# Alternating group without $S_n$

I'm going to start introducing my abstract algebra class to a variety of groups soon. Dihedral groups $$D_n$$ arise out of symmetries on polygons. And the Symmetric group $$S_n$$ makes sense as the group of permutations.

Is there a similarly naive or geometric way to give the Alternating group, $$A_n$$?

I know the Alternating group is the subgroup of even parity elements in $$S_n$$. But I thought it'd be nice for students to work with (an isomorphic version of) the group first and then later show it is the even parity elements.

I'm hoping someone out there has a simple (no pun intended) and independent construction of $$A_n$$.

• This answer on Mathematics might be helpful. – Xander Henderson Feb 19 at 16:17
• Can you have them all build tetrahedra and consider all 12 rotations? [Considering these as permutations of the four vertices of the tetrahedron.] – Nick C Feb 19 at 19:03
• True, though these constructions don't generalize obviously to all $n$, which I think is what the OP is asking for. – kcrisman Feb 19 at 22:47
• Also, upvote just for the pun. – kcrisman Feb 19 at 22:47
• Orientation preserving permutations of a vector basis? Or if they've only taken linear algebra the determinant preserving row/column swaps. – Nate Bade Feb 21 at 19:17

For all $$n \geq 3$$ the alternating group $$A_{n}$$ can be realized as the orientation-preserving symmetries of the regular $$n$$-simplex (triangle, tetrahedron, etc.). In this case the full symmetry group is $$S_{n}$$, and the relation between $$A_{n}$$ and $$S_{n}$$ can be seen geometrically. For instance, when $$n = 3$$, $$A_{3}$$ comprises the rotations through an angle of $$2\pi/3$$ of an equilateral triangle about its centroid, while $$S_{3}$$ includes additionally the reflections through the angle bisectors. One can relate these geometric descriptions to the usual descriptions in terms of permutations by looking at what the geometric transformations do to the triangle's vertices. The symmetries of the regular tetrahedron can be examined in a similarly explicit fashion.

The cardinality $$60$$ alternating group $$A_{5}$$ is also the group of rotational (orientation-preserving) symmetries of a regular icosahedron. (This example has the virtue of being visualizable in three-dimensional space. It has the defect of being peculiar to $$n = 5$$.)

Although somewhat complicated for beginning students, this can be written down explicitly in the following way that uses only introductory linear algebra. Take as the $$12$$ vertices of the isosahedron the cyclic permutations of $$(\pm 1, \pm w, 0)$$ where $$w = (1 + \sqrt{5})/2$$. The icosahedron is the convex hull of these $$12$$ points. $$A_{5}$$ is generated by the order $$2$$ map sending $$x = (x_{1}, x_{2}, x_{3}) \to (-x_{1}, -x_{2}, x_{3})$$, the order $$3$$ map sending $$x = (x_{1}, x_{2}, x_{3}) \to (x_{2}, x_{3}, x_{1})$$ (which is rotation by $$2\pi/3$$ about the axis $$(1, 1, 1)$$), and the order $$5$$ map given by rotating by $$2\pi/5$$ around the axis spanned by any one of the vertices. (Adding $$x \to -x$$ yields the cardinality $$120$$ full symmetry group of the icosahedron; note, however, that this is $$A_{5}\times \mathbb{Z}_2$$ and not the symmetric group $$S_{5}$$.) It is instructive for students to think about what is permuted by this action of $$A_{5}$$.

Comparing the two different geometric descriptions of $$A_{5}$$, as the rotational symmetries of an icosahedron and the rotational symmetries of a regular simplex, would be instructive (they are associated with different irreducible representations of $$A_{5}$$, of dimensions $$3$$ and $$4$$).

The 15 puzzle is a nice example of an alternating group. The classic puzzle gives a concrete model for $$A_{15}$$. Generalizations extend it to $$A_{n-1}$$ for any composite $$n$$.

Students could work out the way in which it is a group, and then parity could be introduced as an answer to the question of which configurations of the puzzle board are reachable from the start state.

• Can you give an example of precisely what a puzzle generalizing it would look like? Obviously they can't all be square - presumably Loyd would have had trouble implementing a 16 puzzle as an interesting rectangle :P See e.g. jstor.org/stable/2689069 If there is a nice concrete generalization for all positive $n$, this would be a valuable answer. – kcrisman Mar 20 at 2:50
• @kcrisman For n = 20, for example, it would be a 5x4 rectangle with the same rules of sliding. – John Coleman Mar 20 at 2:53
• Yeah, but what about n=17? – kcrisman Mar 20 at 17:35
• @kcrisman Good question. No straightforward generalization. Perhaps some version of a shunting puzzle or some variation of the oval track puzzle (that one yields the full permutation group -- but what if the turnstile is for three pieces?) – John Coleman Mar 20 at 18:00
• The only way in which 15 puzzle is a group is if you consider the set of positions which have the empty space in the bottom right corner, like when the puzzle is solved. In this case, the group is $A_{15}$, not $A_{16}$ because there are only 15 tiles. If you consider the empty space to be an invisible 16 tile, then the puzzle is not a group because the permutations (15 16) and (12 16) are solvable (in one move) but (12 15) is not solvable. – Ben May 4 at 14:47