Center of the Rubik's Cube Group

I can think of a Rubik's-Cube subgroup with four two elements which commute with all other Rubik's cube elements:

• Identity
• Super-flip (flip all the edges around)
• Super-swap (swap all the edges with their opposite)
• Super-flip + super-swap

Are there any others?

EDIT: As others have pointed out, there are only two. Super-swap is not in the center.

• Is super-swap really in the centre? – Lord Shark the Unknown May 22 '18 at 0:23

The center of the Rubik's Cube Group is only the identity and the super flip. It's not terribly hard to show that any permutation that moves the location of a cubie can't be in the center.

So that only leaves flipping edges/rotating corners. Furthermore it's not terribly hard to show that any permutation that could be in the center must do the same thing to ALL edges and/or corners. ie if it flips one edge it must flip them all, if it rotates one corner (counter)clockwise it must rotate them all (counter)clockwise.

Flipping every edge is the super flip and it works.

Rotating every corner clockwise isn't a possible permutation. This is because there is no way to rotate just one corner. The closest we can do is rotate one corner clockwise and another corner counterclockwise. In other words if we assign a +1 to all clockwise corner turns and -1 to all counterclockwise corner turns, any permutation that only rotates corners must add up to 0 mod 3. And since there are 8 corners, we know it's impossible.

The cube group $C$ can be written as the semidirect product

$$C_o \rtimes C_p$$

Where $C_o$ and $C_p$ are the well-known orientation and permutation groups. If $s$ in the center of $C$, it is equivalent to commuting with all elements in $C_o$ and $C_p$. Assume for the sake of contradiction that $s$ cycles the edge pieces $\mathcal{P}$ as such: $P_1 \mapsto P_2 \mapsto \ldots \mapsto P_n$ (this is just one cycle in $\mathcal{P}$, and $P_k$ refers to the actual piece, not its position).

If $n \neq 2$, let $o \in C_o$ change the orientation of only $P_1$ and $P_2$. Then relative to $s$, $os$ changes the orientation of $P_n$ and $P_1$. Furthermore, relative to $s$, $so$ changes the orientation of $P_1$ and $P_2$. These cannot be reconciled.

If $n=2$, then let $o \in C_o$ change the orientation of $P_1$ and some $P$ not in the $2$-cycle. Then relative to $s$, $os$ just changes the orientation of $P_2$ and some $P' \neq P_1$. Furthermore, relative to $s$, $so$ just changes the orientation of $P_1$ and $P \neq P_2$.

Thus, $s$ cannot cycle any edges. By a similar argument, $s$ cannot cycle any corners. Thus, $s \in C_o$.

Assume for the sake of contradiction that $s$ flips edge piece $P_1$ but does not flip edge piece $P_2$. Then let $p \in C_p$ swap $P_1$ and $P_2$. Then $sp$ flips $P_2$ but not $P_1$, and $ps$ flips $P_1$ but not $P_2$.

Thus, $s$ must flip all edge pieces, or flip none.

By a similar argument, $s$ must rotate all corner pieces in the same direction, or rotate none. This is impossible, because each move on the Rubik's cube affecting the corners has a total parity of $0$, and rotating all corners as a parity of $1$ or $2$. (This mostly depends on how you define parity)

Thus, the only possibility is the superflip or the identity.

Note: To avoid falling into useless technicalities, the solution above is not extremely rigorous.