I was attracted by this question, motivated by a quick possibility to implement the groups in question in the one or the other computer algebra system, and check the wanted property, together with some other bonus information. At the end, i decided to submit, including a small piece of job which addresses the question. To have an answer, a proof is on the first place, i typed the one that comes into mind when having the cube in the hand.
Let $R$ be the Rubik group, and $S$ the "scramble" group.
To fix ideas, we let both groups act on the set of "pieces" of the $3\times3\times3$ Rubik cube.
Each piece comes here with its place, and its orientation.
Below we use the faithfulness of this action, a "move" in the abstract group $S$ is the identity, iff it is acting trivially on the pieces.
Considered now the pieces in a "solved" / unscrambled / clean position.
Corners will be denoted by $c,d,\dots$ with possible further decorations.
For edges we use similarly $e,f,\dots$ - and let us fix some among them.
- $c_*, d_*$ are the corners in the positions top-front-left, and top-front-right.
- $e_*$, $f_*$ are the edge top-front between $c_*$ and $d_*$, and its opposite on the top face.
Scrambling is generated by the following types of "moves" ("move" is shorter than formula, or permutation...):
- $(E)$ The move $t(e)$, exchange the faces of the edge $e$, its order is two.
We use the notation $e'$ for the edge $e$ with opposite orientation, i.e. $e'=t(e)\ (e)$.
- $(EE)$ The move $t(e,f)=t(f,e)$, switch over the edges $e, f$, its order is two.
- $(C)$ The move $s(c)$, change the three faces/stickers of the corner $c$ using an even permutation, to be specific, $s(c)$ keeps the orientation in space. (We are not allowed to keep one sticker, and unstick the other two, exchange them, then glue them back on "wrong places".) It has order three.
We denote by $c'$ the corner $c$ with orientation changed by $s(c)$, i.e. $c'=s(c)\ (c)$. So all orientations of $c$ are $c,c',c''$.
- $(CC)$ The move $t(c,d)$, exchange the corners $c,d$, it has order two.
(Notation: Here $E$ stays for a move affecting one edge, while $EE$ scrambles two edges. Similarly, $C$, $CC$ involve one, respectively two corners.)
To be precise, one can introduce numbers like in
sage: rubik = CubeGroup()
sage: rubik.display2d('')
+--------------+
| 1 2 3 |
| 4 top 5 |
| 6 7 8 |
+------------+--------------+-------------+------------+
| 9 10 11 | 17 18 19 | 25 26 27 | 33 34 35 |
| 12 left 13 | 20 front 21 | 28 right 29 | 36 rear 37 |
| 14 15 16 | 22 23 24 | 30 31 32 | 38 39 40 |
+------------+--------------+-------------+------------+
| 41 42 43 |
| 44 bottom 45 |
| 46 47 48 |
+--------------+
and define $c^*$ to be the triple $(6,11,17)$, $e^*$ the edge $(7,18)$, and let $S$ act (separated)
- on the set of possible corner triples
like $c=(6,11,17)$, $c'=(11,17,6)$, $c''=(17,6,11)$, and so on,
- and on the set of the possible edges doubles $e=(7,18)$, $e'=(18,7)$, and so on.
But the "common sense" should be enough to keep trace on the movements below.
In particular, for a scramble move $s$, we can write expressions like $s(1), s(2), s(3), \dots, s(48)$, like $s(e)=s(\ (7,18)\ )$, like $s(c)=s(\ (6,11,17)\ )$, and so on.
After collecting some experience with (i.e. formulas for) the cube in real life, one knows that
each randomly scrambled configuration can be brought to an "almost solved" configuration, all what is needed to do now to get the
clean position is to use none, one, or more of the following "easy scramble" moves involving only the singled out pieces $c_*, d_*, e_*, f_*$:
$$
s(c_*)\ ,\ t(c_*,d_*)\ ,\ t(e_*)\ ,\ t(e_*, f_*)\ .
$$
In fact, we can eliminate from the list one of $(FF)$, $(CC)$, because there is a regular $R$-move connecting $(FF)$, $(CC)$, and possibly some $(F)$ and $(C)$ later adjustments. We eliminate $t(e_*, f_*)$ below. So we expect an index $S:R$ to be $12$ (or a smaller divisor - but it is twelve).
We address now the given question: Is $R$ normal in $S$?
Yes. It is enough to show that $srs^{-1}$ is in $R$ for all $r\in R$, and all $s$ among the above easy scramble moves (as coset representatives).
We deal with the types $E$, $EE$, and $C$. (A $CC$ move is up to an $R$-move generated by these types.)
Let us fix some "formula" $r\in R$.
$(E)$ What happens if we apply $t(e_*)\cdot r\cdot t(e_*)$?
The movement $r$ brings the edge $e_*$ into some other edge position $f$, $r(e_*)=f$,
so in the same time $r(e'_*)=f'$.
And other edge $e\ne e_*, e'_*$ is invariated, $r(e)=e$.
So the action on edges of $rt(e_*)$ is the same as the one of $t(f)r$:
$$
\begin{aligned}
rt(e_*)\ (e_*) &= r(e'_*)=f'= t(f)\ (f)=t(f)r\ (e_*)\ ,\\
rt(e_*)\ (e'_*) &= r(e_*)=f= t(f)\ (f')=t(f)r\ (e'_*)\ ,\\
rt(e_*)\ (e) &= r(e)=t(f)r\ (e)\ ,\qquad\text{ for all other $e$, $e\ne e_*,e'_*$, since $r(e)\ne f,f'$.}
\end{aligned}
$$
We obtain (corners are invariated by $t(e_*)$):
$$
t(e_*)\cdot r\cdot t(e_*)
=
t(e_*)\cdot t(f)\cdot r
=
\underbrace{t(e_*)\cdot t(r(e_*))}_{\in R}\cdot r\in R\ .
$$
$(C)$ What happens if we apply $s(c_*)^2\cdot r\cdot s(c_*)$?
This is similar to $(E)$.
The movement $r$ brings the corner $c$ into some other corner position $d$, $r(c_*)=d$,
so in the same time (because we have the same space orientation before and after $r$) $r(c'_*)=d'$, $r(c''_*)=d''$, .
And any other corner $c\ne c_*, c'_*, c''_*$ is invariated, $r(c)=c$.
So the action on corners of $rs(c_*)$ is the same as the one of $s(d)r$:
$$
\begin{aligned}
rs(c_*)\ (c_*) &= r(c'_* ) = d' = s(d)\ (d) = s(d)r\ (c_* )\ ,\\
rs(c_*)\ (c'_*) &= r(c''_*) = d'' = s(d)\ (d') = s(d)r\ (c'_*)\ ,\\
rs(c_*)\ (c''_*) &= r(c_* ) = d = s(d)\ (d'') = s(d)r\ (c''_*)\ ,\\
rs(c_*)\ (c) &= r(c) = s(d)r\ (c)\ ,\qquad\text{ for all other $c$, $c\ne c_*,c'_*,c''_*$, since $r(c)\ne d,d',d''$.}
\end{aligned}
$$
We obtain (edges are invariated by $s(c_*)$):
$$
s(c_*)^2\cdot r\cdot s(c_*)
=
s(c_*)^2\cdot s(d)\cdot r
=
\underbrace{s(c_*)^2\cdot s(r(c_*))}_{\in R}\cdot r\in R\ .
$$
$(EE)$ What happens if we apply $t(e_*,f_*)\cdot r\cdot t(e_*,f_*)$?
This is also similar to $(E)$, but in a different manner.
The movement $r$ brings the edges $e_*,f_*$ into some other edge positions $g,h$, $r(e_*)=g$, $r(e_*)=g$,
so in the same time $r(e'_*)=g'$, $r(f'_*)=h'$.
And other edge $e\ne e_*, e'_*,f_*,f'_*$ is invariated, $r(e)=e$.
So the action on edges of $rt(e_*,f_*)$ is the same as the one of $t(g,h)r$:
$$
\begin{aligned}
rt(e_*,f_*)\ (e_* ) &= r(f_* ) = h = t(g,h)\ (g ) = t(g, h)r\ (e_*)\ ,\\
rt(e_*,f_*)\ (e'_*) &= r(f'_*) = h' = t(g,h)\ (g') = t(g, h)r\ (e'_*)\ ,\\
%
rt(e_*,f_*)\ (f_* ) &= r(e_* ) = g = t(g,h)\ (h ) = t(g, h)r\ (f_*)\ ,\\
rt(e_*,f_*)\ (f'_*) &= r(d'_*) = g' = t(g,h)\ (h') = t(g, h)r\ (f'_*)\ ,\\
%
rt(e_*,f_*)\ (e) &= r(e) = e = t(g,h)\ (e) = t(g,h)r\ (e)\ ,\\
&\qquad\qquad\qquad\text{ for all other $e$, $e\ne e_*,e'_*f_*,f'_*$, since $r(e)\ne g,h,g',h'$.}
\end{aligned}
$$
We obtain (corners are invariated by $t(e_*,f_*)$):
$$
t(e_*,f_*)\cdot r\cdot t(e_*,f_*)
=
t(e_*,f_*)\cdot t(g,h)\cdot r
=
\underbrace{t(e_*,f_*)\cdot t(r(e_*),r(f_*))}_{\in R}\cdot r\in R\ .
$$
The elements with an underbrace-in-$R$-mark are known to be in $R$ by folklore formulas, certainly among those applied at the first time Rubik cube success.
$\square$
Computer check:
Here is a small piece of code checking the normality in sage, by using the existing CubeGroup()
functionality. In fact only the generators of $R$.
To define the bigger "scramble" group $S$, we declare the corners and edges, then the generators of $S$ based on them.
corners = [
( 6, 11, 17), ( 8, 19, 25), (24, 43, 30), (16, 41, 22), # front
( 1, 35, 9), ( 3, 27, 33), (32, 48, 38), (14, 40, 46), # rear
]
edges = [
( 7, 18), (21, 28), (23, 42), (13, 20), # front
( 4, 10), ( 5, 26), (31, 45), (15, 44), # mid
( 2, 34), (29, 36), (39, 47), (12, 37), # rear
]
cgens = (
[ [c] for c in corners ] # change orientation of corner c
+
[ [(c1, d1), (c2, d2), (c3, d3)] # exchange corners c, d
for (c1, c2, c3) in corners
for (d1, d2, d3) in corners if c1 < d1 ]
)
egens = (
[ [e] for e in edges ] # change orientation of edge e
+
[ [(e1, f1), (e2, f2)] # exchange edges e and f
for (e1, e2) in edges
for (f1, f2) in edges if e1 < f1]
)
rubik = CubeGroup()
Sc = PermutationGroup(cgens)
Se = PermutationGroup(egens)
S = PermutationGroup(cgens+egens)
R = S.subgroup(rubik.gens())
And we can ask now for the normality of the Rubik group $R$ inside the scramble group $S$, and some further bonus informations:
sage: R.is_normal(S)
True
This is the computer aided answer. Let us check the index of $R$ in $S$,
it should be twelve, this is a hint that the $(E)$, $(EE)$, $(C)$, and $(CC)$ types are not independent. (And the link gives a formula inside $R$.)
sage: S.order().factor()
2^29 * 3^15 * 5^3 * 7^2 * 11
sage: Se.order().factor()
2^22 * 3^5 * 5^2 * 7 * 11
sage: Sc.order().factor()
2^7 * 3^10 * 5 * 7
sage: R.order().factor()
2^27 * 3^14 * 5^3 * 7^2 * 11
sage: (S.order() / R.order()).factor()
2^2 * 3
It may be interesting to note the following information on the commutators:
sage: S.commutator().order().factor()
2^26 * 3^14 * 5^3 * 7^2 * 11
sage: R.commutator().order().factor()
2^26 * 3^14 * 5^3 * 7^2 * 11
So regarding commutators, $R$ and $S$ have the same level of complexity.