Which group is this subgroup isomorphic to?

Question

I am working with SageMath on graphs, in particular on this graph:

I have found that its automorphism group is generated by the elements $(1,3,5)$ and $(2,4,6)$: how do I find to which group is this group isomorphic to? I ask both computationally and mathematically.

Answer 1

Since they are disjoint three-cycles, they commute and have order three; there are no other restrictions on the group; so the group is isomorphic to the group given by the presentation

$$\langle a,b\mid a^3, b^3, ab=ba\rangle,$$

which is $\Bbb Z_3\times\Bbb Z_3$.

Computationally, you could use GAP.

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If the automorphism group is, instead, generated by $(135)(246)$, then it is cyclic and the order of the group is that of the generating element, which is three; hence it is isomorphic to $\Bbb Z_3$.

Answer 2

The permutations $(135)$ and $(246)$ commute, so this automorphism group is just the direct product of the cyclic groups they generate, i.e. is isomorphic to $(\mathbb{Z}/3\mathbb{Z}) \times (\mathbb{Z}/3\mathbb{Z})$.

I don't know a computational way to recognize the isomorphism class of a group from realizing it as a subgroup of $S_n$ (that might even already be an undecidable problem, not sure). It is likely GAP or some such software can attack this problem, though.