Symmetries of a Cube

Date: 10/09/2003 at 02:27:34
From: Johny
Subject: symmetries of cube

Prove that the group of symmetries of a cube is isomorphic to S_4.

Date: 10/09/2003 at 07:10:14
From: Doctor Jacques
Subject: Re: symmetries of cube

Hi Johny,

Note that we must consider the group of rotations of the cube (i.e., 
reflections not included).

Let G be that group.  First, note that G has order 24.  Indeed, if you 
place a cube on a table, you can put each of the 6 faces up, and 
rotate that face in 4 ways.

The cube has 4 diagonals.  If we label the vertices of the top face A, 
B, C, D (clockwise), and the vertices of the bottom face E, F, G, H, 
in such a way that A is above E, B above F, and so on, the diagonals 
are:

  (AG), (BH), (CE), and (DF)

Note that we consider the diagonals as lines--(AB) is the same object 
as (BA).

Every rotation of the cube maps a diagonal to a diagonal, i.e. 
permutes the diagonals between themselves.  This shows that there is a 
homomorphism from G to S4. We must show that the homomorphism is 
bijective.

As both groups are finite and have the same number of elements, the 
homomorphism will be injective (one-to-one) if and only if it is 
surjective (onto).  Is is somewhat easier to show that it is 
surjective, i.e. that, for any permutation of the diagonals, there is 
a rotation of the cube the induces that permutation.

Now, S4 is generated by the transpositions (the permutations that 
exchange two elements).  Therefore, it is enough to show that any 
transposition of diagonals can be achieved by means of a rotation.

Assume that we want to exchange the diagonals (AG) and (DF).  Let M be 
the midpoint of AD and N the midpoint of GF.

Can you figure out the effect of a rotation of 180 degrees with the 
line MN as axis?

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/09/2003 at 14:39:59
From: Johny
Subject: symmetries of cube

Thank you for the answer.  However, I still have some doubts.  It is 
still not clear to me that every rotation of the cube permutes the
diagonals.  How do you show that? 

Moreover, why is it a group?  It might be that I don't have a good 
spatial imagination, but it still not clear to me.  I would be 
grateful if you could explain it in greater detail. 

Thank you very much!

Date: 10/10/2003 at 02:36:13
From: Doctor Jacques
Subject: Re: symmetries of cube

Hi again Johny,

A rotation of the cube maps a vertex to a vertex.  Also, a rotation 
preserves distances.  The distance between opposite vertices (the 
length of a diagonal) is not equal to any other distance between 
vertices (it is the largest such distance).  The conclusion is that 
the image of a diagonal must be a diagonal.

This means that any rotation permutes diagonals within themselves.  
The product of two rotations is obtained by executing the rotations 
one after the other.  This will execute the diagonal permutations in 
sequence (this is just composition of functions).  This means that, if 
r1 and r2 are rotations, and s1 and s2 the corresponding permutations, 
the permutation corresponding to (r2 o r1) will be (s2 o s1)--this is
a homomorphism.

Permutations of the diagonals are a group under composition--this is 
simply the group of permutations of 4 objects (in this case, the 
diagonals), i.e. S4.

To see the effect of the particular rotation I told you, you should 
try this with a "physical" cube (a die, for example).

With the labeling I used, the rotation through MN will induce the 
following permutation of the vertices:

  (AD)(BH)(CE)(FG)

(You can verify this by noting that the points M and N are fixed by 
the rotation, and that distances are preserved.)

If we consider the 4 diagonals AG, BH, CE, DF, we note that BH and CE 
will remain fixed *as diagonals* (the vertices will be exchanged, but 
the diagonals will remain the same), and AG and DF will be 
interchanged.  This shows that the rotation induces a transposition of 
these two diagonals, and similar rotations can be used to generate 
any transposition, and therefore any permutation of the diagonals can 
be acheived by a composition of such rotations.

This shows that the homomorphism is surjective (onto), and, as the 
groups have the same order, this is an isomorphism.

Does this make sense?

- Doctor Jacques, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 10/16/2003 at 01:07:42
From: Johny
Subject: Thank you (symmetries of cube)

Thank you very much for the answer!  You helped me a lot!