An upper bound for the order of an element of the Rubik's cube group

The maximum order of an element of the Rubik's cube group R is 462 might be 462, or might be 1260, or might be something else.

Why? The Rubik's cube group is a subgroup of S₈ × S₁₂ × Z₃⁸ × Z₂¹²; the four factors here arise from possible permutations of the edge and corner pieces, and the orientations of the edges and the corners. (This is in fact twelve times as large as the actual group.) More concretely, it's the subgroup of this group consisting of quadruples (σ, τ, v, w), where σ ∈ S₈ and τ ∈ S₁₂ have the same parity, v is an element of Z₃⁸ whose coordinates sum up to zero, and w is a similar element of Z₂¹². The group operation in this group is given by working componentwise.

(If I have misunderstood what I've heard about Rubik's cubes, then please ignore the rest of this post.)

[edited, 8:48 pm: turns out I have, as the multiplication isn't exactly componentwise.]

So the order of (σ, τ, v, w) is the least common multiple of the orders of σ, τ, v, w in S₈, S₁₂, Z₃⁸, and Z₂¹² respectively. Call these o(σ), o(τ), o(v), o(w). If σ has one 7-cycle and one 1-cycle, τ has one 11-cycle and one 1-cycle, and v and w are nontrivial, then o(σ) = 7, o(τ) = 11, o(v) = 3, o(w) = 2, and so the order of (σ, τ, v, w) is lcm(7, 11, 3, 2) = 462. (This corresponds to a sequence of moves, whatever it may be, that permutes seven of the corners and eleven of the edges cyclically, and disrupts the orientations of both the corners and the edges.) Note that σ and τ are both even permutations.

Furthermore, we can't do better. I don't think there's a particularly elegant proof, but it's not hard to check by brute-force computation that no choice of the cycle types for σ and τ which gives both the same parity allows us to do better. If you let σ have one cycle of order 8, and τ have one cycle of order 7 and one of order 5, and let v, w both be nontrivial, then you'd think the order of (σ, τ, v, w) was lcm(8, 35, 3, 2) = 840, but σ is odd and τ is even. In fact, that's how I got the number 462 -- take lcm(s, t, 3, 2) where s ran over possible orders of elements of S₈ and t ran over possible orders of elements of S₁₂; the largest number you get this way is 840, but it can only come about in the way I described. 462 is the next-largest.

In case you're wondering how I came to this question -- it's easy to see if you play around with the cube that if you do the same sequence of moves over and over again, you eventually get back where you started. (Theoretically, this is a consequence of the fact that any element of a group has finite order.) So I just started to wonder how many times you might have to do the same sequence of moves on a solved cube to get it back to the solved state.