tetrahedron, group of symmetries

I am supposed to label the vertices of a tetrahedron 1, 2, 3, and 4. I am supposed to explicitly describe the group of symmetries of the tetrahedron as permutations of these vertices. I know there are supposed to be 12 and I have written out all 12, see below. Then I am supposed to write the group table. I do this by composing the permutations, I think: a1.a1, a1.a2, a1.a3....a12.a11, a12.a12.

I don't know how I know that there are 12 permutations in the symmetric group. If 1, 2, 3, 4 are the possibilities how does one know that there are 12 elements in the symmetric group? In other words, if the 5 members in the set, how many elements are in the symmetric group? If there were 12? Also, I was asked to list the conjugacy classes. How do I find those? So far I have completed some of the table but I still have quite a ways to go.

Here are the 12 permutations I found:

a_{1}=(1)

a_{2}=(12)(34)

a_{3}=(13)(24)

a_{4}=(14)(23)

a_{5}=(123)

a_{6}=(243)

a_{7}=(142)

a_{8}=(134)

a_{9}=(132)

a_{10}=(143)

a_{11}=(243)

a_{12}=(124)

Re: tetrahedron, group of symmetries

The number of permutations of a set {1,2,...,n} is n! , so there are 24 permutations in your case.

I wrote more but I edited as I'm unsure if it's right. The permutations of just 2 elements are missing and the permutations of 4 elements are missing.

Re: tetrahedron, group of symmetries

the group of symmetries of a tetrahedron is "larger" than just the "rotations" (the ones we can actually perform with an actual tetrahderon).

for example: suppose i imagine i have a mirror that aligns with one edge and "slices the tetrahedron in half" (let's say the apex and the vertex closest to me form the edge, and the plane of reflection extends through the middle of the opposite face and the base). that is going to have the effect of just exchanging two vertices.

we have 6 edges to choose from, so that gives us 6 more symmetries you did not list:

(1 2)

(1 3)

(1 4)

(2 3)

(2 4)

(3 4)

if you perform the reflection that switches (3 4) and then perform the rotation (1 2 3) that leaves vertex 4 fixed the net result is:

1 (unchanged by the reflection) --> 1 ---> 2 (after rotation)

2 (unchanged by the reflection) --> 2 ---> 3 (after rotation)

3 (gets swapped with 4) ---> 4 (rotation leaves fixed)

4 (gets swapped with 3) ---> 3 ---> 1 (after rotation)

by performing a reflection and then a rotation, we get the following 6 4-cycles:

(1 2 3 4)

(1 4 3 2) <---the "reverse"

(1 2 4 3)

(1 3 4 2)

(1 3 2 4)

(1 4 2 3), which accounts for the 12 you are missing.

the conjugate classes of a symmetric group are exactly the "disjoint cycle types", in this case:

the identity (any and all 1-cycles)

transpositions (2-cycles, "swap")

3-cycles

2 disjoint transposition pairs ("double swaps")

4-cycles

Re: tetrahedron, group of symmetries

I found this in a book:

Symmetries of the tetrahedron

Fix a tetrahedron centered at the origin, with one vertex along the z-axis.

Each edge has an "opposite" edge on the tetrahedron (which is actually

perpendicular to it if you look at it straight on). Each vertex has an

"opposite" face.

There are orientation preserving symmetries (called "rotations") of the

tetrahedron and orientation reversing symmetries of the tetrahedron. The

orientation preserving symmetries of the tetrahedron will be denotes ST.

They are obtained as follows:

* the 4 axes of symmetry through the centers of the faces yield

2 elements each (120 degree clockwise rotation when viewed from

outside and a 240 degree rotation), for a total of 8 elements,

* the 3 pairs of edges (formed by an edge and its opposite)

yield one element each (a 180 degree rotation), for a total

of 3 elements.

These, plus the identity, give 12 elements in ST.

Two books I have consulted tell me that there are 12 elements in the Symmetrical group, not 12. I don;t understand why.

Re: tetrahedron, group of symmetries

the orientation PRESERVING symmetries are called ROTATIONS, and there are 12 of these. THAT is what your book says.

the FULL symmetry group has 24 elements (this includes what are sometimes called "improper rotations" or "orientation-reversing symmetries").

a symmetry (of a geometric object S in euclidean n-space E^{n}) is a mapping f: E^{n}-->E^{n} that maps S "rigidly" onto S (it is an isometry, or "distance-preseving map").

for a line segment, AB, there are exactly two such maps:

AB -->AB (the identity map)

AB -->BA

(if we fix the origin of E^{n} at the midpoint O of AB, we can regard AB as the vector v that starts at A and ends at B. this gives the following two symmetries:

v-->v

v-->-v).

if we adopt coordinate axes so that B = 1 (thus identifying E^{1} with R) then we get the two (1x1) matrices:

I(x) = x (so I = [1]), R(x) = -x (so R = [-1]). the group {I,R} is isomorphic to S_{2}:

I<-->e

R<-->(1 2)

for a triangle: we can fix the center at the origin, and put our vertices (1,2,3) at (1,0), (-1/2,√3/2), (-1/2,-√3/2), giving the following maps from R^{2}-->R^{2}:

$\displaystyle I = \begin{bmatrix}1&0\\0&1 \end{bmatrix};\ A = \begin{bmatrix}\frac{-1}{2}&\frac{-\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}&\frac{-1}{2} \end{bmatrix};\ B = \begin{bmatrix}1&0\\0&-1 \end{bmatrix}$

this yields the group D_{3} = {I,A,A^{2},B,AB,A^{2}B} which is isomorphic to S_{3}:

I<-->e

A<-->(1 2 3)

A^{2}<-->(1 3 2) <---the first three maps you might recognize as rotations of 0,120 and 240 degrees counter-clockwise.

B<-->(2 3) <---reflection across the x-axis

AB<-->(1 2) <---reflection across the line y = (√3)x

A^{2}B<-->(1 3) <---reflection across the line y = -(√3)x

the subgroup of "proper" (orientation-preserving) isometries is {I,A,A^{2}} (the rotations) which is isomorphic to A_{3} = {e, (1 2 3), (1 3 2)}.

for a tetrahedron, we now have to go to at least n = 3 to embed it in. regarding our four vertices (1,2,3,4) as the four points:

(1,0,0), (-1/3,0,2√2/3), (-1/3,-√6/3,-√2/3) and (-1/3,√6/3,-√2/3) we have the 3x3 matrices:

$\displaystyle I = \begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix};\ A = \begin{bmatrix}1&0&0\\0&\frac{-1}{2}&\frac{-\sqrt{3}}{2}\\0&\frac{\sqrt{3}}{2}&\frac{-1}{2} \end{bmatrix}; B = \begin{bmatrix}\frac{-1}{3}&0&\frac{2\sqrt{2}}{3}\\0&1&0\\ \frac{2\sqrt{2}}{3}&0&\frac{1}{3} \end{bmatrix}$

the group generated by A and B can easily be seen to be isomorphic to the subgroup of S_{4} generated by (2 3 4) and (1 2):

the matrix A does this to the vertices:

1-->1

2-->3

3-->4

4-->2

the matrix B does this to the vertices:

1-->2

2-->1

3-->3

4-->4

you may verify for yourself (i urge you to actually DO this) that A^{3} = I, B^{2} = I.

so what is <(1 2),(2 3 4)>? well, clearly we have e, (1 2), (2 3 4) and (2 4 3) as elements.

(1 2)(2 3 4) = (1 2 3 4), so we also have (1 2 3 4)^{-1} = (1 4 3 2). thus we also have:

(1 2 3 4)(1 2) = (1 3 4) (and thus (1 4 3) as well). keep track, that's 8 elements so far.

(2 3 4)(1 3 4) = (1 4)(2 3) (up to 9, now).

(1 3 4)(2 3 4) = (1 3)(2 4) (up to 10)

(1 4)(2 3)(1 3)(2 4) = (1 2)(3 4) (now we have 11). since we also have (1 2) in <(1 2),(2 3 4)>, we also have:

(1 2)(1 2)(3 4) = (3 4) in our subgroup as well. that's 12 elements we've explicitly shown are in <(1 2),(2 3 4)>.

finally, we see that:

(3 4)(2 3 4) = (2 4) is a 13th element. but wait! S_{4} has 4! = 24 elements, and we've found a subgroup that contains more than half of S_{4}.

by lagrange, the order of <(1 2),(2 3 4)> has to be a divisor of 24, and there are no divisors of 24 larger than 12 except...24.

so (1 2) and (2 3 4) generate ALL of S_{4}, which means that the subgroup generated by A and B above must also generate a group of order 24.

on the other hand, it's easy to see that ANY symmetry of our tetrahedron can be represented as a permutation of the vertices.

so the symmetry group of a regular tetrahedron is no greater than 24, and is at least 24, so must be 24.