Non-oriented Rubiks Cube Group Consider the (non-oriented, so we do not care about the center of each side) rubiks cube group, $G$, which is a subgroup of $S_{48}$ and is generated by the letters: $\{F,B,U,D,L,R\}$, i.e. I mean that $G = \langle F,B,U,D,R,L\rangle$ (subject to the canonical relations for the non-oriented $3\times 3$ rubiks cube). I am trying to just manually verify if the superflip is in the center (just by verifying if it commutes with each of the generators) using https://rubikscu.be/#cubesolver , but this is exceptionally cumbersome and time consuming. Let $h\in G$ denote the superflip. I have seen other websites, MSE pages, and pdfs describe the center as simply only containing $1_G$ and $h$, but these pages very rarely distinguish if they mean the oriented cube (the subgroup of $S_{54}$) or non-oriented, I am not really sure if it matters either (but furthermore, I am really sure how one would rigorously prove this in any case, the arguments I can find are mostly fairly ambiguous).
 A: As aschepler mentioned in the comments, it's irrelevant how the superflip is realized in terms of the generators. Since the Rubik's Cube group is formally just a subgroup of $S_{48}$, an element like $h$ is literally just a permutation.
But first, let's choose something other than $\{1,\ldots,48\}$ to serve as the set whose permutations we consider.
Let's label the faces as Front, Bottom, Up, Down, Right, Left, and denote the 48 faces

*

*for corners: like “FUR” to denote the Front sticker (hence the first letter) of the front-up-right cubelet

*for edges: like “FU” to denote the Front sticker of the front-up cubelet.

To avoid ambiguity, let's further deviate from the standard notation and add commas so that a cycle looks like $(FU,UF)$.
For any term representation of the superflip, we will be able to calculate
\begin{align*}
h = &(FU, UF)(FR,RF)(FD,DF)(FL,LF)\\
    &(UR,RU)(UL,LU)(DR,RD)(DL,LD)\\
    & (BU,UB)(BR,RB)(BD,DB)(BL,LB).
\end{align*}
Note in particular that $h$ is of order two.
To prove that it commutes with every cycle, let's refresh our memories and see how the generators are defined:
\begin{align*}
R &:=(\text{corner cycles})(RU, RB, RD, RF)(UR,BR,DR,FR)\\
U &:=(\text{corner cycles})(UR,UF,UL,UB)(RU,FU,LU,BU)\\
F &:=(\text{corner cycles})(FU,FR,FD,FL)(UF,RF,DF,LF)\\
L &:=(\text{corner cycles})(LU,LF,LD,LB)(UL,FL,DL,BL)\\
D &:=(\text{corner cycles})(DF,DR,DB,DL)(FD,RD,BD,LD)\\
B &:=(\text{corner cycles})(BU,BL,BD,BR)(UB,LB,DB,RB).
\end{align*}
It is now easy to verify that each of the commutators $R^{-1}h^{-1}Rh$ etc. vanish, or, equivalently (and perhaps easier to verify if you know what conjugation does in terms of permutations :-) ), that $h^R := R^{-1}hR = h$ etc.  Notice in particular that since $h$ leaves corner faces invariant, the respective corner cycles in the generators don't contribute to the commutator, explaining why I left them out above.
As you correctly stated, if $h$ commutates with all generators, it has to lie in the center.
