# Rubik's cube, elements of order $7$ and $11$

The order of the Rubik's cube group is $$43 252 003 274 489 856 000 = 2^{27} \times 3^{14} \times 5^3 \times 7^2 \times 11$$ Cauchy's theorem guarantees an element of order 7, as well as one of order 11.

An element of order 2 is R2. An element of order 3 is a U permutation. An element of order 4 is R. An element of order 5 is R U R' U.

But I am unable to find an element of order 7 or one of order 11. Does any one know of such elements?

• There is even an element of order $1260$. See this post, for example. And $o(RU)=105$, see here. May 19 at 19:33
• btw it should be $7^2$ but it won't let me edit :-/ May 19 at 19:53
• So $(RU)^{15}$ is of order $7$ May 19 at 19:54
• @timidpueo done :) May 19 at 20:03

Once you know how to permute $$3$$ edges without changing the orientation, you can make permutations of order $$7$$ and $$11$$ easily.

A U-permutation will preserve orientations when applied on four of the faces, say R, U, L, or D, but two edges will change orientation when used on F or B. In the latter case, just re-flip those edges after the U-permutation.

Now the above algorithm is just a cyclic permutation of three edges with everything else left untouched. So numbering the edges $$1$$ to $$12$$, we can string them together with an overlap of one between each (here I use cycle permutation notation evaluated right to left): $$(1\,2\,3)(3\,4\,5)(5\,6\,7) = (1\,2\,3\,4\,5\,6\,7)$$ and $$(1\,2\,3)(3\,4\,5)(5\,6\,7)(7\,8\,9)(9\,10\,11) = (1\,2\,3\,4\,5\,6\,7\,8\,9\,10\,11)$$ I'm (probably) not gonna sit down and convert this to a full algorithm in Rubik's notation, but it's perfectly doable.

The shortest element of order 7 has length 4 (by shortest I mean takes the least moves in half turn metric). An example is: $$R\,U'\,F'\,U$$ Here are a couple other examples of short elements of order 7. Consider $$R\,U'$$, which has order 63, therefore $$(R\,U')^{9}$$ has order 7. Also consider $$R\,U$$, which has order 105, therefore $$(R\,U)^{15}$$ has order 7.

The shortest element of order 11 I could find (there may exist a shorter one, but I couldn't find it) has length 10: $$(R\,U\,D'\,F\,L^2)^2$$ Here are some other interesting elements of order 11. Consider $$R\,L'\,F'\,U$$, which has order 33, therefore $$(R\,L'\,F'\,U)^3$$ has order 11. Also consider $$R\,U\,F^2\,L^2\,U^2$$, which has order 33, therefore $$(R\,U\,F^2\,L^2\,U^2)^3$$ has order 11.

A fun thing to note about elements of order 11 is that, since 11 is prime and is greater than the number of corners, an element of order 11 must act on only the edges (exactly 11 of them, to be more precise).

An interesting website I found (https://www.jaapsch.net/puzzles/cubic3.htm#p34) has counted the number of elements of every order. It turns out that elements of order 11 are the rarest! (Not counting the identity.) Only about 0.000000103% of elements in the Rubik's Cube Group are of order 11. Compare that to the most common order (60), which accounts for 10.6% of all elements.