Hi!

Firstly let's revise the definition of subgroup:

Let be a group ( is the binary operation, is unary operation of taking an inverse and is the identity element). If a set satisfies following conditions:

1)

2) contains the identity element of ,

3) is closed under the operation , that is, for every also

4) is closed under the operation , that is, for every also ,

then we say that is asubgroupof .

This means that the set together with operation restricted to and with the identity element forms a group.

In Z18, for example is a subgroup, or whole Z18 is a subgroup, or is a subgroup,... (check it!)

Then, the definition of a generator of a group:

LetGbe a group and be a set. We say thatSis agenerating set of G,if every element ofGcan be expressed as the product of finitely many elements ofSand their inverses. We write this as <S> =G. When there is only one element inS, , we write <x> =Gand we say that the groupGis generated byx, or thatxis its generator.

Now to your example. What are the generators of Z18? We see that 1 is surely a generator (every element of Z18 can be expressed as 1+1+..+1)

Are there any other generators? Yes, exactly those stated in your book, let's see why. Firstly, can you see why the remaining elements are not generators?

Spoiler:

Next, let's verify that every of the elements 1,5,7,11,13 and 17 generates Z18. If you have heard of Bezout's identity, it is easy. It says that ifaandbare nonzero integers with greatest common divisor d, then there exist integersxandysuch that .

In our case, take one of those 1,5,7,11,13 and 17, take and, of course, . Can you finish from here?