# What kind of “symmetry” is the symmetric group about?

There are two concepts which are very similar literally in abstract algebra: symmetric group and symmetry group. By definition, the symmetric group on a set is the group consisting of all bijections of the set (all one-to-one and onto functions) from the set to itself with function composition as the group operation. When the set is finite, the group is sometimes denoted as $S_n$.

The Dihedral group $D_n$, which is a special case of the symmetry group, has a very strong geometric intuition about symmetry as the picture shows.

I know nothing about the relation between these two concepts but the fact that $D_3$ and $S_3$ are actually the same. For me, symmetric group is more about "permutations". And actually its subgroups are also called permutation groups.

Here are my questions:

• What's the relation between these two concepts: "symmetric group" and "symmetry group"?
• What kind of "symmetry" is the symmetric group about?
• Where is the name "symmetric group" from?
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Reminds me a little of this question on MathOverflow: mathoverflow.net/questions/993/… – grautur Jun 20 '11 at 3:37
I am persuaded by DaG's comment about the complete graph, along with Vladimir Sotirov's point about needing more dimensions to evenly space out more vertices. The n+1 vertices and (n)(n+1)/2 edges of a regular n-dimensional simplex form a complete graph, and I guess that is the most symmetric discrete object one can construct in n dimensions. Every point is connected to every other point by equal-length edges. Though we have not seen a complete explanation of how the symmetric group got its name, at least some of the ingredients of the story are beginning to appear. – user12884 Jul 3 '11 at 2:32
How about the dodecahedron or icosahedron or even a cube, these strike me as "much more symmetric" than a tetrahedron... – t.b. Jul 3 '11 at 2:38
@Ralpgh: please do not use answers to post comments. – Qiaochu Yuan Jul 3 '11 at 2:56
@Theo, it depends on how you measure symmetry. I would guess that the smallest $n$ such that $S_n$ contains the symmetry groups of the dodecahedron and icosahedron is pretty high relative to their dimension, but perhaps low relative to the number of their vertices? – Vladimir Sotirov Jul 3 '11 at 6:54

Mathematically, a symmetry is a bijection of a set onto itself. In some cases, we restrict to bijections that preserve some structure (e.g. isometries, which preserve distances), but in the case of the symmetric group $S_n$ we just have a set with $n$ elements, and no additional structure to preserve, so all bijections (permutations) are allowed.

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I admit to be ignorant of the actual history of the term "symmetric", but I do have a geometric explanation of why $S_n$ represents unrestricted (finite) symmetries.

First, you should ask yourself what is the special property of the (equilateral) triangle that makes its distance-preserving symmetries be $S_3$? The answer is that in the equilateral triangle there are no special relationships between the vertices, as the vertices are linearly independent in the sense that they determine a highest-dimensional object three points could possibly define, and also as they are equally spaced out (the distance between any two vertices is the same) and hence you cannot distinguish them based on distance. (It takes a little bit of slightly more rigorous and technical work to show that this fact gives you that the distance-preserving symmetries form $S_3$, which work will depend on whether you use the Coxeter presentation or the permutation definition of $S_n$; consider that homework)

For a square, one of the vertices lies in the plane determined by the other three, which means that the fourth vertex is a linear combination of the other three, so in particular the four vertices cannot possibly be equally spaced out by the triangle inequality, hence the vertices have special relationships that can be determined by distance, which is why you get the smaller $D_4$ rather than the larger $S_4$.

To get the full $S_n$ in general, you need the $n$ vertices to determine a figure of the largest possible dimension (otherwise the triangle inequality prevents the possibility of the vertices being equally spaced out), and this turns out to be sufficient for you to be able to equally space out the vertices. So, $S_4$ would be the symmetry group of distance-preserving transformations of the $3$-dimensional tetrahedron, and $S_5$ would be the symmetry group of the $4$-dimensional simplex, and so on.

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This is just speculation, but I suspect that it might have something to do with symmetric functions. A function $f$ of $n$ variables is said to be symmetric if its value is unchanged for any permutation of the variables: $$f(x_{\pi(1)},\dots,x_{\pi(n)}) = f(x_1,\dots,x_n)$$ for all $\pi \in S_n$. For example, the polynomial $f(x_1,x_2,x_3)=x_1 x_2 + x_1 x_3 + x_2 x_3$ is symmetric: $f(x_1,x_2,x_3)=f(x_1,x_3,x_2)=f(x_2,x_3,x_1)=\dots$.

Alfred Young wrote a famous series of papers about such things under the title "On Quantitative Substitutional Analysis". In the first part (from 1900), he uses the name "symmetric group" without comment, so the terminology must be older than that.

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+1. I like your "symmetric functions" point of view. Your example reminds me of the symmetric polynomials. – Jack Jun 20 '11 at 3:24
I believe this is the historical reason. Galois initiated the study of permutation groups in the context of solvability of polynomial equations. He was discussing permutations of the roots of these equations. – Ben Blum-Smith Dec 27 '11 at 15:58

I've always thought the terminology came from the pre-history of abstract group theory, which is basically summarized by Cayley's Theorem. http://en.wikipedia.org/wiki/Symmetric_group

This is the theorem that says every finite group is a subgroup of a symmetric group.

This is how the first groups were studied, before the terminology of "abstract groups" existed. For example, the dihedral group. Instead of thinking of it as an abstract group, you can think of it as the collection of permutations of the vertices of an $n$-gon, induced by isometries.

Similarly with any other group of symmetries of anything. For example, the symmetry group of an icosahedron? Think of it as permutations of the vertices.

i.e. every "symmetry group" is a subgroup of this universal group. So it's called the "symmetric group" because of that.

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+1...Note: From my understanding of Cayley's Theorem, his claim was much stronger: that every group (not merely finite groups) is isomormphic to some group consisting of permutations under permutation multiplication. – amWhy Jun 20 '11 at 0:08

Just to add to the examples that have been given of "most symmetric possible" structures, i.e., admitting as their symmetry group the full symmetric group, it is sufficient to consider the so-called complete graph on $n$ vertices, $K_n$. This is simply the graph with $n$ vertices and all possible edges, that is, if we call the vertices $v_1$, $v_2$, ..., $v_n$, it has all edges $(v_i, v_j)$ for all pairs $i\ne j$.
A symmetry (or automorphism) of a graph is in general a permutation $\varphi$ of the set of vertices that preserves the adjacency (i.e., $(v,w)$ is an edge if and only if $(\varphi(v), \varphi(w))$ is). A generic graph admits a limited set of symmetries; in other words, not all permutations of its vertices are automorphism.
But for a complete graph, every permutation of the vertices is an automorphism.
So in a sense $K_n$ is the "most symmetric" graph on $n$ vertices (and, by the way, each group of permutations on $n$ elements, i.e. each subgroup of $S_n$, is the symmetry group of some $n$-vertex graph).

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This is a good example – Andrea Mori Jun 19 '11 at 22:18
+1. I really appreciate your "graph" point of view. I have never noticed such connection between symmetry and graph. – Jack Jun 20 '11 at 3:28

As a matter of fact, there is at least one case in which the symmetric group is a group of symmetries, although "symmetry" must be taken here in an ampler sense. Namely, the symmetric group ${\cal S}_n$ does appear as Galois group of certain polynomials and tells something about the nature of the set of roots.

In general, groups of symmetries of regular polyhedra can be realized as subgroups of the symmetric group. Since any symmetry of a polyhedron will send vertices to vertices, a numbering of the $N$ vertices will induce an embedding of its groups of symmetries into ${\cal S}_N$.

The dihedral group $D_n$ is the group of symmetries of the regular $n$-sided polygon (planar polyhedra).

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