# Groups of symmetries

What are the symmetries of a solid rectangular box whose length, width and height are all different? I get a group of order 4 by rotation 180, flipping along a vertical and horizontal axis and itself.

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Is this of some help to you? –  user21436 Jan 31 '12 at 3:52
Those are the physical symmetries that you can perform on the space. But what about reflecting about a plane? Say, if you place the box so its center is at the origin, then you can reflect about the $yz$-plane, and this does not correspond to any of the rotations/flips. –  Arturo Magidin Jan 31 '12 at 4:05
@Shannon: Note that you can only formally "accept" one answer (I mention it because a few minutes ago it was Kannapan Sampath's answer that you had marked as accepted, and as I type this it seems to be mine which is so marked). –  Arturo Magidin Jan 31 '12 at 4:40

## 2 Answers

If you are only allowing yourself "physical symmetries" (symmetries that can be realized by manipulation of the box in our familiar 3-space):

Place the center of the box at the origin. You get one symmetry by rotation by half a turn using the $z$-axis as pivot; you get one symmetry by rotation by half a turn using the $x$-axis as a pivot; you get one symmetry by rotation of 180 degrees with the $y$-axis as pivot.

If we number the corners of the box with $1,2,3,4$ on top and $5,6,7,8$ in the bottom, with $i+4$ under $i$, then we can identify the four rotations with permutations of the vertices. One of the rotations corresponds to $\sigma= (1,3)(2,4)(5,7)(6,8)$; another to $\tau=(1,8)(4,5)(2,7)(3,6)$; and the third one as $\rho=(1,6)(2,5)(4,7)(3,8)$.

Since \begin{align*}\ \sigma\circ\tau=(1,3)(2,4)(5,7)(6,8)&\circ(1,8)(4,5)(2,7)(3,6)\\ & = (1,8)(4,5)(2,7)(3,6)\circ(1,3)(2,4)(5,7)(6,8)\\ &= (1,6)(2,5)(3,8)(4,7)\\&=\rho,\end{align*} these three together with the identity form a Klein $4$-group. These are the four you've described.

Are there any others? Once you decide where vertices 1 and 4 map, everything else gets forced. 1 has only four possible locations, since orientation cannot change: either 1 maps to itself and 4 to itself; or 1 maps to 3 and 4 maps to 2; or 1 maps to 8 and 4 maps to 5; or 1 maps to 6 and 4 to 7. These are the four permutations given above, so that is all there is; it's not a cyclic group, because it has three elements of order two, so it is the Klein 4-group.

However: 'Symmetries' often includes reflections as well (so the dihedral groups include not only the rotations, but also the reflections). If we include reflections, then in addition to the four possibilities above, we can also exchange 1 and 4; map 1 to 2 and 4 to 3; map 1 to 5 and 4 to 8; or map 1 to 7 and 4 to 6. The first three are accomplished by reflecting about the coordinate planes (the $xy$-plane, the $xz$-plane, and the $yz$-plane, in some order depending on how you are picturing your box), the fourth one by reflection about the origin. This gives four other symmetries, and so the group corresponds to a group of order $8$. It cannot be cyclic, since it has the Klein $4$-group as a subgroup; it is a semidirect product of the Klein $4$-group and the cyclic group of order $2$.

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All the minute details in here, compared to my hand wavy answer that has no group theory in it. +1. –  user21436 Jan 31 '12 at 4:23
I used the same way as you did labeling vertices. I thought it was going to be a big permutaion but turns out its so simple. Thank you for all your information. –  Shannon Jan 31 '12 at 4:41

Your contention that you'll have cyclic subgroup of order $4$ is quite natural but needs a little rethinking. Since, you consider the symmetries of a mattress like object (where symmetry is to mean any rigid motion in 3-space which will move a copy of the mattress in any fashion and place the copy back on the original), a rotation of a "scalene" object by $\frac{\pi}{2}$ will not keep this object in the same place. So, rotation by $\pi$ is a valid symmetry.

Technically, the mattress group is called the Klein $4$-group.

$\hskip 2.5in$

$H$ stands for a horizontal flip about the axis parallel to $13$ through the midpoint of $12$; $V$ for a vertical flip about the axis parallel to $12$ and through the mid point of $13$ and $R$ for rotation through the mid-point of the mattress.

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