# Symmetry group question

a) Draw shapes in the plane which have the symmetry groups:

i. the dihedral group D8,

ii. the dihedral group D4,

iii. the cyclic group C5,

iv. the cyclic group C6.

b) Are there any shapes in the plane which have the symmetry group S4, the symmetric group on 4 letters? Either give an example, or explain why there are not any. A full mathematical proof is not needed.

My attempt:

I'm fine with part (a) which the dihedral group D8 is a square, D4 is a line, C5 and C6 are point groups. But i'm not sure about part (b), I know D8 is definitely in but not sure about others.

Any help would be appreciated. Thanks.

• A circle has a much larger symmetry group than $C_k$. Hint: decorate a regular $k-gon$ so no reflections are symmetries. – Ethan Bolker Dec 10 '17 at 18:29
• I have deleted my answer because I’ve come across a serious flaw. I’ll undelete it if I can see how to patch it. – Stella Biderman Dec 10 '17 at 18:35
• This definitely isn’t the tricky part of the question, but I would reconsider using the line segment — even though the reflection across the line containing the segment is a symmetry of the line segment, it acts as the identity transformation of the segment. A rectangle would work a little better. – pjs36 Dec 11 '17 at 13:58

Assuming the symmetries are to be distance preserving then $S_4$ cannot appear as the group of symmetries of a planar object.

• I would begin by justifying why a distance preserving symmetry of order three must be a rotation by 120 degrees. Any distance preserving transformation of the plane is affine. It has a fixed point (barycenter of an orbit of size three) and the rest should follow easily.
• $S_4$ has a total of eight elements of order three, split into four subgroups $P_1,P_2,P_3,P_4$ each of order three. If a generator of any of these has some point as a fixed point (about which the subgroup is rotating the world), then that same fixed point is shared by all the powers of the said generator. In other words, we can associate a unique fixed point $Q_i$ to subgroup $P_i$, $i=1,2,3,4.$
• The points $Q_i$ must be pairwise distinct. For otherwise two elements of $S_4$ would have the same effect on the plane.
• The subgroups $P_i$ are conjugates of each other. In fact, if $\sigma$ is a generator of $P_1$, then we can (after relabeling if need be) assume that $\sigma P_2\sigma^{-1}=P_3$, $\sigma P_3\sigma^{-1}=P_4$, and $\sigma P_4\sigma^{-1}=P_2$.
• Because $P_2$ has $Q_2$ as a fixed point we see that $\sigma P_2\sigma^{-1}$ has $\sigma(Q_2)$ as a fixed point. Therefore, by the previous bullet, $\sigma(Q_2)=Q_3$, $\sigma(Q_3)=Q_4$ and $\sigma(Q_4)=Q_2$. We can conclude that $\Delta Q_2Q_3Q_4$ is an equilateral triangle with barycenter $Q_1$.
• But, a generator $\tau$ of $P_2$ can play a role similar to $\sigma$. Therefore also $Q_2$ is the barycenter of the triangle formed by the other three. This is absurd.

The following remarks are due.

1. This is absolutely a planar argument. A rotation in 3D would be about an axis as opposed to a point, and the arguments with barycenters lose their footing.
2. In fact, in 3D we can realize $S_4$ as the group of isometries of a regular tetrahedron (rotations form a subgroup isomorphic to $A_4$, and we get the odd permutations by composing rotations with a reflection).
3. I also assumed (and used) that distinct elements of $S_4$ should act on the shape in distinct ways, i.e. the action should be faithful. Otherwise we can certainly make $S_4$ act on a planar shape by isometries by using the quotient group $S_4/V_4\simeq S_3$ where $V_4$ is the normal subgroup containing all the permutations of cycle type $(2,2)$. If we allowed $V_4$ to act trivially, then there would be no objections to two distinct 3-cycles acting on the shape in the same way, and the points $Q_i,i=1,2,3,4,$ could all coincide.
• We could also argue by looking at commutators of two distinct clockwise 120 degree rotations. Those will be non-trivial translations, and have infinite order. – Jyrki Lahtonen Dec 10 '17 at 19:28
• Or, more simply, if $\sigma$ and $\tau$ are two distinct clockwise rotations by 120 degrees, then $\sigma\tau^{-1}$ is a translation, and thus has infinite order. – Jyrki Lahtonen Dec 10 '17 at 23:04