for the multiplicative group Z*27, I was checking to find our the if generators exist Investigate how many and list of these generators
to put in perspective I so far understood 27=3^3 which tells me generator do exist How many probable generators phi ((phi(27)) which comes to 6 So now I know there are 6 Generators, i could see Elements of Z27 would be 1 2 4 5 7 8 10 11 13 14 15 17 19 20 22 23 25 26
and this is where we are stuck
On
Apply to $p=r=3$ the general method to find a generator of $\mathbb Z_{p^r}^\times$ for $p$ (prime) $\ne2$ and $r>1$: let $\theta$ be a generator of $\mathbb Z_p^\times$ (here $\theta=-1$), then some power $\theta^k$ has order $p-1$ in $\mathbb Z_{p^r}^\times$ (here $k=1$), and a generator of $\mathbb Z_{p^r}^\times$ is $c=(p+1)\theta^k$ (here $c=-4$). The 6 generators are $c^k$ for $k$ coprime to $18$, i.e. $\pmod{27}$
$$c^1\equiv23,c^5\equiv2,c^7\equiv5,c^{11}\equiv11,c^{13}\equiv14,c^{17}\equiv20.$$
You may check e.g. in this table on Wikipedia. See also "Primitive Root" on MathWorld.
This theorem can be of assistance: in $\Bbb Z_n^×$, $g$ is primitive iff $$g^{\frac {\phi(n)}{q_i}}\not\equiv1\pmod n$$, where $q_i$ are the distinct primes in $\phi(n)'$s prime factorization.
So here $n=27$, so $\phi(27)=18=2\cdot 3^2$. So try $2$. Check $2^6\equiv 64\equiv 10\not\equiv 1\pmod {27}$. And $2^9\equiv 2^3\cdot2^6\equiv 8\cdot 10\equiv-1\not \equiv 1\pmod {27}$.
Thus $2$ is primitive.
Now the other $5$ are of the form $2^k$ for $(k,18)=1$.