Group generated by pile shuffles

Question

I can find plenty of questions relating to riffle shuffles, but none about the simple pile shuffle. My question is about the permutations possible on eight cards by pile shuffling.

Eight cards are dealt into two piles. Cards $1,3,5$ and $7$ are dealt into the first pile; cards $2,4,6$ and $8$ into the second pile. The two piles are stacked on top of each other. If the first pile is placed on top of the second, the cards are rearranged according to the permutation: $$\alpha := (1)(253)(467)(8)$$ If the second pile is placed on top of the first, the cards are rearranged according to: $$\beta := (157842)(36)$$

Let $G$ be the group generated by these two permutations. My questions: what is $|G|$? What is $G$ isomorphic to?

Both $\alpha$ and $\beta$ are even, so $G \le A_{8}$. I think $|G| = 24$, by working through permutations by hand (and noting identities such as $\alpha \beta \alpha \beta = \beta^{2} \alpha^{2}$, $\alpha \beta^{2} \alpha = \beta^{4}$ and $\alpha \beta^{3} = \beta^{3} \alpha$ ). But I'm not certain that I haven't missed something.

The permutations have an interesting property that a composition of any three (except for $\alpha^{3} = e$) gives a permutation comprised of disjoint transpositions, for example: $$\alpha \beta^{2} = (14)(23)(58)(67)$$ ... and any transposition (on eight elements) can be found in one of these compositions. (This relates to Gergonne's card trick - any of the cards can be moved to any position with three pile shuffles.)

Answer 1

Thanks to the comment by user1729, I've come to an answer. $$G \cong A_4 \times Z_2$$ I used the 'PermGroup' app from Université Côte d'Azur to confirm that $|G|=24$. The app also provided details of the normal subgroups of $G$. I compared these details with the subgroup diagrams provided by this app, for each of the fifteen groups of order 24, to find that $G$ had the same subgroup structure as $A_4 \times Z_2$.

Some other details... The center of $G$ is $\{ e,\beta^{3} \} $. The largest subgroup in $G$ is: $$H=\{e, \alpha, \alpha^2, \beta^2, \beta^4, \beta^2 \alpha^2, \alpha \beta^4, \alpha^2 \beta^2, \beta^4 \alpha, \alpha \beta^2, \beta \alpha \beta, \beta^2 \alpha \}$$ (all elements with an even number of $\beta$s in their compositions). $H$ is normal and $H \cong A_4$.

My task now is to relate this back to the cards: does this give me any 'intuitive' understanding of the permutations of the deck possible by pile shuffling? Also, how might it generalise to other deck sizes?

Edit: gathering together some bits from the comments below this answer (thanks to Jaap Scherphuis)...

'Intuition': $\alpha$ and $\beta$ (hence all elements of $G$) permute the card pairs $\{1,8\}$, $\{2,7\}$, $\{3,6\}$ and $\{4,5\}$, but never intermingle these pairs. For example, $1 \mapsto 2$ iff $8 \mapsto 7$. The permutations of these pairs are given by $H \cong A_4$. The element $\beta^3 = (1,8)(2,7)(3,6)(4,5) \in G \backslash H$ swaps cards within each pair, leaving all pairs in place; this is 'responsible' for the $Z_2$ part of $G \cong A_4 \times Z_2$.

Generalising: Writing a subscript for the deck size (so the original $G$ becomes $G_8$), I have found: $$\begin{aligned} G_2 &\cong Z_2 \\ G_4 &\cong D_4 \\ G_6 &\cong S_4 \\ G_8 &\cong A_4 \times Z_2 \end{aligned}$$ ... and then they get big. $|G_{10}|=1920$ and $|G_{12}|=7680$.

I searched for the sequence of orders of these groups ($2,8,24,24,1920,7680$) in the OEIS, and found the 'Order of group generated by perfect shuffles of 2n cards'! This led me to a (now seemingly obvious) realisation... that the pile shuffle is just the inverse of a perfect shuffle. If pile shuffles permute cards by $\alpha$ and $\beta$, then perfect shuffles permute cards by $\alpha^{-1}$ and $\beta^{-1}$. The groups generated by pile shuffling are hence exactly the groups generated by perfect shuffling.

This paper (Diaconis, Graham & Kantor, 1983) gives the groups generated by perfect shuffles (hence also pile shuffles), for all decks of size $2n$, in terms of semidirect products. It gives: $$\begin{aligned} G_8 &\cong (Z_2)^3 \rtimes Z_3 \\ G_{12} &\cong (Z_2)^6 \rtimes PGL(2,5) \end{aligned}$$ (I'm not proficient enough with semidirect products to understand why $(Z_2)^3 \rtimes Z_3 \cong A_4 \times Z_2$.)

Geometrical interpretations: Given $G_4 \cong D_4$, pile shuffling $4$ cards can easily be interpreted as symmetries of a square (with cards mapped to vertices). Jaap Scherphuis, in the comments, gives an interpretation of $G_6 \cong S_4$, with cards mapped to faces of a cube. $G_8 \cong A_4 \times Z_2$ relates to pyritohedral symmetry; cards can be mapped to the vertices of a cube, where each face of the cube is painted with a line segment that divides the face into two equal rectangles, such that the line segments of adjacent faces do not meet at the edges. Permutations of the cards correspond to symmetries of the cube.