Generator of a group G Assume that $g$ is a generator for a group $G$ of order n and assume $x = g^m$ for some value of $m$ , my question is , is $x$ a generator for group $G$ as well ?
Since $g$ is a generator then $G = \{g^0,g^1,g^2,.....,g^{n-1}\}$ then imagine $x = g^2$ for instance then $<x> = \{g^2,g^4,g^6,...\}$ and so $x$ can not be a generator for $g$ unless $m=1$ maybe ? , am I right here or wrong ?
 A: $x=g^m$ is a generator of the group iff $\gcd(m,n)=1$. In other words, $m$ must be relatively prime to $n$. You can see why as follows. Every element of $\{g^0,g^1,\cdots,g^{n-1}\}$ is distinct since the group has order $n$. So you want for each $k=0,1,\cdots,n-1$ a non-negative integer $y$ such that $x^y=g^k$. Writing out the definition of $x=g^m$ gives $my\equiv k\mod n$. If $\gcd(m,n)=1$ then you can always find a $y$. On the other hand, if $\gcd(m,n)>1$, then $m$ has no inverse $\mod(n)$ so there will be some $k$ for which you can't satisfy the equality. In your example of $x=g^2$, you will never get $g^3$ by taking powers of $x$. 
A: Suppose $g$ generates $G$, a group of order $n$, and $x=g^m$ for some $m$. We claim that $x$ will generate $G$ if and only if $\gcd(m,n)=1$.
Suppose $\gcd(m,n)=1$. Then we may write $1=am+bn$ where $a,b\in \mathbb{Z}$. So $g^1=g^{am+bn}=g^{am}g^{bn}=(g^m)^a(g^n)^b=(g^m)^ae=x^a$. Then for every power of $k$ of $g$, take the $ak$th power of $x$.
For the only if part, we will proceed by contradiction. Suppose, for a contradiction $\gcd(m,n)>1$ but, we can generate $G$ with $x$. Then in particular $g=x^k$ for some power $k$. Substitution on this equality yields $g=g^{mk}$ or $e=g^{mk-1}$. This is equivalent to saying $n|(mk-1)$. So there exists $p\in \mathbb{Z}$ such that $np=mk-1$. By a simple subtraction on both sides we see $1=mk-np$. But here we have written $1$ as a linear combination of $m,n$. So $1$ must be the greatest common divisor after all. This contradicts our assumption and proves our assertion.
