Question: Let $G$ be a cyclic group of order 6. How many of its elements generate $G$?
This was an old problem I had. I know that $G=\{1, g, g^2, g^3, g^4, g^5\} $ and have an answer written down that $g$ and $g^5$ are order $6$. But why?

Answer: 
All elements of order $k$ in a cyclic group of order $n$, such that $n$ and $k$ are relatively prime, generate the group. Thus, using your notation, $g$ and $g^5$ generate the group.
Well, $g^2$ cannot generate the whole group, because $g^2\*g^2$ = $g^4$, and $g^2\*g^2\*g^2$ = $g^6$ = $1$. Similarly, $g^3\*g^3$ = $1$ and $g^4\*g^4$ = $g^2$, and $g^4\*g^4\*g^4$ = $1$.

It is clear, from this, that $g^2$, $g^3$ and $g^4$ cannot generate all of the elements of G, because they fail to create $g^5$. You can confirm that $g^5$ and $g$ do actually generate every element.

As another answer has pointed out, this is due to the fact that 1 and 5 are the only numbers less than 6 that are prime to 6.