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
 A: GAP shows that the group is in fact isomorphic to $S_5$. A geometric interpretation was requested for what $5$ things are being permuted. Consider the following $5$ sets of edges between the vertices.
$(1,2,5,4)\leftrightarrow(orange,blue,purple,green)$
$(2,3,6,5)\leftrightarrow(red,blue,purple,green)$

To see that this is all of $S_5$ and not some subgroup, compute some products of elements.
$(orange,blue,purple,green)*(red,blue,purple,green)=(red,purple,orange,blue,green)$ 
which has order $5$, so the order of the group is a multiple of $5$.
$(orange,blue,purple,green)*(red,blue,purple,green)^{-1}=(red,orange,blue)$ 
which has order $3$, so the order of the group is a multiple of $3$.
And $(orange,blue,purple,green)$ has order $4$ so the order of the group is a multiple of $4$.
Now the order of the group must be a multiple of $3*4*5=60$, so either $S_5$ or $A_5$. But we have elements of order $4$ in our group, which leaves only $S_5$.
A: 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.
