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Sequel to Group Theory
We'll start with reminding the reader about the "Important Remark" near the end of Part III, a Remark concerning the direction of reading products of group elements, like, say, apq. We read such products (from that Remark onwards) from back to front. Thus (with respect to apq) first q, then p, and then a.
Cosets
We have already dealt with cosets on many occasions. What follows is a brief recap about what cosets really are, and some deepening of this important concept.
Consider the set of permutations in the following list. They form the group D_{6}. The period of each permutation has been indicated, and they have been expressed in terms of generators p, x, a, b and c (by no means a minimum set of generators, since in fact the group D_{6} requires only two).
permutation | period | cycle | |
1 | A B C P Q | 1 | (A)(B)(C)(P)(Q) |
p | C A B P Q | 3 | (CBA) |
p^{2} | B C A P Q | 3 | (BCA) |
x | A B C Q P | 2 | (PQ) |
px | C A B Q P | 6 | (ACB)(PQ) |
p^{2}x | B C A Q P | 6 | (ABC)(PQ) |
a | A C B P Q | 2 | (BC) |
b | C B A P Q | 2 | (CA) |
c | B A C P Q | 2 | (AB) |
ax | A C B Q P | 2 | (BC)(PQ) |
bx | C B A Q P | 2 | (CA)(PQ) |
cx | B A C Q P | 2 | (AB)(PQ) |
The arrangement is that the first six are the product of the disjoint cycles of the two sets {A, B, C} and {P, Q} forming the subgroup C_{6}. The last six are single or double transpositions, and all of them keep the sets {A, B, C} and {P, Q} distinct.
Consider those permutations which do not move A.
1 | A B C P Q |
x | A B C Q P |
a | A C B P Q |
ax | A C B Q P |
These clearly form the subgroup D_{2} (D_{6} does, in fact, possess three subgroups D_{2}) :
1 | x | a | ax | |
1 | 1 | x | a | ax |
x | x | 1 | ax | a |
a | a | ax | 1 | x |
ax | ax | a | x | 1 |
And we may associate with this subgroup the property that A is not moved.
p | C A B P Q |
px | C A B Q P |
c | B A C P Q |
cx | B A C Q P |
But they do form what is called a Coset of the subgroup {1, x, a, ax}.
The elements of this coset {p, px, c, cx} may be written {p, px, ap, apx}, since c = ap, or {p, xp, ap, axp} because px = xp, or {1, x, a, ax}p, where we accept the latter notation to mean that post-multiplication by p distributes p over the whole set. In fact, if we had denoted the subgroup by the single capital letter H, the coset which corresponds to those permutations which send A to the second position would be denoted Hp. This is called the Right Coset of the subgroup H by the element p.
Consider next the Left Coset of the subgroup H by the element p :
pH = p{1, x, a, ax} = {p, px, pa, pax} = {p, px, b, bx},
p | C A B P Q |
px | C A B Q P |
b | C B A P Q |
bx | C B A Q P |
a different set from Hp, and consisting of the permutations which evidently all do the job of sending C to the first position.
We may form the coset (left or right) of a given subgroup by any element, e.g. the left coset of H by b, i.e. bH = {b, bx, ba, bax} = {b, bx, p, px}, and this turns out to be the same coset as pH. But if we look at cH = {c, cx, ca, cax} = {c, cx, p^{2}, p^{2}x}, we obtain a new coset, the remaining elements of the group that were neither in H, nor in pH.
Without proof we give here some properties of cosets :
a{1, ..., b, ...}
In this left coset the elements a and ab occur.
{1, ..., b, ...}a
In this right coset the elements a and ba occur, and if a and b do not commute, ab is different from ba. And if ab does also not occur in some other form (i.e. as the product xa = ab) in the right coset, it is present in the left coset, but absent in the corresponding right coset.
The above enumerated properties of cosets will now be proved, and with it we are at the same time in a position to prove Lagrange's Theorem which reads :
If H is a subgroup of a group G, and if the order of G (i.e. its number of elements) is equal to n, and the order of H is m, then m is a factor of n.
The following proofs of the mentioned properties of cosets are (as are many other considerations in the present series of documents) taken from BUDDEN, F. The Fascination of Groups, 1978. (p. 346/7).
In the proofs we make use of the so-called cancellation law (in fact already used on many occasions) : If two elements of a group, pq and pr, are equal, i.e. when we have pq = pr, then obviously also p^{-1}pq = p^{-1}pr, which is the same as 1q = 1r, and which means that q = r. So we may cancel p from pq = pr. The same applies when have qp = rp : qpp^{-1} = rpp^{-1}, which implies q = r.
Further it is evident that any coset with respect to a subgroup H has as many elements as this subgroup : for if we form a left coset of the subgroup {1, p_{1}, p_{2}, p_{3}, . . . , p_{r-1}}, which has r elements, by the element q, we get q{1, p_{1}, p_{2}, p_{3}, . . . , p_{r-1}} = {q, qp_{1}, qp_{2}, qp_{3}, . . . , qp_{r-1}}, which also has r elements. The same applies to the corresponding right coset.
Further, we know that a subgroup must always contain the identity element (which we denote by the symbol 1).
Finally, we know that any subgroup H from which cosets are formed is itself also a coset, namely 1H ( = H1 = H).
Well, now the proofs.
Suppose H is a subgroup of the group G, and let the elements of H be {1, h_{1}, h_{2}, h_{3}, . . . , h_{m-1}} where H is of order m. Take an element x of G which is not in H, and form the left coset xH :
x, xh_{1}, xh_{2}, xh_{3}, . . . , xh_{m-1}
Now the elements of this coset must all be distinct, for if xh_{r} = xh_{s}, then the cancellation law would lead to h_{r} = h_{s} which is false since the elements of the subgroup must be distinct. Moreover, H and xH cannot possibly have any element in common. For suppose that xh_{r} = h_{s}. Then xh_{r}h_{r}^{-1} = h_{s}h_{r}^{-1}, or x = h_{s}h_{r}^{-1}. But h_{r}^{-1} is in H, and so is the product h_{s}h_{r}^{-1} since H is a subgroup. But we deliberately chose x not to be in H, so we have reached a contradiction. Therefore H and xH are fully disjoint.
Now it is possible that the union of H and xH is equal to G, i.e. to the whole group, which would mean that all the elements of G have appeared either in H or in xH. If this is the case, then the order of G would be 2m, since both H and xH each contain m elements, and Lagrange's Theorem would then be proved (i.e. the order of the subgroup H, which stands for any subgroup of the group G, is a factor of the order of the whole group (G)).
If, however, there are still elements in G which are not in H and not in xH, choose any one of these, y, and form the left coset yH :
y, yh_{1}, yh_{2}, yh_{3}, . . . , yh_{m-1}
The cancellation law again convinces us that these are all distinct, but we must still guard against appearances of elements already listed in the first two cosets H and xH. An argument similar to the one which we used to show that there was no overlapping between H and xH will be used : Suppose if possible that xh_{r} = yh_{s}. Then y = xh_{r}h_{s}^{-1}, and since h_{r}h_{s}^{-1} is an element of H, we have : y is an element of xH, which is contrary to our certain knowledge that y is not in xH. Thus xH and yH are fully disjoint. Also H and yH are fully disjoint, because when we suppose the contrary, namely that h_{r} = yh_{s}, then, again, h_{r}h_{s}^{-1} = y, and since h_{r}h_{s}^{-1} is an element of H, y would be in H, which also is contrary to our knowledge that it is not so. Thus there is no overlapping between these three cosets H, xH and yH -- they are disjoint, and contain 3m distinct elements of the group.
If by now the group is exhausted, its order is 3m, and Lagrange's Theorem is proved (i.e. the order of the subgroup H is a factor of the order of the whole group (G)).
If not, choose an element z not in H, xH, nor yH, and form the left coset zH ....
The argument proceeds as above, at each stage it being necessary to prove
(a) no repetitions in the newly formed coset, and
(b) no duplications in the new coset of any elements in the previously formed cosets.
Finally, if the group is finite, and, after, say r cosets, H, xH, yH, zH, . . . have been formed, there remain no unused elements of G anymore from which to form new cosets, we conclude that G contains exactly rm elements, so that m must be a factor of the order n of the group, and Lagrange's Theorem is proved.
The number of cosets, r ( = n / m ), is called the index of the subgroup in the group G.
Of course, when (above) we selected x, y, z, ... to form the new cosets, we had a free choice on each occasion from all the remaining elements of the group, and had we made different choices, we should have arrived at the same cosets, but in a different order. We might, for example, have selected y first, then x. In this case the coset yH would have been second (H, yH, ...), instead of third (H, xH, yH, ...) on the list. Or we might have selected the element zh_{7} with which to form our second coset. In that case, we should have had
{zh_{7}, zh_{7}h_{1}, zh_{7}h_{2}, zh_{7}h_{3}, . . . , zh_{7}h_{m-1}}, i.e.
z{h_{7}, h_{7}h_{1}, h_{7}h_{2}, h_{7}h_{3}, . . . , h_{7}h_{m-1}}.
But the set between { } now obtained is simply a rearrangement of the elements of the subgroup H, so that the left coset of H by zh_{7} is in fact zH.
As a final assurance -- not only that the cosets are distinct -- but also that every element in the group appears in some coset, this is obvious, since if g is any element whatever of G, then it most certainly appears in the coset gH, being the product of itself with the identity, which is undoubtedly in H. The above consideration points to the general result that if y is any element in xH, then xH and yH are the same set. Indeed, y must be in yH (where it is multiplied with the identity), and because, as we have seen, left cosets cannot overlap, so they must be identical. The same goes for right cosets.
The main lines of the above argument are still sound when applied to infinite groups so far as the proof of the disjoint property of the cosets is concerned. But of course either r or m or both must now be infinite, and we do not have any result for infinite groups corresponding to Lagrange's Theorem.
(BUDDEN, p. 347)
Note that the non-overlapping (disjoint) property of cosets, i.e. that xH and yH do not have any element in common, is entirely dependent upon H being a subgroup. For taking the group table for D_{6} (See below), suppose H = {1, p} (which is not a subgroup, because it is not closed : pp is not in H). Then Ha = {1, p}a = {a, pa} = {a, b}, and Hc = {1, p}c = {c, pc} = {c, a}, and we see that the element a occurs in both sets, i.e. these sets are not disjoint.
1 | p | p^{2} | a | b | c | x | y | z | u | v | w | |
1 | 1 | p | p^{2} | a | b | c | x | y | z | u | v | w |
p | p | p^{2} | 1 | b | c | a | y | z | x | v | w | u |
p^{2} | p^{2} | 1 | p | c | a | b | z | x | y | w | u | v |
a | a | c | b | 1 | p^{2} | p | u | w | v | x | z | y |
b | b | a | c | p | 1 | p^{2} | v | u | w | y | x | z |
c | c | b | a | p^{2} | p | 1 | w | v | u | z | y | x |
x | x | y | z | u | v | w | 1 | p | p^{2} | a | b | c |
y | y | z | x | v | w | u | p | p^{2} | 1 | b | c | a |
z | z | x | y | w | u | v | p^{2} | 1 | p | c | a | b |
u | u | w | v | x | z | y | a | c | b | 1 | p^{2} | p |
v | v | u | w | y | x | z | b | a | c | p | 1 | p^{2} |
w | w | v | u | z | y | x | c | b | a | p^{2} | p | 1 |
Table 14.2
On the other hand, {1, p, p^{2}} is a subgroup, which can be seen by the fact of its closure (the other group requirements are then automatically fulfilled, because it is part of an existing overall group structure). The cosets are indicated by colors, and we see that they are disjoint.
x | y | z |
y | z | x |
z | x | y |
Note that the order of elements in a set, and thus also in a coset, does not make any difference, so {x, y, z} = {y, z, x} = {z, x, y}.
This coset can be written as x{1, p, p^{2}} = y{1, p, p^{2}} = z{1, p, p^{2}}.
Cosets and Equivalence Classes
Given a subgroup H of a group G, two elements of G (not in H) may or may not lie in the same coset of H. For example, taking H to be the subgroup {1, a, b, c, d, f} of the group S_{4} (See for this group Table 13.2 of the previous document), the elements p and s are in the same right coset, whereas p and w are not. Afgain, in the same group, we have the subgroup {1, i, m} with the right coset {a, k, t} and the left coset {a, j, n}. So for example, a and t do occur in the same right coset, but are not in the same left coset.
The relationship between two elements of belonging to the same coset of a particular subgroup, is an important one. It is an equivalence relation.
An equivalece relation between two elements of a set means that the elements have something in common, by reason of which they belong to the same class. If we would classify all words according to the number of letters of which they are built, then we can say that the word "see" belongs to the Class 3 of all words. The word "house" then belongs to the Class 5. Also can we say that the words "bad", "sad" and "red" belong, with the word "see", in same Class. We say that the set of words has been partitioned into equivalence classes, membership of these classes being described by the equivalence relation defined thus :
Two words belong to the same equivalence class if they have an equal number of constituent letters. The relationship between the words is that of consisting of an equal number of letters, and we shall denote this relationship by the symbol ~, so that in the present case BAD ~ SAD, but SAD /~HOUSE (in which " / " is a negation).
These three requirements are also sufficient, and constitute a definition for an equivalence relation.
The following Theorem can be easily proved :
An Equivalence Relation partitions a set into disjoint subsets.
It can be demonstrated that, in a group, membership of the same coset (left or right) relative to a given subgroup is an equivalence relation. Further it is useful to show that we can express membership of the same coset in a slightly different manner.
Suppose a and b belong to the same left coset xH (Note that a coset is not necessarily a group or subgroup), then it is possible to find elements h_{r}, h_{s} in H such that a = xh_{r}, and b = xh_{s}. Therefore
a^{-1}b = (xh_{r})^{-1}(xh_{s}) = h_{r}^{-1}x^{-1}xh_{s} [generally (pq)^{-1} = q^{-1}p^{-1}] = h_{r}^{-1}h_{s} [becuse x^{-1}x = 1], so that, since h_{r} and h_{s} belong to the subgroup H, so does h_{r}^{-1} , and so does the product h_{r}^{-1}h_{s} . Thus a^{-1}b is an element of H.
So we have shown that if a and b belong to the same left coset, then a^{-1}b is an element of H.
And, now the other way around : This condition is also sufficient to guarantee membership of the same left coset. For a^{-1}b being an element of H implies that a^{-1}b = h, (where h is an element of H). Now we can say that the statement a^{-1}b = h is the same as the statement aa^{-1}b = ah, and this is the same as the statement b = ah, so that b is in aH. And because H is a subgroup it contains the identity element, which implies that aH contains a (because it contains a1, which is equal to a ), and we just saw that b also is in aH, thus both elements a and b are in the same left coset.
In the same way a necessary and sufficient condition for a and b to belong to the same right coset is : ab^{-1} is an element of H.
Multiplication of subsets
If A = {a_{1}, a_{2}, a_{3}, ..., a_{r}} and B = {b_{1}, b_{2}, b_{3}, ..., b_{s}} are two subsets of a group G, we shall use the notation AB to denote the set consisting of all possible products
{a_{1}b_{1}, a_{2}b_{1}, ..., a_{r}b_{1}, a_{1}b_{2}, a_{2}b_{2}, ... a_{r}b_{2}, ..., a_{r}b_{s}} the number of which is clearly rs, though some values may be repeated. We can set out these products as an array in a multiplication table :
b_{1} | b_{2} | b_{3} | ..... | b_{s} | |
a_{1} | a_{1}b_{1} | a_{1}b_{2} | a_{1}b_{3} | ..... | a_{1}b_{s} |
a_{2} | a_{2}b_{1} | a_{2}b_{2} | a_{2}b_{3} | ..... | a_{2}b_{s} |
. | ..... | . | |||
. | ..... | . | |||
. | ..... | . | |||
a_{r} | a_{r}b_{1} | a_{r}b_{2} | a_{r}b_{3} | ..... | a_{r}b_{s} |
The object of the notation AB is for the purpose of abbreviation. We shall call AB the product set of the subsets A and B.
Products of subsets from a finite group
We now take some examples (BUDDEN, pp. 357) from the group Q_{6}, whose table is shown below (In order to clearly distinguish the numeral 1 from the letter l, we indicate the latter as l* ).
1 | a | b | c | d | f | g | h | j | k | l* | m | period | |
1 | 1 | a | b | c | d | f | g | h | j | k | l* | m | 1 |
a | a | l* | 1 | d | f | g | h | j | c | m | k | b | 6 |
b | b | 1 | m | j | c | d | f | g | h | l* | a | k | 6 |
c | c | j | d | k | l* | a | 1 | b | m | g | h | f | 4 |
d | d | c | f | m | k | l* | a | 1 | b | h | j | g | 4 |
f | f | d | g | b | m | k | l* | a | 1 | j | c | h | 4 |
g | g | f | h | 1 | b | m | k | l* | a | c | d | j | 4 |
h | h | g | j | a | 1 | b | m | k | l* | d | f | c | 4 |
j | j | h | c | l* | a | 1 | b | m | k | f | g | d | 4 |
k | k | m | l* | g | h | j | c | d | f | 1 | b | a | 2 |
l* | l* | k | a | f | g | h | j | c | d | b | m | 1 | 3 |
m | m | b | k | h | j | c | d | f | g | a | 1 | l* | 3 |
Table 14.3
Let A be the subset {a, d, m, h}, then A^{2} ( = AA) is given by the elements in the table below, i.e. A^{2} is the set {l*, f, b, j, c, k, g, 1}, and we may note that, even if B were limited to the elements {a, d}, the product AB would still be the same set, the first two rows of the table containing all eight elements of A^{2}. (Also here : In order to clearly distinguish the numeral 1 from the letter l, we indicate the latter as l* ).
a | d | m | h | |
a | l* | f | b | j |
d | c | k | g | 1 |
m | b | j | l* | f |
h | g | 1 | c | k |
Again, if we take C = {a, b, k} and D = {d, g, j}, then the product of C and D contains the elements in the two tables below :
d | g | j | |
a | f | h | c |
b | c | f | h |
k | h | c | f |
a | b | k | |
d | c | f | h |
g | f | h | c |
j | h | c | f |
In this case, it seems that we get the minimum possible number of elements -- only three, and that although Q_{6} is non-Abelian (so that, in fact, none of the above pairs of elements commute), yet here we have the same set, namely {c, f, h} in both cases with CD equal to DC. Is this a fluke, or is it inevitable, we may ask?
d | f | g | |
a | f | g | h |
b | c | d | f |
c | l* | a | 1 |
a | b | c | |
d | c | f | m |
f | d | g | b |
g | f | h | 1 |
We can see that EF = {1, a, c, d, f, g, h, l*} whereas FE = {1, b, c, d, f, g, h, m}.
a | b | d | m | |
a | l* | 1 | f | b |
b | 1 | m | c | k |
k | m | l* | h | a |
a | b | k | |
a | l* | 1 | m |
b | 1 | m | l* |
d | c | f | h |
m | b | k | a |
This time, the product of the subsets is commutative, but here we get nine different elements of the possible twelve products, three of the latter having been repeated ( l*, 1, m in GH, and l*, 1, m in HG ).
Products of cosets
Some of the subsets discussed above were cosets of some of the subgroups of the group Q_{6} (e.g. A, C, D, G, H). There was one case, CD, when only three -- the minimum possible number -- different elements were present in the product set (The maximum number of different elements in this case is nine).
That the minimum number of different elements is three in this case (and not two, or one) is beause of the 'Latin Square Property' of groups, namely that each individual row of a group table contains all the elements of the group, without repetitions. The same goes for the columns. Here, however, we have to do with only fragments of the group table, so not all elements of the group are present in a row or column, but the fact of no repetitions in a row, and no repetitions in a column, still stands. So indeed in th present case (CD) the minimum number of different elements appearing as products in the table is three.
We saw that the product of a subgroup with itself has this property, i.e. that the product set contains the minimum number of different elements (this minimum number is here equal to the number of elements of that subgroup). Yet neither C, nor D is a subgroup.
The answer to the question "when will the product of two sets each of m elements contain only m distinct elements? is : "When the two sets are cosets of a normal subgroup".
In the above case, C and D were cosets of the normal subgroup {1, l*, m}.
A normal subgroup can be defined as follows :
A subgroup H of a group G is n o r m a l when, for every element g of G holds : gH = Hg, i.e. that in all cases the right coset, of H is equal to the corresponding left coset of H.
In the present case we see for example that
a{1, l*, m} = {a, k, b} (Which can all be read off from the group table of Q_{6}).
{1, l*, m}a = {a, k, b}.
etc.
Normal subgroups will be discussed more fully later on.
You will appreciate also that the product set, {c, f, h} is another of the cosets of this same subgroup. It all belongs to the theory of homomorphisms, discussed earlier, and to which we will return later on.
The notation A^{2} suggests that we may invent a meaning for A^{-1}.
If A is the set {a, b, c, d, . . . }, then we may define A^{-1} to be the set
{a^{-1}, b^{-1}, c^{-1}, d^{-1}, . . . }.
In the group Q_{6}, with the notation above we had,
A = {a, d, m, h}
E = {a, b, c}
D = {d, g, j}
Then
A^{-1} = {a^{-1}, d^{-1}, m^{-1}, h^{-1}} = {b, h, l*, d} (To find the element a^{-1}, we realize that aa^{-1} = 1, so while reading off the group table of Q_{6} (See Table 14.3, we look for the element which gives 1, when multplied with the element a ).
E^{-1} = {a^{-1}, b^{-1}, c^{-1}} = {b, a, g}.
D^{1} = {d^{-1}, g^{-1}, j^{-1}} = {h, c, f} = CD or DC.
So if we consider the SET of cosets of the normal subgroup (1, l*, m}, then, at least for the element D, there is an inverse, CD ( = DC) which is also a coset of that same normal subgroup.
Finally, if J = {1, d, h, k} (a subgroup), then, J^{-1} = {1, h, d, k} = J.
So J = J^{2} = J^{-1}.
In the case of a subgroup, it is plain that, since the inverse of every element of the subgroup must also lie in the subgroup, that no new elements can arise. In the present case we have : d^{-1} = h, h^{-1} = d, and k^{-1} = k.
As we saw already in previous documents, and will discuss further later on, the cosets of a normal subgroup themselves form a group under the product set law of composition described above. This makes possible to set up a mapping between the group to the group of cosets of one of its normal subgroups. Such a mapping generally is a mapping from a group to a smaller group, the telescopic image. Products are preserved in such a mapping, which is called a homomrphism.
In the next document we will discuss and explain the conjugacy of group elements.
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