# The group of symmetries of the regular pentagram is isomorphic to what. (Math GRE exam 0568 Q.60) For me this type of question I have not ever seen it, my understanding to the solution is not clear, and I do not understand why the other choices are wrong, could anyone explain this for me please? Also any recommendation for books containing this type of questions would be greatly appreciated.

• I like the solution. "There's no reason to believe this answer isn't right." Anyway, you can eliminate the other choices, (B) being the hard one to get rid of. (To me, at least.) It can't be cyclic because it has too many reflections (elements of order 2). It can't be $S_5$ because some of those symmetries are "discontinuous." – Randall Aug 14 '17 at 19:30
• $A_5$ is order 60, but pentagram has only reflection and rotation, $5+5=10$. And the pentagram symmetry is either single fixed point or derangement, but $A_5$ has permutations with two or three fixed points (you can't change just two points between themselves like $(21)(345)$ would. ). – Nij Aug 14 '17 at 19:37
• @Randall why elements of order 2 means reflections? why some of those symmetries are "discontinuous."? – Emptymind Aug 15 '17 at 19:28
• @Intuition: if you do a reflection twice ($g^2$) you leave the original figure unchanged (the identity). Thus reflections are group elements of order 2. Nij's last sentence explains better what I was trying to say by "discontinuous." – Randall Aug 15 '17 at 19:30
• Wikipedia has some excellent introductions to the basics of group theory, and there are thousands of good questions in Math SE that will help (or of course, asking new ones will work). Look at the cyclic, alternating and symmetric group pages too. Try physically drawing the smaller groups like $C_4$, $D_4$, $A_4$, $S_3$ and see how you can move points to points while following the rules for each type of group (cycle only, reflection and rotation, then symmetries and total choice). – Nij Aug 16 '17 at 0:27

The group of symmetries of the pentagram consists of all reflections and rotations mapping the pentagram to itself. Any rotation doing this will map the regular pentagon to itself (notice the small regular pentagon in the middle of the figure). Also, any reflection doing this must be a reflection about a line cutting the figure into two congruent pieces. Obviously, any such line will cut the pentagon into two congruent pieces as well. This "proves" that the symmetry group $G$ of the pentagram is isomorphic to a subgroup of $D_5$, the symmetry group of the pentagon (by mapping each symmetry of the former to its restriction to the pentagon). By looking at the diagram we can detect at least $10$ elements of $G$, so $G$ is actually isomorphic (equal) to $D_5$.

Hopefully this was sufficiently non-hand-wavy.

• From where did you know that "The group of symmetries of the pentagram consists of all reflections and rotations mapping the pentagram to itself." could you please recommend a reference for me ? – Emptymind Aug 15 '17 at 20:34
• you have a geometrical understanding to the meaning of groups which I do not have at all :( , I really need a book. – Emptymind Aug 15 '17 at 20:39
• @Intuition don't fret! It's not difficult. You just need to see a few examples at work. First, a symmetry of a $2D$ figure $X$ must either be a rotation, reflection, translation or a combination of translation and reflection (aka "glide reflection"). Obviously no translation (well, except the translation by $0$, which is the identity symmetry) can map our figure to itself, and so obviously no glide reflection can do that. What we have left is reflections and rotations. Now we just have to observe the pentagon in the middle to completely determine our group. – Cauchy Aug 16 '17 at 12:09
• @Intuition I don't have references in mind. I mostly read about these stuff (way long ago) on the internet, and every so often I saw an example here or there and I got used to them. Also note that I am by no means an expert in geometry or algebra or any connection between them, neither do I have that kind of crazy sharp intuition you think I have. – Cauchy Aug 16 '17 at 12:11

The automorphism group of the cycle graph $C_5$ is the Dihedral group of order 10. Now the pentagram is the graph complement of $C_5$. So the automorphism group is preserved.

• Upon further inspection, the five-cycle is its own complement. Moving the vertices around, it is possible to draw a planar version of the pentagram, which is just the five cycle. – ml0105 Aug 15 '17 at 16:34
• "The automorphism group of the cycle graph$C_{5}$ is the Dihedral group of order 10 " why?..... "the pentagram is the graph complement of $C_{5}$" why?....."So the automorphism group is preserved" and so what ? – Emptymind Aug 15 '17 at 20:42
• Most of these are standard statements about the dihedral group and graph isomorphism. The dihedral group of order 2n is introduced as the group of symmetries of the regular n-gon, which is the cycle graph on n vertices. Symmetry is a fancy way of saying automorphism. An isomorphism of two graphs is also an isomorphism of their complements, and vice versa. An isomorphism preserves the adjacency relation, and therefore the non-adjacency relation (the complement). – ml0105 Aug 15 '17 at 23:20