Date: Fri, 30 Dec 94 15:17:00 WET
From: Martin Schoenert <Martin.Schoenert@math.rwth-aachen.de >
Subject: Re: Normal Subgroups of G
```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>.

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
```