Unit - 3

Properties of cosets

Definition of cosets

Coset of H in G-

Let G be a group and let H be a subset of G. For any a G, the set {ah | h H} is denoted by aH. Analogously, Ha = {ha | h H} and aH = {ah | h H}. When H is a subgroup of G, the set aH is called the left coset of H in G containing a, whereas Ha is called the right coset of H in G containing a. In this case, the element a is called the coset representative of aH (or Ha). We use |aH| to denote the number of elements in the set aH, and |Ha| to denote the number of elements in Ha.

Example: Let H 5 {0, 3, 6} in Z9 under addition. In the case that the group operation is addition, we use the notation a + H instead of aH. Then the cosets of H in Z9 are

Properties of Cosets

Let H be a subgroup of G, and let a and b belong to G. Then,

1. a aH,

2. AH = H if and only if a H,

3. AH = bH if and only if a bH

4. AH = bH or aH bH = ,

5. AH = bH if and only if b [ H,

6. |aH| = |bH|,

7. AH = Ha if and only if H = aH1,

8. AH is a subgroup of G if and only if a H.

Proof of the above properties

1. a = ae aH.

2. To verify property 2, we first suppose that aH = H. Then a = ae aH = H. Next, we assume that a H and show that aH H and H aH. The first inclusion follows directly from the closure of H. To show that HaH, let h H. Then, since aH and hH, we know that h H. Thus, h = eh = (a)h = a(h) aH.

3. If aH = bH, then a = ae aH = bH. Conversely, if a bH we have a = bh where h H, and therefore aH = (bh)H = b(hH) = bH.

4. Property 4 follows directly from property 3, for if there is an element c in aH y bH, then cH = aH and cH = bH.

5. Observe that aH = bH if and only if H = bH. The result now follows from property 2.

6. To prove that |aH| = |bH|, it suffices to define a one-to-one mapping from aH onto bH. Obviously, the correspondence ah → bh maps aH onto bH. That it is one-to-one follows directly from the cancellation property.

7. Note that aH = Ha if and only if (aH)1 = (Ha) = H(a) =

H—that is, if and only if aH = H.

8. If aH is a subgroup, then it contains the identity e. Thus, aH eH 2 0/; and, by property 4, we have aH = eH = H. Thus, from property 2, we have a H. Conversely, if a H, then, again by property 2, aH = H.

Note-

- Property-1 means that the left coset of H containing a does contain a.
- Property-2 means that the H “absorbs” an element if and only if the element belongs to H.
- Property-3 means that a left coset of H is uniquely determined by any one of its elements. In particular, any element of a left coset can be used to represent the coset.
- Property-4 means that two left cosets of H are either identical or disjoint.
- Property 5 shows how we may transfer a question about equality of left cosets of H to a question about H itself and vice versa.
- Property-6 means that all left cosets of H have the same size.
- Property 7 is analogous to property 5 in that it shows how a question about the equality of the left and right cosets of H containing a is equivalent to a question about the equality of two subgroups of G. The last property of the lemma says that H itself is the only coset of H that is a subgroup of G.

Definition of subgroup:

Let (G, .) be the group and let H be the subset of the group G.

H is called the subgroup of G, written H < G, if H is a group relative to the binary operation in G.

For any group G the singleton (e) and G itself are subgroup of G, called trivial subgroups.

A subgroup H of G is said to be proper subgroup if H is not equal to e and H and G are unequal.

It is seen that the identity element of a subgroup of a group must be same as that of the group.

Theorem: Let G be a group. A be non empty subset H of G is a subgroup of G iff either of the following rules holds

- For all a, b , ab and inverse of a
- For all a, b,

Proof:

If H is a subgroup the both the conditions above are true. Conversely suppose H satisfies the condition first then for any , .

Hence, e = . Therefore, H is a subgroup.

Now suppose that H follows the second rule.

Let . Then .

Hence .

Therefore , which proves that H is a subgroup of G.

Theorem: let be a homomorphism of groups, then ker is a subgroup of G and Im is a subgroup of H.

Proof:

Ker and Im both are non empty. Let then where e’ is the identity of G.

Hence . This proves that ker is a subgroup of G.

Now let then for some x, y.

Hence , which proves that Im is the subgroup of H.

Definition of centre of a group:

The centre of a group G is the set of those elements in G that commute with every element in G, that is

Theorem: the centre of a group G is a subgroup of G.

Proof:

Since ex= x = xe for all . Let a, b. Then for all

Hence, . Therefore, Z(G) is the subgroup of G.

Theorem: let H and K be subgroups of a group (G, .), then HK is a subgroup of G iff HK = KH.

Proof:

Let HK = KH. Since e = ee , HK is not empty.

Let then for some . Hence

Where .

Now hence for some , therefore

Where hence .

This proves that HK is a subgroup.

Conversely, suppose that HK is a subgroup, let , so that a = kh for some . Then . Hence .

Therefore, KH is subgroup of HK.

Now let . Then . Hence for some . Therefore . Hence HK

Theorem: Let G be a group and

- If for some integer then o(a)|n
- If o(a) = m, then for all integers I, , where r(i) is the remainder of I modulo m.
- [a] is of order m iff o(a) = m

Proof:

Part-1: if then , hence for some i > 0. Therefore by the well ordering property of N, there is a least +ve integer m = 0(a) such that .

By using division algorithm, n = mq + r, .

Hence . Therefore r = 0 and m|n.

Part-2: again using division algorithm, for any , i = mq + r, . Hence , where r = r(i) is the remainder of i modulo m.

Part-3: Let o(a) = m then are distinct, for otherwise for some I, j, . Hence, , a contradiction.

Let H = [a] be the cyclic subgroup generated by a.

For any , . This implies that H has exactly m elements .

Conversely, suppose H is of finite order. Then are not distinct for all

Hence for some . Then . Hence a is of finite order, say m but then H has exactly m elements as proved earlier.

Left coset and right coset:

Let H be a subgroup of G. Given , the set

Is called the left coset of H determined by a.

A subset C of G is called a left coset of H in G if C = aH for some a in G.

The set of all cosets of H in G can be written as G/H.

A right coset Ha us defined similarly.

The set of all right cosets of H in G is written as H.

Definition: Let H be a subgroup of G. The cardinal number of the set of left and right cosets of H in G is called the index of H in G and it is denoted by [G:H].

Note- Let G be a finite group and let H be any subgroup of G. Let |G| = n and |H| = m then every left coset of H has m elements.

Since the distinct left coset of H are pairwise disjoint and their union is G, we must have n = km, where k is the number of left cosets of H in G.

Key takeaways

- Let (G, .) be the group and let H be the subset of the group G. H is called the subgroup of G, written H < G, if H is a group relative to the binary operation in G.
- Let be a homomorphism of groups, then ker is a subgroup of G and Im is a subgroup of H.
- The centre of a group G is the set of those elements in G that commute with every element in G, that is

4. The centre of a group G is a subgroup of G.

5. Let H be a subgroup of G. Given , the set

Is called the left coset of H determined by a.

This theorem first stated by Lagrange in 1770 and the complete proof was given by Pietro Abbati.

Theorem- If G is a finite group and H is a subgroup of G, then |H| divides |G|. Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|.

Proof:

Let H, H, . . . , H denote the distinct left cosets of H in G. Then, for each a in G, we have aH = H for some i. Also, by property 1 of the lemma, a aH. Thus, each member of G belongs to one of the cosets aiH. In symbols,

G = H . . . . H.

Now, property 4 of the lemma shows that this union is disjoint, so that

|G| = |H| + |H| + ……+ |H|.

Finally, since |H| = |H| for each i, we have |G| = r|H|.

Note- The converse of Lagrange’s Theorem is false. For example, a group of order 12 need not have a subgroup of order 6.

The index of a subgroup H in G is the number of distinct left cosets of H in G. This number is denoted by |G:H|. As an immediate consequence of the proof of Lagrange’s Theorem, we have the following useful formula for the number of distinct left (or right) cosets of H in G.

Corollary- If G is a finite group and H is a subgroup of G, then |G:H| 5 |G|/|H|.

Corollary- In a finite group, the order of each element of the group divides the order of the group.

Corollary- A group of prime order is cyclic.

Corollary- Let G be a finite group, and let a G. Then, = e.

Fermat’s Little Theorem- For every integer a and every prime p, mod p = a mod p.

Proof:

By the division algorithm, a = pm + r, where 0 r , p. Thus, a mod p = r, and it suffices to prove that mod p = r. If r = 0, the result is trivial, so we may assume that r U(p). [Recall that U(p) = {1, 2, . . . , p 2 1} under multiplication modulo p.] Then, by the above corollary, mod p = 1 and, therefore, mod p 5 r.

Note-

We use Fermat’s Little Theorem in conjunction with computers to test for primality of certain numbers. One case concerned the number p = . If p is prime, then we know from Fermat’s Little Theorem that mod p = 10 mod p and, therefore, mod p 5 100 mod p. Using multiple precision and a simple loop, a computer was able to calculate mod p = mod p in a few seconds. The result was not 100, and so p is not a prime.

Theorem- An application of cosets to permutation groups

Stabilizer of a Point

Let G be a group of permutations of a set S. For each i in S, let stablizerG(i) = { G | (i) = i}. We call stablizerG(i) the stabilizer of i in G.

Orbit of a Point

Let G be a group of permutations of a set S. For each s in S, let orbG(s) = { (s) | G}. The set orbitG(s) is a subset of S called the orbit of s under G. We use |orbG(s)| to denote the number of elements in orbG(s).

Example: Let G 5 {(1), (132)(465)(78), (132)(465), (123)(456), (123)(456)(78), (78)}.

Then

OrbG(1) 5 {1, 3, 2}, stabG(1) 5 {(1), (78)},

OrbG(2) 5 {2, 1, 3}, stabG(2) 5 {(1), (78)},

OrbG(4) 5 {4, 6, 5}, stabG(4) 5 {(1), (78)},

OrbG(7) 5 {7, 8}, stabG(7) 5 {(1), (132)(465), (123)(456)}.

Key takeaways

- If G is a finite group and H is a subgroup of G, then |H| divides |G|. Moreover, the number of distinct left (right) cosets of H in G is |G|/|H|.
- The converse of Lagrange’s Theorem is false. For example, a group of order 12 need not have a subgroup of order 6.
- If G is a finite group and H is a subgroup of G, then |G:H| 5 |G|/|H|.
- A group of prime order is cyclic.

External Direct Product

Let G1, G2, . . . , Gn be a finite collection of groups. The external direct product of G1, G2, . . . , Gn, written as G1 G2 … Gn, is the set of all n-tuples for which the ith component is an element of Gi and the operation is componentwise.

In symbols,

Where (g1, g2, . . . , gn)(g1’, g2’, . . . , gn’) is defined to be (g1g1’, g2g2’, . . . , gngn’). It is understood that each product gigi ‘ is performed with the operation of Gi.

Example:

U(8) U(10) = {(1, 1), (1, 3), (1, 7), (1, 9), (3, 1), (3, 3), (3, 7), (3, 9), (5, 1), (5, 3), (5, 7), (5, 9), (7, 1),(7, 3), (7, 7), (7, 9)}.

The product (3, 7)(7, 9) 5 (5, 3), since the first components are combined by multiplication modulo 8, whereas the second components are combined by multiplication modulo 10.

Example: Classification of Groups of Order 4

A group of order 4 is isomorphic to Z4 or Z2 Z2. To verify this, let G = {e, a, b, ab}. If G is not cyclic, then it follows from Lagrange’s Theorem that |a| = |b| = |ab| = 2. Then the mapping e (0, 0), a (1, 0), b (0, 1), and ab (1, 1) is an isomorphism from G onto Z2 Z2.

Theorem- Order of an Element in a Direct Product

The order of an element in a direct product of a finite number o finite groups is the least common multiple of the orders of the components of the element.

Symbolically

Proof:

Denote the identity of Gi by ei. Let S =

And

t =

Because s is a multiple of each |gi| implies that

We know that ts. On the other hand, from

We see that t is a common multiple of . Thus .

Theorem- Criterion for GH to be Cyclic

Let G and H be finite cyclic groups. Then G H is cyclic if and only if |G| and |H| are relatively prime.

Proof:

Let |G| = m and |H| = n, so that |G H| 5 mn. To prove the first half of the theorem, we assume G H is cyclic and show that m and n are relatively prime. Suppose that gcd(m, n) = d and (g, h) is a generator of G H. Since = = (e, e), we have mn = |(g, h)| mn/d. Thus, d = 1. To prove the other half of the theorem, let G = <g> and H = <h> and suppose gcd(m, n) = 1. Then, |(g, h)| = lcm(m, n) = mn = |G H|, so that (g, h) is a generator of G H.

Corollary-1: Criterion for … to Be Cyclic

An external direct product … of a finite numberof finite cyclic groups is cyclic if and only if || and || are relatively prime when I and j are unequal.

Key takeaways

- Let G1, G2, . . . , Gn be a finite collection of groups. The external direct product of G1, G2, . . . , Gn, written as G1 G2 … Gn, is the set of all n-tuples for which the ith component is an element of Gi and the operation is component wise.
- Let G and H be finite cyclic groups. Then G H is cyclic if and only if |G| and |H| are relatively prime.

Definition of normal subgroup-

A subgroup H of a group G is called a normal subgroup of G if aH = Ha for all a in G. We denote this by H G.

In other words we can define it as-

Let G be a group. A subgroup N of G is called a normal subgroup of G, written as N G, if for every .

Note- If G is abelian then every subgroup of G is a normal subgroup, but the converse is not same.

A group in which every subgroup is normal is not necessarily abelian.

Note- “H is normal in G” means ah = ha for a G and h H. This is not what normality of H means; rather, it means that if a G and h H, then there exist elements h’ and h’’ in H such that ah = h’a and ha = ah’’.

Theorem- (normal subgroup test)- A subgroup H of G is normal in G if and only if H for all x in G.

Proof:

If H is normal in G, then for any x [ G and h [ H there is an h’ in H such that xh = h’x. Thus, = h’, and therefore H.

Conversely, if H for all x, then, letting x = a, we have H or aH Ha. On the other hand, letting x = , we have = H or Ha aH.

Note- Let N be a subgroup of a group G, then the following are equivalent-

- N G
- for every
- XN = Nx for every
- (xN)(yN) = xyN for all x, y

Theorem: let N be a normal subgroup of the group G. Then G/N is a group under multiplication. The mapping , given by , is a surjective homomorphism and

Proof:

By the properties above (xN)(Yn) = (xy)N for all . Hence, G/N is closed under multiplication, because multiplication is associative in G, multiplication is also associative in G/N.

The coset eN = N is the identity for multiplication in G/N, and for any ,

This proves that G/N is a group.

The mapping is a homomorphism. Further, xN = eN iff .

Hence ker .

Theorem: Let G be a group. For any non-empty subset S of G, N(S) is a subgroup of G, further, for any subgroup H of G,

- N(H) is the largest subgroup of G in which H is normal
- If K is a subgroup of N(H), then H is a normal subgroup of KH.

Proof:

Clearly if , if x, y, then

Hence , therefore, N(S) is a subgroup of G.

Let H is a subgroup of G, then for all , therefore H is a subset, hence a subgroup of N(H).

Further by the definition for all.

Hence H N(H).

Let K be any subgroup of G such that H .

Then for all .

Hence K is subset of N(H). This proves that N(H) is the largest subgroup of G containing H as normal subgroup.

Let K be a subgroup of N(H), then for all hence kH = Hk, therefore KH = HK.

Hence by the theorem KH is a subgroup of N(H), and H, consequently H HK.

Theorem: Let G be a group, and let G’ be the derived group of G.

Then

- G’ G.
- G/G’ is abelian
- If H G, then G/H is abelian iff G’.

Proof:

- Suppose be any commutator in G. Then is also a commutator, moreover for any g in G,

Now any element y in G’ is a product of a finite number of commutators,

Say

Where are commutators. Then for any

Hence, G’ is a normal subgroup of G.

2. For all

Hence (aG’)(bG’) = (bG’)(aG’). Therefore G/G’ is abelian.

3. Suppose G/H is abelian. Then for all

Hence, . This proves that .

We can prove the converse similarly.

Example: If G is a group and H is a subgroup of index 2 in G, then H is a normal subgroup of G.

Sol:

If a does not belongs to H then by the hypothesis G = H and aH.

Also G = H. Thus aH = Ha, .

But clearly aH = Ha, for every , hence for every , proving that H is normal in G.

Example: If A < G and B G, then A and AB < G.

Sol:

Of course A in order to prove the normality for this, suppose . Then , because B G, trivially, . Thus

To prove AB < G, let a, , b, , then ab(, because for some . Thus AB < G.

Example: If N and M are normal subgroups of G such that N , then nm = mn, .

Sol:

If , then

And also

Thus,

Hence, nm = mn.

Definition of quotient group: Let N be a normal subgroup of G. The group G/N is called the quotient group of G by N.

The homomorphism given by is called the canonical or natural homomorphism of G onto G/N.

Definition of normalizer: let G be a group and S be a non-empty subset of G, the normalize of S in G is the set

The normalizer of a singleton {a} is written as N(a).

Definition of commutator subgroup:

Let G be a group, for any , is called a commutator in G. The subgroup of G generated by the set of all commutators in G is called the commutator subgroup of G. It is denoted by G’.

Example: Every subgroup of an Abelian group is normal. (In this case, ah = ha for a in the group and h in the subgroup.)

Example: The center Z(G) of a group is always normal. [Again, ah = ha for any a G and any h Z(G).

Factor group:

When the subgroup H of G is normal, then the set of left (or right) cosets of H in G is itself a group—called the factor group of G by H (or the quotient group of G by H).

Theorem: Let G be a group and let H be a normal subgroup of G. The set G/H = {aH | a G} is a group under the operation (aH)(bH) = abH

Proof:

First we need to show that the operation is well defined; that is, we must show that the correspondence defined above from G/H G/H into G/H is actually a function. To do this we assume that for some elements a, a’, b, and b’ from G, we have aH = a’H and bH = b’H and verify that aHbH = a9Hb’H. That is, verify that abh = a’b’H. (This shows that the definition of multiplication depends only on the cosets and not on the coset representatives.) From aH = a’H and bH = b’H , we have a’ = ah1 and b’ = bh2 for some h1, h2 in H, and therefore

a’b’H = ah1bh2H = ah1bH = ah1Hb = aHb = abH. Here we have made multiple use of associativity, and the fact that H G. EH = H is the identity; H is the inverse of aH; and (aHbH)cH = (ab)HcH = (ab)cH = a(bc)H = aH(bc)H = aH(bHcH). This proves that G/H is a group.

Theorem: (Cauchy’s Theorem for Abelian Groups) - Let G be a finite Abelian group and let p be a prime that divides the order of G. Then G has an element of order p.

Proof:

The statement is true for the case in which G has order 2. We prove the theorem by using the Second Principle of Mathematical Induction on |G|. That is, we assume that the statement is true for all Abelian groups with fewer elements than G and use this assumption to show that the statement is true for G as well. Certainly, G has elements of prime order, for if |x| = m and m = qn, where q is prime, then |; | = q. So let x be an element of G of some prime order q, say. If q = p, we are finished; so assume that q in not equals p. Since every subgroup of an Abelian group is normal, we may construct the factor group = G/<x>. Then is Abelian and p divides ||, since | | = |G|/q. By induction, then, has an element—call it y<x> —of order p.

Definition Internal Direct Product of Hand K

We say that G is the internal direct product of H and K and write G = H K if H and K are normal subgroups of G and

G = HK and H K = {e}.

Note- We call G the internal direct product of H and K if H and K are subgroups of G, and if G is naturally isomorphic to the external direct product of H and K.

Key takeaways

- A subgroup H of a group G is called a normal subgroup of G if aH = Ha for all a in G. We denote this by H G.
- If G is abelian then every subgroup of G is a normal subgroup, but the converse is not same.
- A group in which every subgroup is normal is not necessarily abelian.
- Let N be a normal subgroup of G. The group G/N is called the quotient group of G by N.
- The normalizer of a singleton {a} is written as N(a).

References:

1. Contemporary abstract algebra, Joseph A. Gallian

2. Schaum’s abstract algebra

3. Basic abstract algebra by P.B. Bhattacharay, SK jain, S.R. Nagpaul

4. Modern abstract algebra by Shanti narayan, S.Chand & Co.