TITLE: Center of the Rubik's Cube Group
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.