About a PermutationGroup with generators 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.
 A: *

*It is not true that if you write $$G = \langle (1\;2\;3\;4\;5\;6),
    (2 \; 6)(3\; 5)\rangle$$ that there are "no relations", because the
elements $(1\;2\;3\;4\;5\;6)$ and $(2 \; 6)(3\; 5)$ are interpreted
as elements of $S_6$, which does have relations. Indeed, the
relations $r^n = ()$, $t^2 = ()$, $(rt)^2 = ()$ are satisfied in
$S_n$ when letting $r = (1\;2\;3\;4\;5\;6)$ and $t = (2 \; 6)(3\;
5)$.

*The resulting group is finite because $S_n$ is finite, and the group generated by elements of a group is a subgroup of the ambient group.

*It's not true that if the generators of a group have the same order, then the groups are isomorphic. Consider Your example: $D_{12}$ is generated by elements of order $6$ and $2$. But so is $\mathbb Z_6 \times \mathbb Z_2$, which is abelian, but $D_{12}$ is not. Thus, they are not isomorphic. You actually need more relations than just the orders of elements. It is not even true that if the group is generated by elements of finite order, then the group is finite. There are many examples. My favorite one is the group 
$$\text{PSL}(2, \mathbb Z) = \text{SL}(2, \mathbb Z)/\{\pm 1\},$$
which is generated by the equivalence classes of the matrices
$$ 
\begin{pmatrix}
1 & -1 \\ 1 & 0
\end{pmatrix}
\text{ and }
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}.
$$
These two matrices have order $3$ and $2$, respectively, but the group $\text{PSL}(2, \mathbb Z)$ is infinite.

*No, this is not always true. Consider $S_6$, which is not a semidirect product, but it is the group generated by $(1\; 2\; 3 \; 4 \; 5 \; 6)$ and $(1 \; 2)$.

