Let i= $\sqrt{-1}$ $\in$ Complex Numbers and H=$\{$a+bi : a,b $\in$ Z$\}$. Show that H is not cyclic. The operation is (Complex Numbers(C),+).
In my opinion,
Suppose, (C,+) is cyclic.
Then , there exist a x $\in$ C s.t $\lt$x$\gt$= C.
Note, x=a+bi
Note, C=$\lt$x$\gt$=$\{$n.x : n $\in$ N$\}$
Note, (1/2)x $\in$ C ,but (1/2)x is not element of $\lt$x$\gt$. If it were, (1/2)x = nx for some n $\in$ N
Note, 1/2=n a contradiction.
Hence, (C,+) is not cyclic.
Is this proof correct ? I'm confused about we can choose a,b $\in$ Z.
If this proof whether true or not could you please give an another proof of this?
Suppose $H$ is cyclic with generator $z = a+bi$.
Note that $i = 0+i \in H$. Thus, if $H$ is cyclic, $i = na + (nb)i$ for some $n \in \Bbb Z^+$. This leads to:
$na = 0, nb = 1$.
Since $n \neq 0$, we must have $a = 0$ (because $\Bbb Z$ is an integral domain). Since $nb = 1$, we must have $b = 1$ (since $n > 0$).
But $1 = 1 + 0i \in H$, as well, and $mi = 1$ has no solutions for $m \in \Bbb Z^+$.