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In many elementary group theory books there are pictures of symetries for objects ( Cube, Square etc. ) and some group that represents that symmetries.

My question is what property of a group is related to symmetries, and if there are more than one type of symmetry (e.g. translation, rotation, miroring respective to some axis etc. ) then could the existence of number of types of symmetry be inferred directly from the given group?

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4 Answers 4

up vote 5 down vote accepted

When you look at a figure $F$ with some symmetries you don't see a group but certain obvious symmetries: reflections with respect to various axes, rotational symmetries centered on various points, or translations, all of them being maps $\phi:\ F\to F$ respecting incidences, distances or colorings present in $F$.

Such a figure $F$ is erected on a "ground set" $E$, e.g. the euclidean plane, and any symmetry $\phi:\ F\to F$ carries with it a bijection of $E$. It follows that such a $\phi$ has an inverse which is also a symmetry.

Now comes the second step, namely the composition $\psi\circ\phi$ of symmetries. It follows from general principles about maps that any finite composition of symmetries $\phi$ is again a symmetry in the sense that it leaves incidences, distances, and colorings invariant, and that the set of all symmetries of $F$ obtained in this way forms a group $S_F\ $.

If $F$ is a simple figure, e.g., a rosetta with a center $O$ and $n$ leaves, then it is easy to describe $S_F\ $: It is either the cyclic group ${\mathbb Z}_n$ of $n$ elements generated by a rotation of ${2\pi\over n}$ around $O$, or it is the dihedral group $D_n$ which in addition contains $n$ reflections.

If $F$ is a complicated figure, say a tessellation of ${\mathbb R}^2$ by equilateral triangles, then you can immediately see a lot of symmetries, but you don't get a quick overview over $S_F$. In particular it is not obvious whether finite products of rotations, reflections, etc. again can be viewed as such "elementary" symmetries (with other pivots or axes), or whether there are new kinds of motions present in $S_F$. It turns out that $S_F$ may contain "glide reflections" which maybe you didn't see at the outset.

What I'm trying to say is that the connection between a figure $F$ with some symmetries and a particular group is not at all trivial. In many treatments of the subject (or of examples, like the symmetries of the $3$-cube) the question whether any product of "obvious" symmetries is again an "obvious" symmetry is not sufficiently addressed.

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A group acts on a set in a symmetric way. It is the group which rotates or flips the square. It is the formal definition of the symmetries of the square.

We can say that "Groups are symmetries" because of a theorem of, I believe, Max Dehn, which says that given a group $G$ generated by a set $S$, the group has a Cayley Graph, $\Gamma(S)$, and that $G$ is the set of symmetries of this graph.

Unfortunately, because of the negative answer to the word problem for groups, one cannot always construct the Cayley Graph of a group...(because loops in the graph correspond to words equal to $1$ on the generators.)

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Why the downvote? –  user1729 Jul 15 '11 at 10:48
If anything, I'd say that group consists of symmetries. But this is already some model of the group, while I prefer to think about groups as abstract objects that act on other objects (usually in a way that preserves some structure). –  Marek Jul 15 '11 at 12:45

All groups are groups of permutations: that's Cayley's theorem. The permutations (aka bijections) of a set are its symmetries.

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Let $G$ the set of all the transformations (rotations, mirrorings, etc.) leaving an object $X$ unaltered.

  • $G$ is endowed with a natural associative operation: the transformation composition.
    Indeed if $T_1$, $T_2\in G$ then $T_1 T_2 X = T_1 X = X$, so $T_1 T_2\in G$.

  • $\operatorname{id} X = X$ implies $\operatorname{id}\in G$

  • If $T\in G$ then $T^{-1} X = T^{-1} T X = X$, so also $T^{-1}\in G$.

The above properties show $G$ can be regarded as a group.

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What do you mean when you write $\operatorname{id} X = X$? That the identity is surjective? Why is it a transformation? How do you define a transformation? –  Rasmus Jul 15 '11 at 10:10
@Rasmus. For "tranformation", I mean an automorphism of a family of objects, $\mathcal F$, which $X$ belongs to. $\operatorname{id}$ is the identity map of $\mathcal F$. For examples $F$ can be the set of all the squares of the plane and the transformations, the maps generated on $\mathcal F$ by rotations around a point of the plane. –  AlbertH Jul 15 '11 at 10:54

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