With the start of the proof, the author JOSEPH A. GALLIAN first claim that every element of $H$ can be written in the form $a^n$ where $n$ is positive. For this: Since $G=[a]$, every element of $H$ has the form $a^t$; and when $a^t$ belongs to $H$ with $t<0$, then $a^{-t}$ belongs to $H$ also and $-t$ is positive. Thus, our claim is verified.
But now I have a doubt in the same.
Take 4th root of unity under multiplication and take $i^{-3}$ instead of $i$ as both of them are equal i.e.
$$ H =(\{i^{-3}, i^2, i^3,i^4\},\times )$$ Here in $i^2,i^2,i^4\to 2,3,4$ are positive so take it as it is. In $i^{-3} \to -3$ is negative take inverse of $i^{-3}$ i.e. $ i^3$. Now the obtained element so far is in some positive power of $i$ and belongs to $H$ (i.e. $\{i^3,i^2,i^3,i^4\}$ but are not forming group.
Am I wrong or the author means something else than this.
I think you misunderstood the proof.
The author wants to claim that there exists at least one $a^t$ in the subgroup $H$ such that $t>0$. (Not that every element in $H$ is of the form $a^t$ where $t>0$.)
We may assume that $H$ is non-trivial subgroup.
Choose a non-identity element $x\in H$.
If $x=a^t$ where $t>0$, then we are done.
If $x=a^t$ where $t<0$, then note that $a^{-t}=x^{-1}\in H$. Since $-t>0$, so $x^{-1}$ is the desired element.