Is the group cyclic? Introduction
You can skip this part if you already know what a cyclic group is.
A group is defined by a set and an associative binary operation \$\times\$ (that is, \$(a \times b) \times c = a \times (b \times c)\$. There exists exactly one element in the group \$e\$ where \$a \times e = a = e \times a\$ for all \$a\$ in the group (identity). For every element \$a\$ in the group there exists exactly one \$b\$ such that \$a \times b = e = b \times a\$ (inverse). For every two elements \$a, b\$ in the group, \$a \times b\$ is in the group (closure).
We can write \$a^n\$ in place of \$\underbrace{a\times a\times a\times ...\times a}_{n \text{ times}}\$.
The cyclic subgroup generated by any element \$a\$ in the group is \$&lt;a&gt; = \{e, a, a^2, a^3, a^4, ..., a^{n-1}\}\$ where \$n\$ is the order (size) of the subgroup (unless the subgroup is infinite).
A group is cyclic if it can be generated by one of its elements.
Challenge
Given the Cayley table (product table) for a finite group, determine whether or not it's cyclic.
Example
Let's take a look at the following Cayley table:
1 2 3 4 5 6
2 3 1 6 4 5
3 1 2 5 6 4
4 5 6 1 2 3
5 6 4 3 1 2
6 4 5 2 3 1

(This is the Cayley table for Dihedral Group 3, \$D_3\$).
This is 1-indexed, so if we want to find the value of \$5 \times 3\$, we look in the fifth column on the third row (note that the operator is not necessarily commutative, so \$5 \times 3\$ is not necessarily equal to \$3 \times 5\$. We see here that \$5 \times 3 = 6\$ (also that \$3 \times 5 = 4\$).
We can find \$&lt;3&gt;\$ by starting with \$[3]\$, and then while the list is unique, append the product of the last element and the generator (\$3\$). We get \$[3, 3 \times 3 = 2, 2 \times 3 = 1, 1 \times 3 = 3]\$. We stop here with the subgroup \$\{3, 2, 1\}\$.
If you compute \$&lt;1&gt;\$ through \$&lt;6&gt;\$ you'll see that none of the elements in the group generate the whole group. Thus, this group is not cyclic.
Test Cases
Input will be given as a matrix, output as a truthy/falsy decision value.
[[1,2,3,4,5,6],[2,3,1,6,4,5],[3,1,2,5,6,4],[4,5,6,1,2,3],[5,6,4,3,1,2],[6,4,5,2,3,1]] -&gt; False (D_3)
[[1]] -&gt; True ({e})
[[1,2,3,4],[2,3,4,1],[3,4,1,2],[4,1,2,3]] -&gt; True ({1, i, -1, -i})
[[3,2,4,1],[2,4,1,3],[4,1,3,2],[1,3,2,4]] -&gt; True ({-1, i, -i, 1})
[[1,2],[2,1]] -&gt; True ({e, a} with a^-1=a)
[[1,2,3,4,5,6,7,8],[2,3,4,1,6,7,8,5],[3,4,1,2,7,8,5,6],[4,1,2,3,8,5,6,7],[5,8,7,6,1,4,3,2],[6,5,8,7,2,1,4,3],[7,6,5,8,3,2,1,4],[8,7,6,5,4,3,2,1]] -&gt; False (D_4)
[[1,2,3,4,5,6],[2,1,4,3,6,5],[3,4,5,6,1,2],[4,3,6,5,2,1],[5,‌​6,1,2,3,4],[6,5,2,1,‌​4,3]] -&gt; True (product of cyclic subgroups of order 2 and 3, thanks to Zgarb)
[[1,2,3,4],[2,1,4,3],[3,4,1,2],[4,3,1,2]] -&gt; False (Abelian but not cyclic; thanks to xnor)

You will be guaranteed that the input is always a group.
You may take input as 0-indexed values.
Husk, 11 10 9 bytes

VS≡`ȯU¡!1


1-based.
Returns the index of a generator if one exists, 0 otherwise.
Try it online!

Explanation

V          Does any row r of the input satisfy this:
      ¡!    If you iterate indexing into r
   `    1   starting with 1
    ȯU      until a repetition is encountered,
 S≡         the result has the same length as r.