Order of a permutation group I'm playing with a permutation group, with generators $(1,2,3)(6,5,4)$ and $(2,5,7)(8,6,3)$, in cycle notation. After careful counting, I believe its order is 24. But I have no real method, except computing products until I exhaust all possibilities.
Is there a general method to compute order of a permutation group, which I could apply here ?
 A: There is a general computer algorithm for such calculations (Schreier-Sims), but it is tedious to do it by hand.
Carrying on with DonAntonio's calculations, you might notice that $(ab)^2 = (1,4)(3,5)(2,6)(7,8)$ commutes with both $a$ and $b$, so the subgroup $H = \langle (ab)^2 \rangle$ of $G$ has order 2 and is in the centre of $G$.
Now, if we let $\overline{a}$ and $\overline{b}$ be the images $aH$ and $bH$ of $a,b$ in the quotient group $G/H$, we have $\overline{a}^3 = \overline{b}^3 = (\overline{ab})^2 = 1$.
So, $G/H$ is a quotient group of the group $\langle x,y \mid x^3=y^3=(xy)^2=1 \rangle$, which is well known to be a presentation of the alternating group $A_4$ of order 12. So we have $|G/H| \le 12$ and hence $|G| \le 24$.
A: I don't think there is a general method for all cases but work each one separatedly. Let us put
$$a:=(123)(465)\;,\;\;b:=(257)(386)$$
Then we clearly have that $\,G:=\langle\,a,b\,\rangle\le A_8\,$ and the following relations hold:
$$a^3=b^3=1\;,\;\;ab=(1246)(3857)\neq(1543)(2867)=ba$$
and thus the group is non-abelian, and also:
$$a^{-1}ba=(167)(284)$$
$$b^{-1}ab=(176)(248)$$
Thus we already have four elements of order $\,3\,$, each generating a subgroup of $\,G\,$ of order three. IF $\,|G|=24=2^3\cdot 3\,$ then the number of Sylow $\,3-$subgroups is either one or four, so if there one more element of order $\,3\,$ which is not a power of any of the above elements then it cannot be $\,|G|=24\,$ , for example.
Try to continue from here with other ideas.
A: The general method to compute the order of a permutation group involves is called the Schreier-Sims algorithm, and involves computing a so-called Base and Strong Generating Set. That's a fairly tricky procedure which is best done by a computer. In a nutshell, and in your case, it boils down to the following observations: 


*

*The only element in your group that fixes the points 1 and 2 at the same time is the identity. That's the base.

*The pointwise stabiliser of 1 is generated by $(2 5 7)(8 6 3)$ and has order three.

*The index of the pointwise stabiliser of 1 has index 8 in the full group (I'm not sure of how to explain that nicely, and how you can figure that out, as I'm pretty new to this stuff as well).


So: yes, there is a procedure which a lot faster than enumerating all the possible products, but you'll probably need either brute force (the size of 24 is correct, by the way!) or some smart observations (like DonAntonio and Derek Holt provided). 
