has no multiplicative inverse (modulo 17). Since 16 isn't prime, you can't infer from general group theory that this group is cyclic, but it is a theorem that the multiplicative group of a finite field (like $\mathbb Z/17$) is cyclic. So the multiplicative group in your situation is a cyclic group of order 16.
In such a group, a cyclic group of non-prime order, some of the elements are generators (that's what it means to be cycic), but not all of them. For example, 6 happens to be a generator modulo 17. If you start with 6 and repeatedly multiply by it (reducing modulo 17 at every step), you get (in order) 6, 2, 12, 4, 7, 8, 14, 16, 11, 15, 5, 13, 10, 9, 3, 1. So 6 is a generator of the multiplicative group. On the other hand, if you do the same thing starting with 4, you get 4, 16, 13, 1 and then the numbers just repeat. So 4 is not a generator of your multiplicative group. Similarly, if you tart with 2, you get the computation in your question, showing that 2 is not a generator of the multiplicative group. (It generates a subgroup of order 8.)
The bottom line is that you did a computation in the multiplicative group, which has order 16, but then quoted a theorem that applies to groups of prime order like the additive group of order 17. If you keep straight which of the two groups you're talking about, and the fact that their orders are slightly different, your difficulties should vanish.