Unit – 3

Properties of cosets

Q1) What are the cosets?

A1)

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.

Q2) Give the properties of cosets.

A2)

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.

Q3) 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

1. For all a, b , ab and inverse of a
2. For all a, b,

A3)

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.

Q4) let be a homomorphism of groups, then prove that ker is a subgroup of G and Im is a subgroup of H.

A4)

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.

Q5) Let H and K be subgroups of a group (G, .), then HK is a subgroup of G iff HK = KH.

A5)

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

Q6) Let G be a group and

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

A6)

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.

Q7) Define left and right coset.

A7)

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.

Q8) State and prove Lagrange’s theorem.

A8)

Statement- 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.

Q9) For every integer a and every prime p, mod p = a mod p.

A9)

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.

Q10) Explain orbit of a point with the help of example.

A10)

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)}.

Q11) What do you understand by the external direct product?

A11)

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)}.

Q12) What is an order of an element in a Direct Product.

A12)

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 .

Q13) Define normal subgroup.

A13)

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.

Q14) 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

A14)

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 .

Q15) 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,

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

A15)

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.

Q16) Let G be a group, and let G’ be the derived group of G.

Then

1. G’ G.
2. G/G’ is abelian
3. If H G, then G/H is abelian iff G’.

A16)

1. 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.

Q17) 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

A17)

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.

Q18) State and prove the Cauchy’s Theorem for Abelian Groups

A18)

Statement- 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.