I read online that the generator for the cyclic group $(\Bbb Z_4^*,\times)$, where $\Bbb Z_4^*:=\{1,2,3,4\}$ are 2 and 3 .
I tried working this out but it doesn't make sense to me , here's what I did ;
$$\langle2\rangle=\{2^0,2^1,2^2,...\}=\{1,2,0\}$$, But three is not included in this set ?
$$\langle3\rangle:=\{3^0,3^1,3^2,3^3,...\}=\{1,3\}$$.
What am I doing wrong ?
The ring of integers modulo $\;4\;$ is the set of equivalence classes of residues modulo $\;4\;$ in the integers $\;\Bbb Z\;$, meaning: all the possible residues one can obtain when dividing any integer by four.
The set of representatives for the four equivalence classes is usually taken to be $\;\{0,1,2,3\}\;\pmod 4\;$, yet this does not have to be this way. For example, the set you took $\;\{1,2,3,4\}\;$ is also a full set of representatives of residues modulo $\;4\;$ .
So we can take $\;\Bbb Z_4=\{0,1,2,3\}=\{1,2,3,4\}=\{-3,-1,2,12\}\;$ , etc. Any set with four integers which are different residues modulo $\;4\;$ can do it, so by simplicity we assume $\;\Bbb Z_4=\{0,1,3,4\}\pmod4\;$
Now, for any (commutative, for simplicity) unitaru ring $\;R\;$ , the set $\;R^*\;$ is the set of all units in the ring, meaning:
$$R^*:=\{r\in R\;|\;\exists\,s\in R\;\;s.t.\;\;rs=1\}$$
In your case, $\;\Bbb Z_4^*=\{1,3\}\;$ , as you can easily check. As this is an abelian (multiplcative) group of prime order, it is cyclic...and etc.