Good day,.,.is it possible for a cyclic group to have only one generator???
It's not just possible. That's the definition of a cyclic group. Cyclic groups are groups generated by a single element.
So far as I know, the phrase "one generator" and "generated by a single element" mean the same thing.
[EDIT]: See below for a correction of the understanding here.
I'm not an algebraist, but that's my understanding. The wiki page to which I linked before has a precise definition of the term. It says, "In group theory, a cyclic group is a group that can be generated by a single element, in the sense that the group has an element g (called a 'generator' of the group) such that, when written multiplicatively, every element of the group is a power of g (a multiple of g when the notation is additive)."
Does that answer your question?
Oops. I have to be careful here. A cyclic group can be generated by only one generator. However, it may not be the case that every element in a cyclic group generates the whole group. In addition, more than one element in the group may generate the group. In $\displaystyle \mathbb{Z}_{5},$ for example, every element except the identity generates the group.
,.,.thnx a lot sir,.,now i understand,.,.one last question sir,.,.i find it easy now to find examples of cyclic groups with more than one generator such as Z7 and Z6,.,.but im having a hard tym looking for a cyclic group with one generator,.,can u give me an example??thnk u so much for ur help sir,.,
I would write that group additively as {0,1} using addition modulo 2, or multiplicatively as {-1,1} using regular multiplication. I think {e,1} might be a bit confusing unless it's understood that you're talking about a multiplicative group, and e = 0.
You're very welcome for whatever help I could provide.
the number of generators of a cyclic group of order $\displaystyle n$ is $\displaystyle \varphi(n)$ and the number of generators of an infinite cyclic group is $\displaystyle 2$. to see this, suppose $\displaystyle G = \langle x \rangle$ is a cyclic group of order $\displaystyle n$. then we know that the order of an element of $\displaystyle G$, say $\displaystyle x^k$, is $\displaystyle \frac{n}{\gcd(k,n)}$. so $\displaystyle x^k$ is a generator of $\displaystyle G$ if and only if $\displaystyle \gcd(k,n)=1$ and this proves what i said.
if you want your group to have one generator only, it means $\displaystyle \varphi(n)=1$ which has two solutions: $\displaystyle n = 1,2.$