#### dougreed

##### Member

I was reading a book I just picked up called Adventures in Group Theory (available here) when I decided to look at a few of my OLL and PLL algs and see which ones I could easily break into conjugates and commutators.

Here is one I quickly discovered:

My OLL: R2 D' R U2 R' D R U2 R

My thought process:

This alg doesn't look like a commutator [A,B] simply because the inverse of the first move, R2, doesn't appear in that form anywhere in the rest of the algorithm. Even though this could be attributed to move cancellation, I decided to see what I could find out about this algorithm by assuming it is a conjugate of a commutator and some move C, e.g. [A,B]^C == C [A,B] C'.

My first thought is that C' is probably only going to be one face-turn, so I say C'=R which implies that C = R'.

The first move of the algorithm is R2, which can be represented as R R or R' R'. By using R' R', we come up with an acceptable move for C. So, now we have:

[C] R' D' R U2 R' D R U2 [C']

C=R

First, we know that U2 == (U2)' because U2 is its own inverse. Then, we can rewrite the above:

[C] R' D' R U2 R' D R (U2)' [C']

Now, we can easily see that this algorithm, sans the C and C', is a commutator [A,B] if A = R' D' R and B = U2.

C A B A' B' C' == C [A,B] C' == [A,B]^C

if A = R' D' R, B = U2, C = R

Here are a few more algorithms I regularly use that can be broken down similarly, for anyone who is interested:

R2 B2 R F R' B2 R F' R

R' U2 R U R' U R (2 commutators)

-Doug