In the 10th edition of Gallian, is the proof of theorem 7.5, to show that the symmetry group of the cube is isomorphic to $S_4,$ when the $4$ long diagonals are used to represent the $24$ symmetries.
It states that there the rotations contain an $8$ element subgroup $$\{e, \alpha, \alpha^2, \alpha^3,
\beta^2, \beta^2\alpha, \beta^2\alpha^2, \beta^2\alpha^3\}$$
It states the cube's long diagonals' based symmetries about the midpoints of the anti-podal faces, parallel to the $x-z$ direction as:$\alpha=(1234),$ while states:
'another $90^o$ rotation about an axis perpendicular to the first axis, to yield the permutation :$\beta= (1423).$'
$\beta$ is $90^o$ clockwise (ck) rotation about the anti-podal faces, parallel to the $x-y$ axis.
Hence, $\beta^2= 180^o= (12)(34)$ cck rotation; and $\beta^3= 270^o= (1324)$ cck rotation.
$\alpha=90^o$ cck rotation in one plane, while $\beta^3=90^o$ ck rotation in a perpendicular plane ($x-y$). Or, $\beta$ is $270^o$ cck rotation in the same perpendicular plane ($x-y$).
The orientation of the symmetry $\beta$ does not make difference to the subgroup of $8$ symmetries, even if it were $90^o$ cck rotation, as $\beta^2$ is still the same.
Say, if $\beta= (1324)=90^o$ cck rotation, still $\beta^2= 180^o$ rotation, in either cck or ck direction.
Next, state the $4$ long diagonals' based permutations:
$\alpha=(1234),$
$\alpha^2=(13)(24),$
$\alpha^3=(1432),$
$\beta^3 = (1324),$
$\beta^2= (12)(34),$
Next, get the remaining symmetries of the subgroup as:
$\beta^2\alpha = (12)(34)(1234)= (24),$
$\beta^2\alpha^2 = (12)(34)(13)(24)= (14)(23),$
$\beta^2\alpha^3 = (12)(34)(1234)(13)(24)= (13).$
This leads to the subgroup being given as:
$\{e, \alpha=(1234), \alpha^2=
(13)(24), \alpha^3=
(1432), \beta^2=
(12)(34), \beta^2\alpha=
(24), \beta^2\alpha^2=
(14)(23), \beta^2\alpha^3=(13)\},$
or:
$$\{e, (1234), (13)(24), (1432), (12)(34), (24), (14)(23), (13)\}$$
Next, states the order of element $\alpha\beta= (1234)(1423)= (142),$ an order $3$ element.
In the given subgroup, the types of rotations, given by the permutations is of two different types:
- $(24), (13),$ give rotations about anti-podal edges,
- Rest are rotations, but of different amount of rotations, and along different axis formed by all three different anti-podal faces.
a) $(1234)$ is a $90^o$ counter-clockwise rotation symmetry, along the axis formed by the midpoints of anti-podal faces along the $x-z$ axis; while $(1432)$ is $270^o$ rotation, along the same axis.
b) rest are $180^o$ rotations along each of the three anti-podal faces.
Doubt #1: The significance of taking an $8$ element subgroup, and an order $3$ element is unclear.
It makes sense only if, have $3$ elements of the cube of a certain type, and each can have $8$ symmetries.
There are six faces, and $3$ pairs of anti-podal faces.
An axis through such a pair of anti-podal faces has $4$ rotation symmetries in the $0^o, 90^o, 180^o, 270^o$ cck direction; rather than $8.$
Also, an order $3$ element is taken as the permutation corresponding to $\pm120^o$ rotation about a pair of anti-podal vertices.
So, request why the above subgroup, and an order $3$ element is taken?
Doubt #2: The proof states : "But, we must be careful not to assume that different rotations correspond to different permutations."
But, seems not elaborates it.
There are $24=4!$ permutations/symmetries, but it is "not" possible to have same rotations by different symmetries.
Say, to get a $180^o$ rotation about the anti-podal edges, can "not" get it by two different permutations.
Taking the cube labelled as per the vertices, as below. Let, the initial configuration is denoted by: $[1234][5678].$
An alternative (and, easier) example is that the symmetry $(12)(34)$ brings the final configuration $[3412][7856].$
I don't see if any two symmetries can bring the same configuration, of the initial cube.
Edit: For the $8$ element group, let $G,$ there are four group properties:
1.Closure wrt the group operation, here composition (let, be denoted by $\star$) of the group symmetries. Formally stated, $\forall a,b\in G,$ then $a\star b\in G.$
2. Associativity wrt the group operation. $\forall x,y,z\in G, x\star (y \star z)= (x\star y)\star z.$
3. Identity element, $e,$ exists in the group. $\exists e\in G,\, \text{s.t.}\, \forall x\in G, e\star x= x\star e = x.$
4. Inverse exists for all elements in the group. $\forall x \in G, \exists y\in G, \, \text{s.t.}\, x\star y = y\star x = e.$
The easiest to prove is the property of having the identity element. It would need only $n$ operation, in the worst case, for an $n$ element group. Though, here it is obvious that $\epsilon= e.$
Next need find inverse.
The members are:
$$\{e=\epsilon, \alpha=(1234), \alpha^2=
(13)(24), \alpha^3=
(1432), \beta^2=
(12)(34), \beta^2\alpha=
(24), \beta^2\alpha^2=
(14)(23), \beta^2\alpha^3=(13)\}.$$
$(1234)$ is of order $4,$ so cannot be its own inverse.
Get $(1234)(1432)= e,$ or $\alpha\star \alpha^3= e.$
$\alpha^2$ is its own inverse.
Also, $\beta^2$ is its own inverse, as an order $2$ symmetry.
Rest of the non-trivial elements are their own inverse, as all are order $2$ symmetries.
Next need show closure.
Seems it is better to simply construct the Cayley table, as all four properties are shown by that alone.
Request if there is a shortcut to the same.
The group table is:
\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline
e&e=\epsilon&\alpha=(1234)& \alpha^2=
(13)(24)& \alpha^3=
(1432)& \beta^2=
(12)(34)& \beta^2\alpha=
(24)& \beta^2\alpha^2=
(14)(23)& \beta^2\alpha^3=(13)\\ \hline
\alpha
&\alpha&\alpha^2& \alpha^3&e&
\alpha\beta^2=(1234)
(12)(34)=(13)=\beta^2\alpha^3& \alpha\beta^2\alpha=
(1234)(24)=(12)(34)=\beta^2& \alpha\beta^2\alpha^2=(1234)
(14)(23)=(24)=\beta^2\alpha& \alpha\beta^2\alpha^3 =(1234)(13)=(14)(23)=\beta^2\alpha^2\\ \hline
\alpha^2&\alpha^2=(13)(24)& \alpha^3=(1432)=\alpha^3&e&\alpha&\alpha^2\beta^2=(13)(24)
(12)(34)=(14)(23)=\beta^2\alpha^2 &\alpha^2\beta^2\alpha=
(13)(24)(24)=(13)=\beta^2\alpha^3& \alpha^2\beta^2\alpha^2=(13)(24)
(14)(23)=(12)(34)=\beta^2&\alpha^2\beta^2\alpha^3 =(13)(24)(13)=(24)=\beta^2\alpha\\ \hline
\alpha^3&\alpha^3=(1432)=\alpha^3& e&\alpha=(1234)&\alpha^2=(13)(24)&\alpha^3\beta^2=(1432)
(12)(34)=(24)=\beta^2\alpha &\alpha^3\beta^2\alpha=
(1432)(24)=(23)(14)=\beta^2\alpha^2& \alpha^3\beta^2\alpha^2=(1432)
(14)(23)=(13)=\beta^2\alpha^3 &\alpha^3\beta^2\alpha^3 =(1432)(13)=(12)(34)=\beta^2\\ \hline
\beta^2&\beta^2=
(12)(34)&\beta^2\alpha=
(12)(34)(1234)=(24)=\beta^2\alpha&\beta^2\alpha^2=(12)(34)(13)(24)=(14)(23)&\beta^2\alpha^3=(14)(23)(1234)=(13)&e &\alpha=
(1234)& \alpha^2=(13)(24) &\alpha^3 =(1432) \\ \hline
\beta^2\alpha&\beta^2\alpha=
(24)&\beta^2\alpha\alpha=
(24)(1234)=(14)(23)=\beta^2\alpha^2& \beta^2\alpha\alpha^2=(24)(13)(24)=(13)=\beta^2\alpha^3
& \beta^2\alpha\alpha^3=(24)(1432)= (12)(43)=\beta^2& \beta^2\alpha\beta^2
=(24)(12)(34)=(1432)=\alpha^3 &\beta^2\alpha\beta^2\alpha=e&\beta^2\alpha\beta^2\alpha^2=(24)(14)(23)=(2341)=\alpha &\beta^2\alpha\beta^2\alpha^3 =(24)(13)=\alpha^2 \\ \hline
\beta^2\alpha^2&\beta^2\alpha^2=(14)(23)=\beta^2\alpha^2&\beta^2\alpha^2\alpha=(14)(23)(1234)=(13)=\beta^2\alpha^3 & \beta^2\alpha^2\alpha^2=(14)(23)(13)(24)= (12)(43)=\beta^2&\beta^2\alpha^2\alpha^3=(14)(23)(1432)=(24)=\beta^2\alpha&\beta^2\alpha^2\beta^2=(14)(23)(12)(34)=(13)(24) =\alpha^2&\beta^2\alpha^2\beta^2\alpha=(14)(23)(24)= (1432)=\alpha^3&\beta^2\alpha^2\beta^2\alpha^2= e &\beta^2\alpha^2\beta^2\alpha^3=(1234)=\alpha\\ \hline
\beta^2\alpha^3&\beta^2\alpha^3=(13)&\beta^2=(12)(34)&
\beta^2\alpha=(24)&\beta^2\alpha^2=(14)(23)&\beta^2\alpha^3\beta^2=(13)(12)(34)=(1234)=\alpha&\beta^2\alpha^3\alpha=(13)(24)=\alpha^2&\beta^2\alpha^3\beta^2\alpha^2=(13)(14)(23)=(1432)=\alpha^3 &\beta^2\alpha^3\beta^2\alpha^3 =e \\ \hline
\end{array}
But, still not helped as the two doubts still remain, particularly the first one.
Best Answer
The text could be improved by using more precise language, but perhaps the author did want to burden the reader with pedantic verbiage.
Regarding doubt 2: Let $G$ be the group of rotations of the cube and let $S_4$ denote the permutation group of the diagonals We can associate to each $g\in G$ a permutation $\phi(g)\in S_4$. Let $\phi(G)=G'\subset S_4$ denote the associated subgroup of $S_4$. One issue raised in the sentence "we must be careful not to assume that different rotations correspond to different permutations" can be restated as the possibility that $\phi$ need not be 1-1. That is, there might be a nontrivial $g \in G$ for which $\phi(g)=e$ is the identity permutation on diagonals. In other words there might be a rotation of the cube $g$ that maps each diagonal line to itself. The other issue to worry about is that $\phi$ need not be onto. That is, perhaps there is a permutation of the diagonals that cannot be accomplished using only rotations.
Regarding doubt 1: If a group $G'$ has subgroups of orders $a$ and $b$ then a general theorem asserts that $a$ and $b$ must be divisors of the order of $G'$. Since in our case $a=3$ and $b=8$ are relatively prime, $24$ divides the order of $G'$. Thus $G'$ has order 24 or 48 or 72 or . But $G'\subset S_4$ so 24 is the only possibility. This is the roundabout method by which the author shows that all 24 permutations of the diagonals can be accomplished by rotations. Thus $\phi$ is onto. Since the cardinality of the domain and range of $\phi$ are both 24, that ensures that $\phi$ is also 1-1. Thus both issues suggested in the quoted passage are resolved.