I'm studying Dixon's Permutation Group and I came across a PermutationGroup with these generators: $$G=\langle(1 2 3 4 5 6),(2 6)(3 5)\rangle$$ (page 15).
I almost immediately recognised this could be related on symmetries of the Hexagon, and turned out to be isomorphic to $D_{12}$, i.e. the symmetry group of a regular Hexagon.
However, the way in which is defined actually highlighted some ignorance of mine, on that.
So here we are dealing with a group generated by a couple of generators, let's call them $\langle a,b\rangle$. Actually also SAGE Math is used to define permutations group in this way, so it's a common way.
But some doubts came to my mind:
1) this presentation is without any relation. Can we consider it as a free group, since no relations?
2) assuming 1) true, how we could be sure the resulting group is finite? I mean, this case is simple, since you enumerate first the 6 elements in $C_6$ and then you got the other 6 ones by multiplying $C_6$ by $(2 6)(3 5)$. But in a general case, may be with more than two generators but without any relations, how you can get it's finite?
3) assuming you can tell that (I'm sure there's a way, but I had no formal studies on groups and I'm studying them for fun since years), is there a way to compute the group order given the order of its generators? I know these latter should divide the order of the group, by Lagrange theorem, but I don't think it's a simple product of them always.
4) actually $D_{12}$ can be seen as a semi-direct product of $C_2$ and $C_6$. Is this always true? I mean, getting a couple of generators without relations, composing them "naturally", would always give back a group made a semi-direct product?
Actually, you can present a dihedral group by $$\langle r, t \; | \; r^n, t^2, {(rt)}^2\rangle.$$ Here we are getting the same result, without specifying any relations.
Thank for you support and sorry if I asked some obvious question, from a trained eye.
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