Few days ago, I found a problem that you can use Cayley table to see group is cyclic. I also use this site but I did clear about it. As they state that any element just below the identity element in row is generator. But when I take group of fourth root of unity then it is not true. So please give me what is actual idea ?
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$\begingroup$A cyclic group of order $n$ has exactly $\phi(n)$ generators. In your case, $G=\{1,-1,i,-i\}$ is a cyclic group under multiplication of order $n=4$, see here. Hence only $\phi(4)=2$ elements are generators, i.e., $\pm i$. Once we have found a generator, we know that the group is cyclic. This can be easily seen from the Cayley table, by reproducing the Cayley table with the powers of a generator $a^0,a^1,\ldots,a^{n-1}$, whose product is given according to the rule $a^ia^j=a^{i+j}$.
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