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Click on this link if you are not familiar with Rubik's cube notation

Suppose I want to repeat some arbitrary pattern of moves on a solved Rubik's cube, until it is solved again, i.e. F R F R F R... and so on, may be the simplest non-trivial option. Is there a way to determine how many moves, or equivalently iterations of the base sequence until it returns to solved again? (This question applies to any arbitrary base pattern, not just F R).

I think this can be characterised as a group theory problem where if we call for example, the action of F as $f$, and the action of R as $r$, then we want to solve for $(f*r)^n=e$, but I wouldn't know how to go about solving this, as I'm a physics undergrad, and haven't formally covered any group theory.

The sequence above (with base F R), I believe has a cyclic number of 115, if that is of any use.

FizzKicks
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  • This older question may be interesting to you: https://math.stackexchange.com/questions/2392906/determine-the-highest-order-of-an-element-of-a-rubiks-cube-group/3029900#3029900 – DreiCleaner Jun 15 '20 at 17:39

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Any action of a group $G$ on a set $X$ gives a homomorphism $G \to \mathrm{Aut}(X)$. If $X = \{1, 2, \ldots, n\}$ then this is a homomorphism $G \to S_n$. The nice thing about $S_n$ is that once a permutation is written down in cycle notation it's trivial to find it's order, it's the LCM of the cycle lengths.

If you can find a faithful action of $G$ on some $\{1, \ldots, n\}$, i.e., an action for which no group element acts as the identity, then the resulting homomorphism $G \to S_n$ is injective and you can compute orders by first mapping into $S_n$.

One obvious faithful action would be to number all the sides of the cubes and then the Rubiks cube group acts by moving those numbers around.

Jim
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