g is a generator for G, is also g^n a generator?

Question

The title is general, but I have a specific problem:

$2$ is a generator for $\mathbb Z_{179}^*$. Is $2^{1977769}$ also a generator?

Is there a general way to solve this kind of problem?

Answer 1

Since $179$ is a prime number, the group $\mathbb{Z}_{179}^*$ is cyclic of order $178$. Compute $\gcd(1977769,178)$: if this is $1$ then you have a generator, otherwise you have not a generator.

Now, $\gcd(1977769,178)= 1$, so $2^{1977769}$ generates $\mathbb{Z}_{179}^*$.

Answer 2

In general, the answer is no: If $g$ generated the group $G$ of order $n$, then we have $g^n = 1$, and unless $n=1$, $1$ will not generate $G$.

What can be said is the following: If $n$ and the order $d$ of $G$ are relatively prime, $g^n$ will also generate $G$: Suppose that $(g^n)^k = 1$, then $g^{nk} = 1$, hence $nk$ is a multiple of $d$, say $nk = md$. As $d$ and $n$ are relatively prime, $d \mid k$. Hence the order of $g^n$ is $d$, and therefor $\left<g^n\right>=G$.

Answer 3

Assuming you are only talking about finite cyclic groups then you can use this - $g^n$ is a generator of $G$ iff gcd$(n, m) $ = 1 where $m$ is the order of $G$