Subdividing a group

28.7 SUBDIVIDING A GROUP
(i) the set of elements H in G that are images of the elements of G forms a
subgroup of G ;
(ii) the set of elements K in G that are mapped onto the identity I in G forms
a subgroup of G.
As indicated in the previous section, the subgroup K is called the kernel of the
homomorphism.
To prove (i), suppose Z and W belong to H , with Z = X and W = Y , where
X and Y belong to G. Then
ZW = X Y = (XY )
and therefore belongs to H , and
Z −1 = (X )−1 = (X −1 )
and therefore belongs to H . These two results, together with the fact that I belongs to H , are enough to establish result (i).
To prove (ii), suppose X and Y belong to K; then
(XY ) = X Y = I I = I (closure),
I = (XX −1 ) = X (X −1 ) = I (X −1 ) = (X −1 )
and therefore X −1 belongs to K. These two results, together with the fact that I
belongs to K, are enough to establish (ii). An illustration of this result is provided
by the mapping Φ of
→ U(1) considered in the previous section. Its kernel
consists of the set of real numbers of the form 2πn, where n is an integer; it forms
a subgroup of R, the additive group of real numbers.
In fact the kernel K of a homomorphism is a normal subgroup of G. The
defining property of such a subgroup is that for every element X in G and every
element Y in the subgroup, XY X −1 belongs to the subgroup. This property is
easily verified for the kernel K, since
(XY X −1 ) = X Y (X −1 ) = X I (X −1 ) = X (X −1 ) = I .
Anticipating the discussion of subsection 28.7.2, the cosets of a normal subgroup
themselves form a group (see exercise 28.16).
28.7 Subdividing a group
We have already noted, when looking at the (arbitrary) order of headings in a
group table, that some choices appear to make the table more orderly than do
others. In the following subsections we will identify ways in which the elements
of a group can be divided up into sets with the property that the members of any
one set are more like the other members of the set, in some particular regard,
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than they are like any element that does not belong to the set. We will find that
these divisions will be such that the group is partitioned, i.e. the elements will be
divided into sets in such a way that each element of the group belongs to one,
and only one, such set.
We note in passing that the subgroups of a group do not form such a partition,
not least because the identity element is in every subgroup, rather than being in
precisely one. In other words, despite the nomenclature, a group is not simply the
aggregate of its proper subgroups.
28.7.1 Equivalence relations and classes
We now specify in a more mathematical manner what it means for two elements
of a group to be ‘more like’ one another than like a third element, as mentioned
in section 28.2. Our introduction will apply to any set, whether a group or not,
but our main interest will ultimately be in two particular applications to groups.
We start with the formal definition of an equivalence relation.
An equivalence relation on a set S is a relationship X ∼ Y , between two
elements X and Y belonging to S, in which the definition of the symbol ∼ must
satisfy the requirements of
(i) reflexivity, X ∼ X;
(ii) symmetry, X ∼ Y implies Y ∼ X;
(iii) transitivity, X ∼ Y and Y ∼ Z imply X ∼ Z.
Any particular two elements either satisfy or do not satisfy the relationship.
The general notion of an equivalence relation is very straightforward, and
the requirements on the symbol ∼ seem undemanding; but not all relationships
qualify. As an example within the topic of groups, if it meant ‘has the same
order as’ then clearly all the requirements would be satisfied. However, if it meant
‘commutes with’ then it would not be an equivalence relation, since although A
commutes with I, and I commutes with C, this does not necessarily imply that A
commutes with C, as is obvious from table 28.8.
It may be shown that an equivalence relation on S divides up S into classes Ci
such that:
(i) X and Y belong to the same class if, and only if, X ∼ Y ;
(ii) every element W of S belongs to exactly one class.
This may be shown as follows. Let X belong to S, and define the subset SX of
S to be the set of all elements U of S such that X ∼ U. Clearly by reflexivity
X belongs to SX . Suppose first that X ∼ Y , and let Z be any element of SY .
Then Y ∼ Z, and hence by transitivity X ∼ Z, which means that Z belongs to
SX . Conversely, since the symmetry law gives Y ∼ X, if Z belongs to SX then
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28.7 SUBDIVIDING A GROUP
this implies that Z belongs to SY . These two results together mean that the two
subsets SX and SY have the same members and hence are equal.
Now suppose that SX equals SY . Since Y belongs to SY it also belongs to SX
and hence X ∼ Y . This completes the proof of (i), once the distinct subsets of
type SX are identified as the classes Ci . Statement (ii) is an immediate corollary,
the class in question being identified as SW .
The most important property of an equivalence relation is as follows.
Two different subsets SX and SY can have no element in common, and the collection
of all the classes Ci is a ‘partition’ of S, i.e. every element in S belongs to one, and
only one, of the classes.
To prove this, suppose SX and SY have an element Z in common; then X ∼ Z
and Y ∼ Z and so by the symmetry and transitivity laws X ∼ Y . By the above
theorem this implies SX equals SY . But this contradicts the fact that SX and SY
are different subsets. Hence SX and SY can have no element in common.
Finally, if the elements of S are used in turn to define subsets and hence classes
in S, every element U is in the subset SU that is either a class already found or
constitutes a new one. It follows that the classes exhaust S, i.e. every element is
in some class.
Having established the general properties of equivalence relations, we now turn
to two specific examples of such relationships, in which the general set S has the
more specialised properties of a group G and the equivalence relation ∼ is chosen
in such a way that the relatively transparent general results for equivalence
relations can be used to derive powerful, but less obvious, results about the
properties of groups.
28.7.2 Congruence and cosets
As the first application of equivalence relations we now prove Lagrange’s theorem
which is stated as follows.
Lagrange’s theorem. If G is a finite group of order g and H is a subgroup of G of
order h then g is a multiple of h.
We take as the definition of ∼ that, given X and Y belonging to G, X ∼ Y if
X −1 Y belongs to H. This is the same as saying that Y = XHi for some element
Hi belonging to H; technically X and Y are said to be left-congruent with respect
to H.
This defines an equivalence relation, since it has the following properties.
(i) Reflexivity: X ∼ X, since X −1 X = I and I belongs to any subgroup.
(ii) Symmetry: X ∼ Y implies that X −1 Y belongs to H and so, therefore, does
its inverse, since H is a group. But (X −1 Y )−1 = Y −1 X and, as this belongs
to H, it follows that Y ∼ X.
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(iii) Transitivity: X ∼ Y and Y ∼ Z imply that X −1 Y and Y −1 Z belong to H
and so, therefore, does their product (X −1 Y )(Y −1 Z) = X −1 Z, from which
it follows that X ∼ Z.
With ∼ proved as an equivalence relation, we can immediately deduce that it
divides G into disjoint (non-overlapping) classes. For this particular equivalence
relation the classes are called the left cosets of H. Thus each element of G is in
one and only one left coset of H. The left coset containing any particular X is
usually written XH, and denotes the set of elements of the form XHi (one of
which is X itself since H contains the identity element); it must contain h different
elements, since if it did not, and two elements were equal,
XHi = XHj ,
we could deduce that Hi = Hj and that H contained fewer than h elements.
From our general results about equivalence relations it now follows that the
left cosets of H are a ‘partition’ of G into a number of sets each containing h
members. Since there are g members of G and each must be in just one of the
sets, it follows that g is a multiple of h. This concludes the proof of Lagrange’s
theorem.
The number of left cosets of H in G is known as the index of H in G and is
written [G : H]; numerically the index = g/h. For the record we note that, for
the trivial subgroup I, which contains only the identity element, [G : I] = g and
that, for a subgroup J of subgroup H, [G : H][H : J ] = [G : J ].
The validity of Lagrange’s theorem was established above using the far-reaching
properties of equivalence relations. However, for this specific purpose there is a
more direct and self-contained proof, which we now give.
Let X be some particular element of a finite group G of order g, and H be a
subgroup of G of order h, with typical element Yi . Consider the set of elements
XH ≡ {XY1 , XY2 , . . . , XYh }.
This set contains h distinct elements, since if any two were equal, i.e. XYi = XYj
with i = j, this would contradict the cancellation law. As we have already seen,
the set is called a left coset of H.
We now prove three simple results.
• Two cosets are either disjoint or identical. Suppose cosets X1 H and X2 H have
an element in common, i.e. X1 Y1 = X2 Y2 for some Y1 , Y2 in H. Then X1 =
X2 Y2 Y1−1 , and since Y1 and Y2 both belong to H so does Y2 Y1−1 ; thus X1
belongs to the left coset X2 H. Similarly X2 belongs to the left coset X1 H.
Consequently, either the two cosets are identical or it was wrong to assume
that they have an element in common.
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28.7 SUBDIVIDING A GROUP
• Two cosets X1 H and X2 H are identical if, and only if, X2−1 X1 belongs to H. If
X2−1 X1 belongs to H then X1 = X2 Yi for some i, and
X1 H = X2 Yi H = X2 H,
since by the permutation law Yi H = H. Thus the two cosets are identical.
Conversely, suppose X1 H = X2 H. Then X2−1 X1 H = H. But one element of
H (on the left of the equation) is I; thus X2−1 X1 must also be an element of H
(on the right). This proves the stated result.
• Every element of G is in some left coset XH. This follows trivially since H
contains I, and so the element Xi is in the coset Xi H.
The final step in establishing Lagrange’s theorem is, as previously, to note that
each coset contains h elements, that the cosets are disjoint and that every one of
the g elements in G appears in one and only one distinct coset. It follows that
g = kh for some integer k.
As noted earlier, Lagrange’s theorem justifies our statement that any group of
order p, where p is prime, must be cyclic and cannot have any proper subgroups:
since any subgroup must have an order that divides p, this can only be 1 or p,
corresponding to the two trivial subgroups I and the whole group.
It may be helpful to see an example worked through explicitly, and we again
use the same six-element group.
Find the left cosets of the proper subgroup H of the group G that has table 28.8 as its
multiplication table.
The subgroup consists of the set of elements H = {I, A, B}. We note in passing that it has
order 3, which, as required by Lagrange’s theorem, is a divisor of 6, the order of G. As in
all cases, H itself provides the first (left) coset, formally the coset
IH = {II, IA, IB} = {I, A, B}.
We continue by choosing an element not already selected, C say, and form
CH = {CI, CA, CB} = {C, D, E}.
These two cosets of H exhaust G, and are therefore the only cosets, the index of H in G
being equal to 2.
This completes the example, but it is useful to demonstrate that it would not have
mattered if we had taken D, say, instead of I to form a first coset
DH = {DI, DA, DB} = {D, E, C},
and then, from previously unselected elements, picked B, say:
BH = {BI, BA, BB} = {B, I, A}.
The same two cosets would have resulted. It will be noticed that the cosets are the same groupings of the elements
of G which we earlier noted as being the choice of adjacent column and row
headings that give the multiplication table its ‘neatest’ appearance. Furthermore,
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if H is a normal subgroup of G then its (left) cosets themselves form a group (see
exercise 28.16).
28.7.3 Conjugates and classes
Our second example of an equivalence relation is concerned with those elements
X and Y of a group G that can be connected by a transformation of the form
Y = G−1
i XGi , where Gi is an (appropriate) element of G. Thus X ∼ Y if there
exists an element Gi of G such that Y = G−1
i XGi . Different pairs of elements X
and Y will, in general, require different group elements Gi . Elements connected
in this way are said to be conjugates.
We first need to establish that this does indeed define an equivalence relation,
as follows.
(i) Reflexivity: X ∼ X, since X = I −1 XI and I belongs to the group.
−1 −1
−1
(ii) Symmetry: X ∼ Y implies Y = G−1
i XGi and therefore X = (Gi ) Y Gi .
,
and
it
follows
that
Y
∼
X.
Since Gi belongs to G so does G−1
i
−1
(iii) Transitivity: X ∼ Y and Y ∼ Z imply Y = G−1
i XGi and Z = Gj Y Gj
−1 −1
−1
and therefore Z = Gj Gi XGi Gj = (Gi Gj ) X(Gi Gj ). Since Gi and Gj
belong to G so does Gi Gj , from which it follows that X ∼ Z.
These results establish conjugacy as an equivalence relation and hence show
that it divides G into classes, two elements being in the same class if, and only if,
they are conjugate.
Immediate corollaries are:
(i) If Z is in the class containing I then
−1
Z = G−1
i IGi = Gi Gi = I.
Thus, since any conjugate of I can be shown to be I, the identity must be
in a class by itself.
(ii) If X is in a class by itself then
Y = G−1
i XGi
must imply that Y = X. But
−1
X = Gi G−1
i XGi Gi
for any Gi , and so
−1
−1
−1
X = Gi (G−1
i XGi )Gi = Gi Y Gi = Gi XGi ,
i.e. XGi = Gi X for all Gi .
Thus commutation with all elements of the group is a necessary (and
sufficient) condition for any particular group element to be in a class by
itself. In an Abelian group each element is in a class by itself.
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28.7 SUBDIVIDING A GROUP
(iii) In any group G the set S of elements in classes by themselves is an Abelian
subgroup (known as the centre of G). We have shown that I belongs to S,
and so if, further, XGi = Gi X and Y Gi = Gi Y for all Gi belonging to G
then:
(a) (XY )Gi = XGi Y = Gi (XY ), i.e. the closure of S, and
(b) XGi = Gi X implies X −1 Gi = Gi X −1 , i.e. the inverse of X belongs
to S.
Hence S is a group, and clearly Abelian.
Yet again for illustration purposes, we use the six-element group that has
table 28.8 as its group table.
Find the conjugacy classes of the group G having table 28.8 as its multiplication table.
As always, I is in a class by itself, and we need consider it no further.
Consider next the results of forming X −1 AX, as X runs through the elements of G.
I −1 AI
= IA
=A
A−1 AA
= IA
=A
B −1 AB
= AI
=A
C −1 AC
= CE
=B
D−1 AD
= DC
=B
E −1 AE
= ED
=B
Only A and B are generated. It is clear that {A, B} is one of the conjugacy classes of G.
This can be verified by forming all elements X −1 BX; again only A and B appear.
We now need to pick an element not in the two classes already found. Suppose we
pick C. Just as for A, we compute X −1 CX, as X runs through the elements of G. The
calculations can be done directly using the table and give the following:
X
X −1 CX
: I
: C
A
E
B
D
C
C
D
E
E
D
Thus C, D and E belong to the same class. The group is now exhausted, and so the three
conjugacy classes are
{I},
{A, B},
{C, D, E}. In the case of this small and simple, but non-Abelian, group, only the identity
is in a class by itself (i.e. only I commutes with all other elements). It is also the
only member of the centre of the group.
Other areas from which examples of conjugacy classes can be taken include
permutations and rotations. Two permutations can only be (but are not necessarily) in the same class if their cycle specifications have the same structure.
For example, in S5 the permutations (1 3 5)(2)(4) and (2 5 3)(1)(4) could be in
the same class as each other but not in the class that contains (1 5)(2 4)(3). An
example of permutations with the same cycle structure yet in different conjugacy
classes is given in exercise 29. 10.
In the case of the continuous rotation group, rotations by the same angle θ
about any two axes labelled i and j are in the same class, because the group
contains a rotation that takes the first axis into the second. Without going into
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