When speaking to interested parties lacking formal mathematical background, I've illustrated how different objects can have the same number of symmetries and yet have different types of symmetry with the example of an oriented hexagon (obtained by putting arrows on the edges in cyclic fashion) versus an unadorned triangle. These have cyclic and dihedral symmetry groups, respectively, each with six symmetries. There are multiple reasons these are distinct types of symmetry:

  • (a) the rotations of the hexagon can all be generated by a single symmetry (unlike with the triangle, where flips and rotations cannot be generated by a single symmetry);
  • (b) any two rotations of the oriented hexagon (by $x$ and $y$ degrees) can be performed in either order, but applying a flip and a rotation of the triangle in different orders yield distinct symmetries (they permute the vertices differently);
  • (c) one can tabulate how many applications of each symmetry is required to achieve an initial state, and notice we obtain a different list of numbers for the two symmetry groups.

In other words, cyclicity, commutativity, and order information.

However, when I've asked people to identify a reason the triangle and oriented hexagon have different types of symmetry, they've consistently (and understandably) identified the fact the triangle has reflectional symmetry while the oriented hexagon doesn't. Which is a perfectly valid answer, because my question is ambiguous: they are identifying why the individual symmetries are different types of transformations, instead of looking at the internal structure of the symmetry group. When I explain I am interested in answers that look instead at how the symmetries interact with each other, I am met with confusion, although when I explain (a), (b) and (c) above it is more clear what I intend.

Question: What are some pedagogical examples of distinguishing basic symmetry groups that avoid this pitfall?

To be specific, two types of situations could fit the bill:

  1. Different symmetry groups where the individual symmetries are the same types of transformations of some figure or basic mathematical object.
  2. Different figures or mathematical objects with the same symmetry group but where some individual symmetries do different types of things.

Examples of type (1) would be more powerful in my opinion, but for added effect examples of type (2) may be explored first. Preferably examples should be basic, close to napkin-level math.

Here's the simplest case of (2) which I thought of: (i) on the one hand, consider a kind of "sunset" picture that has reflectional symmetry across the horizon, but say there's a seagull on the right or left side that precludes sliding or vertical reflections from being symmetries, and (ii) on the other hand, the outline of the yin-yang symbol (so, not painting it in), which has no reflectional symmetries but has a single nontrivial $180^{\circ}$ rotational symmetry.

  • 2
    $\begingroup$ It sounds to me that you are encountering a confusion between a group action and an abstract group. Frankly, people may have a better intuitive understand of group actions than they do of abstract groups. Along with this, they may have a better understanding of conjugacy between group actions than they have of isomorphism between abstract groups... $\endgroup$ – Lee Mosher Nov 3 '17 at 19:50
  • $\begingroup$ ... They can "see" that two reflection actions of $\mathbb{Z}/2\mathbb{Z}$ are conjugate to each other; and that two $180^\circ$ rotation actions of $\mathbb{Z}/2\mathbb{Z}$ are conjugate to each other; and that the reflection action is not conjugate to the rotation action. But they cannot "see" that the order 2 reflection group is abstractly isomorphic to the order 2 rotation group. $\endgroup$ – Lee Mosher Nov 3 '17 at 19:50
  • $\begingroup$ How do you get past "why do you even care"? But assuming you do, maybe you can convince your conversation partner that symmetry is not even intrinsically a geometric concept, e.g. shuffles of a 3 card deck $\cong$ symmetries of the triangle. $\endgroup$ – fredgoodman Nov 4 '17 at 16:38
  • $\begingroup$ How about taking the rotations of the tetrahedron (en.wikipedia.org/wiki/Tetrahedron) versus the rotations of the hexagonal prism (en.wikipedia.org/wiki/Hexagonal_prism)? Additionally you can also look at a pyramid with a regular dodecagon (en.wikipedia.org/wiki/Dodecagon) as base to get a cyclic rotation group for your example. $\endgroup$ – j.p. Nov 9 '17 at 7:53
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    $\begingroup$ On second thought, the rotational groups of a triangular prism and a pyramid with hexagonal base might be easier to grasp for your friends. $\endgroup$ – j.p. Nov 9 '17 at 8:16

Consider the group of symmetries of a brick $P$ (a right-angled parallelepiped with side-lengths $a, b, c$, where $a> b> c$; as an option you can also consider $a> b=c$, let's call it $P'$) and the group of symmetries of a cube $Q$.

Most people will immediately identify reflection symmetries of both. With a bit help from you, they will discover order 2 rotational symmetries of the brick (or an order 4 rotational symmetry of $P'$) and the central symmetry of both $P$ and $P'$. Most will have harder time believing (even with your help) that cube has a rotational symmetry of order 3. Then some even might ask you "how do you know that $P'$ has no order 3 symmetry".

Then you explain that all symmetries of $P$ commute, but they will need your help to see noncommuting symmetries of $P'$ and $Q$. Before explaining the noncommutativity phenomenon, give them the following example: $A=$ put on socks; $B=$ put on shoes. Then demonstrate that $AB\ne BA$.

Then notice that the central symmetries of $P, P'$ and $Q$ commute with all other symmetries. After you are done demonstrating this fact, ask them

"can you think of a highly symmetric solid which has no nontrivial symmetries commuting with the rest of the symmetries?"

(The regular tetrahedron is an answer.)

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A uniform triangular prism has as rotational group the dihedral group of order $6$, whereas a right hexagonal pyramid with regular hexagonal base has a cyclic rotational group of order $6$.

I made this answer community wiki in the hope someone nice will add some images for illustration, as this question deserves it (and remove this comment).

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