Group structure of rotating-square puzzle Suppose we arrange the numbers $1$ through $6$ at the "vertices" of the shape formed by aligning the sides of two squares, as shown below:

In this "puzzle," the only moves allowed are rotating the vertices of either square counterclockwise.
I would like to find the group $G$ that represents this puzzle, but I can't figure out how to account for the interaction between the two squares. All I know right now is that $G\subset S_6$, and that $G$ is generated by the permutations $(1254)$ and $(2365)$.
However, I can't figure out how to express $G$ using well-known groups like $S_n$, $A_n$, $D_n$, and $\mathbb Z_n$, the direct product $\times$, and the semidirect product $\rtimes$ (with no corresponding homomorphism specified).
Can someone please show me how to find the group corresponding to this game?
NOTE: To someone who is experienced with group theory, this is probably an easy exercise; however, to a novice like myself, this is quite confusing

Here are a few tips for identifying the isomorphism type of a small group like this one:


*

*Determine generators.

*Determine the size of the group.

*Determine the orders of its elements.

*Determine relations between the generators.


Once you have done this, you can consult a list of small groups to get a short list of candidates. You already know that $G$ is isomorphic to a subgroup of $S_6$, so you can consult a list of subgroups of $S_6$ right away, for example here.
Here's a start from my end; you have observed $G=\langle(1\ 2\ 5\ 4),(2\ 3\ 6\ 5)\rangle$. Because these two $4$-cycles do not commute, this implies $|G|$ is a multiple of $8$. Moreover, the product
$$(1\ 2\ 5\ 4)(2\ 3\ 6\ 5)=(1\ 2\ 3\ 6\ 4),$$
shows that $G$ has an element of order $5$, so its order is a multiple of $40$. Trying out another combination of the generators yields
$$(1\ 2\ 5\ 4)(2\ 3\ 6\ 5)^2=(1\ 2\ 6\ 5\ 3\ 4),$$
so $|G|$ is a multiple of $6$ as well. The linked document then leaves only three isomorphism types for $G$; either $S_5$, $A_6$ or $S_6$. Because both generators are odd permutations, it cannot be $A_6$. Then it is either $S_5$ or $S_6$, and you could ask whether every permutation of the vertices can be achieved.
More simply, since $G$ contains the $6$-cycle 
$$(1\ 2\ 6\ 5\ 3\ 4)=(3\ 6\ 4\ 5)(1\ 2\ 3\ 4\ 5\ 6)(3\ 5\ 4\ 6),$$
and since $(1\ 2\ 3\ 4\ 5\ 6)$ and any transposition together generate $S_6$, you could ask whether $G$ contains any element of the form $(3\ 6\ 4\ 5)\tau(3\ 5\ 4\ 6)$, where $\tau$ is a transposition.

This is clearly a very ad hoc approach, and some familiarity with small groups goes a long way in making the right considerations. Using a software package such as GAP can be a very helpful alternative.
