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Subject:

Jerry Bryan wrote in his e-mail message of 1994/12/27

Recently, there was some discussion of whether the set C of twenty-four

rotations is a normal subgroup of the cube group G=<Q>. It isn't, but

I decided to write up some information about normal subgroups as it

relates to the cube. Most of the following is from Frey and Singmaster.

Any good stuff is theirs. Any crud that sneaks in is mine.

Very good. The lattice of normal subgroups of G is not too complicated

and relates nicely to the structure of this group. Allow me to throw in

my $0.02.

Nitpicking alert! C is not even a subgroup of G=<Q>. It is a subgroup

of CG. But not a normal one.

Jerry continued

It should be noted that H normal does not imply that the elements

h of H commute with the elements x of G. That is, just because

Hx=xH we do not necessarily have hx=xh for every h in H (or even for

any h in H other than the identity). However, I think it is fair

to characterize a normal subgroup as commuting "globally" with G,

even if it does not commute "locally". On the other hand, if a

subgroup H does commute "locally" (i.e., if hx=xh for all h in H

and all x in G), then H is certainly normal.

In group theory there this distinction is made by using the terms

*normal* and *central* (even though globally and locally are perhaps

more descriptive names).

If xH = Hx, then x is said to *normalize* H. The set of all elements

that normalize H is called the *normalizer* of H, usually written N_G(H).

It is easy to see that N_G(H) is a subgroup of G containing H.

If every x of G normalizes H, then H is said to be *normal* in G.

Of course H is normal in G, if and only if N_G(H) = G.

If xh = hx for every h in H, then x is said to *centralize* H. The set

of all elements that centralize H is called the *centralizer* of H,

usually written C_G(H). It is easy to see that C_G(H) is again a

subgroup of G, but it need not contain H (it contains H if and only if H

is abelian). If every x of G centralizes H, then H is said to be

*central* in G. Of course H is central in G, if and only if C_G(H) = G.

Furthermore it is easy to see that H is central, if and only if H is a

subgroup of C_G(G), which is the set of those elements in G that commute

with all elements in G. C_G(G) is called the *center* of G.

And as you say, a central subgroup is also normal, but a normal subgroup

need not be central.

If N1 and N2 are two normal subgroups of G, then it is easy to see that

the intersection of N1 and N2 and the closure of N1 and N2 are both

normal subgroups too. Thus the set of normal subgroups of G is closed

w.r.t. intersection and closure. In other words, the set of normal

subgroups forms a lattice. I shall draw the lattice of normal subgroups

of G below.

Jerry continued

Normal groups serve a function with respect to finite groups analogous

to the function served by prime numbers with respect to natural numbers.

First of all, any finite group always has at least two trivial

normal subgroups, namely the group itself and the group containing

only the identity. Second, a finite group containing normal subgroups

may be "factored" in a fashion analogous to prime numbers factoring

composite numbers. A finite group containing no normal subgroups

is called simple, analogous to numbers with no factors being called

prime.

This is correct. Allow me a few more remarks.

Groups are composed from simple groups, which correspond to primes. The

simple groups have been classified. There are several families (the

alternating groups A_n are one such family), and 26 sporadic simple

groups. This classification is one of the outstanding mathematical

achievements. It is estimated that the complete proof is about 10000

pages long (distributed over several papers, books, Ph.D. thesis, etc.).

And then there are the ways in which those simple groups can be composed.

In the case of natural numbers, this is very simple. The fundamental

theorem tells us, that there is, up to the order, just one way in which

any natural number is composed from primes. In the case of groups it is

much more difficult. There is still a theorem which tells us that the

composition factors of a group are determined up to order. But not any

order will do. For example the symmetric group S_3 of size 6, has a

factor group C_2 over a normal subgroup C_3, but it cannot be decomposed

with a factor group C_3 over a normal subgroup C_2. Furthermore given a

certain set of composition factors and a certain order, there may be

several groups that decompose in this way. For example the cyclic group

C_6 can also be decomposed with a factor group C_2 over a normal subgroup

C_3. Even in the simplest case, groups of prime power size, which

decompose as a sequence of cyclic groups C_p, is so difficult that they

have not been classified (and maybe never will).

Jerry continued

The cube group G does not have very many normal subgroups, but it does

have a few. The first place to look for normal subgroups is to look

for subgroups with index 2. That is, look for subgroups that are

half as big a G. Such a subgroup is the subgroup A of even

permutations. ("A" stands for "Alternating", I think.)It is easy to see that A is normal. If x is even, then Ax=xA=A.

if x is odd, then Ax=xA=Abar, where Abar is the set (not group!)

of odd permutations.Similarly, any subgroup H with index 2 is normal. If the index of H

in G is 2, then H partitions G into two equal size sets H and Hbar.

If x is in H, then Hx=xH=H. If x is in Hbar, then Hx=xH=Hbar.

... and later on ...The factor group G/A contains two elements, and is isomorphic to any

group containing only two elements. We may write it as

<Abar>={A,Abar}, where A is the identity of the group.

This is correct. There is another way to obtain A, which is also very

instructive.

Let G be any group. Let g1 and g2 be two elements of G. Then the

element g1^-1 * g2^-1 * g1 * g2 is called the commutator of g1 and g2,

and is usually written as [g1,g2]. Now let g be any element of G. Then

g^-1 [g1,g2] g = g^-1 g1^-1 g2^-2 g1 g2 g = (g^-1 g1^-1 g) (g^-1 g2^-1 g) (g^-1 g1 g) (g^-1 g2 g) = (g^-1 g1 g)^-1 (g^-1 g2 g)^-1 (g^-1 g1 g) (g^-1 g2 g) = [ (g^-1 g1 g), (g^-1 g2 g) ].

Thus the conjugate of a commutator is again a commutator. It follows

that the subgroup generated by all commutators of all pairs of elements

of G is a normal subgroup. This subgroup is called the *commutator

subgroup* or *derived subgroup* of G, and is usually written G'.

It is the minimal normal subgroup of G, such that the factor group G/G'

is an abelian group. Minimal means that for each normal subgroup N of G

such that G/N is an abelian group, G' is a subgroup of N.

In the case of the cube group G' is A. I shall use G' instead of A.

Jerry continued

If we may digress briefly to the set M of 48 rotations and reflections,

then there are three subgroups of M with index 2. In Dan Hoey's

taxonomy, they are called C, A, and H. We may categorize the elements

of M as even or odd, and as rotations or reflections. There are 12

even rotations, 12 odd rotations, 12 even reflections, and 12 odd

reflections. If we take 12 even rotations and 12 odd rotations,

we have C. So C is a normal subgroup of M, even if it is not a normal

subgroup of G. If we take 12 even rotations and 12 even reflections,

we have A. This A (a subgroup of M) is not to be confused with the

A we have already talked about which is a subgroup of G. But I think

the name derives from the same source ("Alternating") in either case.

If we take 12 even rotations and 12 odd reflections, we have H.

In the case of M, M' is a subgroup of index 4. And the factor group M/M'

is isomorphic to the group { (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) }.

This group is usually called C_2^2, because it is the direct product of

two cyclic groups of size 2. This group has three (normal) subgroups of

index 2, which correspond to the three normal subgroups of M Jerry

described.

Jerry continued

Returning to G, the next two normal subgroups are Ac which leaves

the set of edges fixed, and Ae which leaves the set of corners

fixed. Ac is even on the corners, and Ae is even on the edges, in

order to conserve parity. Note that both Ac and Ae are normal

subgroups of A as well as of G.

... and later on ...The factor group G/Ac is isomorphic to the set of all permutations

on the edges (which we have written as G[E] in the recent past).

The factor group G/Ae is isomorphic to the set of all permutations

on the corners (which we have written as G[C] in the recent past).

This is again typical. If we have a permutation group G that has more

than one orbit, then the stabilizer of each orbit is a normal subgroup

of G (which may or may not be trivial). And the group G is a subdirect

product of the factor groups over those normal subgroups.

Time for the first picture (apologies to all that have no large screen,

but I want to add more to it later without cluttering it too much).

GCE /|\ 2 X G X / \|/ \ 2 / G' \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ GE ~ G[E] / / \ / 2 / / GE' / / / G[C] ~ GC / / 2 \ / / GC' / \ / \ / \ / \ / \ / 12!/2 2^11 \ / 8!/2 3^7 \ / \ / \ / \ / \ / \ / \ / \ / <1>

Let GC be the operation of G on the corners, which is isomorphic to

the factor group G/GE' = G[C], and GE be the operation of G on the

edges, isomorphic to the factor group G/GC' = G[E] (this distinction

between GC and G[C] and GE and G[E] is subtle and not very important).

Let GCE be the direct product of GC and GE. G is a subdirect product

of GC and GE, so it is a subgroup of GCE. It is a subgroup of index 2.

G has a subgroup of index 2, namely G', which Jerry called A above.

The two stabilizers are GC' and GE', which Jerry called Ac and Ae above.

They are in fact the derived subgroups of GC and GE.

Note that GCE has two more (normal) subgroups of index 2, namely

<G',GC> and <G',GE>. And G' is the derived subgroup of CGE.

So GCE/GCE' is again isomorphic to C_2^2.

Jerry continued

Since Ac and Ae are normal subgroups of A, we may write A/Ac and A/Ae

which are isomorphic to Ae and Ac, respectively.

G' (aka A) is in fact the direct product of GC' (aka Ac) and GE' (aka

Ae). And the standard isomorphism theorem tells us that G'/GC' ~ GE'

and G'/GE' ~ GC'.

Jerry continued

We can find normal subgroups of Ac and Ae. The set At of all

permutations in Ac which leave all corner locations fixed except for

twisting some of them is a normal subgroup of Ac. The set Af of

all permutations in Ae which leave all edge locations fixed except

for flipping some of them is a normal subgroup of Ae. (This twists

and flips have to follow the normal rules of conservation of twist

and flip, of course.)This completes the list of normal subgroups. I will now give Frey

and Singmaster's proof that we are done, while interposing some

questions of my own for the cube theory experts out there.

I haven't seen what Frey and Singmaster prove. But this is not true.

Together with the trivial normal subgroup G and <1> you have listed

7 normal subgroups, but G has indeed 13 normal subgroups.

Jerry continued

My first question is that Frey and Singmaster do not state that At

and Af are normal subgroups of G. It seems obvious that they are.

However, is the formal argument that (for example) At is a normal

subgroup of Ac and Ac is a normal subgroup of G; hence, At is a

normal subgroup of G? How analogous is the factoring of groups

by normal subgroups to the factoring of composite numbers by

prime numbers?

This is not true, and Michael Reid gave the smallest counterexample.

Group theory would certainly be a lot easier if this was true, but

probably also a lot less challenging. But there is a similar argument.

If M is a normal subgroup of a group N that is invariant under all

automorphisms of N, then M is called *characteristic*. For example N' is

always a characteristic subgroup of N. If N is a subgroup of a group G,

and M is a characteristic subgroup of N, then M is normal in G. Also if

N is a characteristic subgroup of G, and M is a characteristic subgroup

of N, then M is also a characteristic subgroup of G.

This actually happens here, At and Af are characteristic in GC' and GE',

so they are normal in G. But this is not so easy to prove, it is simpler

to verify directly that At and Af are normal in G (or use the fact that

they are again stabilizers of appropriate operations of G).

Jerry continued:

I guess my questions are as follows: 1) why must we restrict ourselves

to alternating groups? 2) For example, just as we found three

subgroups of M with index 2, might we not find other subgroups of

G with index 2 than the one we found? 3) Might we not find a

normal subgroup of G with some index other than 2, e.g., with index 3?

I got A as the derived subgroup of G. Since this is the minimal normal

subgroup such that the factor group is abelian, there cannot be another

normal subgroup of index 2 or 3. But there can be other normal subgroups

with non-abelian factor groups, and indeed there are.

The rest of this message describes how one can find all normal subgroups

of G. My approach may or may not be equal to Frey and Singmaster's.

Let us first consider GC. This group is a subgroup of index 3

in the wreath product C_3 <wreath> S_8.

This wreath product is isomorphic to the group of the following elements

( c_1, c_2, c_3, c_4, c_5, c_6, c_7, c_8; p )

where the c_i are in {0,1,2} and p is a permutation in S_8.

Multiplication of those elements is defined as follows

(c_1,c_2,...,c_8;p) (d_1,d_2,...,d_8;q) := ( c_1 + d_{1^p}, c_2 + d_{2^p}, ..., c_8 d_{8^p}; p * q )

where d_{i^p} denotes d_j, where j is the image of i under the

permutation p, and c_i + d_j is the sum of c_i and d_j modulo 3.

p * q is simply the product of p and q in S_8.

So you first permute the components of the second element with

the permutation of the first element, then sum componentwise,

and finally multiply the two permutations.

Clearly C_3 <wreath> S_8 has 3^8 8! elements. GC is the subgroup of those elements of C_3 <wreath> S_8 for which c_1+c_2+...+c_8 = 0 (mod 3).

It is easy to see that the set of all 3^7 elements

(c_1,c_2,...,c_8,<identity>) of GC forms a normal subgroup of GC.

I shall call this subgroup VC (V because VC is in fact a vector space),

Jerry called this subgroup At.

Next I shall show that no proper subgroup of VC can be a normal subgroup

of GC. Suppose that H is a normal subgroup of VC, and let

x = (x_1,x_2,...,x_8;<identity>) be any non-trivial element of H.

Because 1+1+...+1 <> 0 (mod 3) and 2+2+...+2 <> 0 (mod 3), there are two

components x_i and x_j which are different. An easy calculation shows

y := (0,0,...,0;(i,j))^-1 * x^-1 * (0,0,...,0;(i,j)) * x

has y_i = 1 and y_j = 2 (or the other way around). Since H is supposed

to be normal, y must also be in H. Furthermore the 6 elements

zk := (0,0,...,0;(j,k))^-1 * y * (0,0,...,0;(j,k))

where k <> j and k <> i all have zk_i = 1 and zk_k = -1.

Again since H is supposed to be normal, they must all lie in H.

But y and those 6 elements are obviously linearly independent,

i.e., form a basis for VC. In particular it follows that H = VC.

So no proper subgroup of VC is a normal subgroup of GC.

Now let N be any normal subgroup of GC. Then the intersection of N and

VC must be a normal subgroup of GC contained in VC. Thus there are only

two possibilities for this intersection; it can be VC (i.e. N contains

VC) or it can be trivial.

Assume first that N contains VC. Then N/VC must be a normal subgroup of

GC/VC. But GC/VC is S_8, which has only three normal subgroups, namely

S_8, A_8 (= S_8'), and <1>. Thus we have three normal subgroups

containing VC, namely GC, GC', and VC.

On the other hand assume that N intersects trivially with VC. Then the

closure M of N and VC must be one of the three normal subgroups given

above. The isomorphism theorem tells us that M/N ~ VC. GC does not have

a factor group isomorphic to VC, since its largest abelian factor group

is GC/GC' of size 2. GC' also does not have a factor group isomorphic to

VC, since it has no non-trivial abelian factor group at all. So M must

be VC, and N must be trivial.

All in all GC has 4 normal subgroups, GC, GC', VC, and <1>.

Basically the same argument works for GE. But there is one exception.

Namely VE has one normal subgroup of size 2 , generated by the element

(1,1,...,1;<identity>). You may not recognize this element, but it is

in fact the superflip, which flips all twelve edges. I shall call this

subgroup Z.

Thus GE has 5 normal subgroups GE, GE', VE, Z, and <1>.

Now we are ready to return to G resp. GCE. The normal subgroups of GCE

are direct or subdirect products of normal subgroups of GC and GE.

For a direct product we take a normal subgroup NC of GC and a normal

subgroup NE of GE and take their direct product, i.e., their closure

in GCE.

For a subdirect product we must take a normal subgroup NC of GC and a

normal subgroup NE of GE and ``glue'' together a common factor group F of

NC and NE. But there are only two cases where a normal subgroup of GC and

a normal subgroup of GE have a common factor group. Once case is where

NC = GC and NE = GE and F = C_2, and we get G as a subdirect product.

The other case is NC = GC and NE = Z and F = C_2.

All in all, we get the following lattice of normal subgroups of GCE.

GCE /|\ X G X / \|/ \ / G' \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ \ / / \ X / / \ / \ / / X \ / / / \ \ X / / \ \ / \ / / \ \ / X / \ GE ~ G[E] / / \ / \ / 2 X / \ / GE' /|\ / \ / / G[C] ~ GC + X \ / / 2 \|/ \ \ / / GC' \ \ / / \ \ \ / / \ \ \ / / 12!/2 \ \ X / 8!/2 \ \ / \ / \ \ / \ / \ \ / \ / \ X \ / \ / \ \ / VC \ VE \ \ / \ \ / 2^10 3^7 \ \ / \ Z \ / 2 <1>

(Looks nice, doesn't it?)

Bear with me, we are almost done. We only need one more step, namely to

show that the normal subgroups of GCE that lie in G are exactely the

normal subgroups of G. One direction is obvious, the normal subgroups of

GCE that lie in G are also normal subgroups of G. But we need to show

that any normal subgroup of G is also a normal subgroup of GCE.

Assume then that N1 is a normal subgroup of G that is not *not* normal in

GCE. G is then the normalizer of N1, and because the index of the

normalizer in GCE is the number of conjugates of N1, it follows that N1

has one more conjugate subgroup in G. Call this subgroup N2, and assume

N2 = x^-1 N1 x for an appropriate element x of GCE. Because

N_GCE(N2) = N_GCE(x^-1 N1 x) = x^-1 NGCE(N1) x = x^-1 G x = G,

it follows that N2 is also a normal subgroup of G.

The closure and the intersection of a whole family of conjugated

subgroups are always normal. Thus the closure and the intersection of N1

and N2 are normal subgroups of GCE (and are therefore subgroups that

appear in the above lattice). Call them N12 and N. Clearly N12 and N are

subgroups of G.

Now the isomorphism theorem tells us that N12/N1 ~ N2/N. Then |N12/N| = |N12/N1| |N1/N| = |N12/N1| |N2/N| = |N12/N1| |N12/N1|. Thus |N12/N| is a square. But the only factor groups with square sizes are VE/Z, (VE*VC)/(Z*VC), (VE*GC')/(Z*GC') (all of size 2^10).

We can intersect the whole situation into VE, so we can without loss of

generality assume that N1 and N2 are subgroups of VE. But if N1 and N2

are normal subgroups of G, then they are certainly normal subgroups of

GE'. But GE' has only the 4 normal subgroups: GE', VE, Z, and <1>.

Thus there cannot be a normal subgroup of G that is not normal in GCE.

So the following picture gives the lattice of normal subgroups of G.

G | 2 G' / \ / \ / \ / \ / \ / \ / \ / \ / X / / \ / / \ / / \ X / \ / \ / \ / \ / GE' / \ / / X \ / / / \ \ / / GC' \ \ / / \ \ \ / / \ \ \ / / 12!/2 \ \ X / 8!/2 \ \ / \ / \ \ / \ / \ \ / \ / \ X \ / \ / \ \ / VC \ VE \ \ / \ \ / 2^10 3^7 \ \ / \ Z \ / 2 <1>

So Jerry had it almost right. But he missed the center Z, and all

the closures of pairs of normal subgroups.

Have a nice day.

Martin.

-- .- .-. - .. -. .-.. --- ...- . ... .- -. -. .. -.- .- Martin Sch"onert, Martin.Schoenert@Math.RWTH-Aachen.DE, +49 241 804551 Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany