Sub-groups

Sub-groups

Theorem 1:

The necessary and sufficient condition that a non-empty subset H of a group G be a subgroup is a Є H, b Є H ⇒ a*b^-1€ H.        [A.U N/D 2012]

Proof :

Necessary condition :

Assume that H is a subgroup of G.

Since H itself is a group.

We have a, b Є H ⇒ a * b Є H (closure)

Also b Є H ⇒ b^-1 Є Н  (inverse)

a, b Є H ⇒ a,b^-1 Є H ⇒ a * b^-1 Є H

Sufficient condition :

Let a * b^-1 Є H, for all a, b Є H …. (1)

and H is a subset of G.

We have to prove: H is a subgroup of G.

(i) Closure: Let b Є H ⇒ b^-1 Є H

For a, b Є H ⇒ a, b^-1 Є H

⇒ a * (b^-1)^-1 Є H

⇒ a * b Є H

H is closed under the operation "*"

(ii) Associative: Since H ≤ G, the elements of H are also the elements of G.

Since * is associative in G, it must also be associative in H.

(iii) Identity: Let a Є H, ⇒ a*a^-1 Є H

⇒ e Є H

e is the identity element of H.

(iv) Existence of inverse: Let e Є H, a Є H

⇒ e * a^-1 Є H

⇒ a^-1 Є H

Every element of H has an inverse in H.

H itself is a group under the operation * in G.

Theorem 2:

Let (G, *) be a finite group, and H is non-empty subset of G and H is closed under *. Then H is a subgroup of G.

Proof: (G, *) is a finite group and H is a subset of G which is closed under *.

i.e., a, b Є H ⇒ a *b Є H.

Let O (G) = n

Now a, a Є H

Then

a * a = a² Є H

a², a Є H. Then a² * a = a³ Є H and so on.

Since G is finite there exists a 'm' with 1 ≤ m ≤ n such that

a^m = e Є H

That is e Є H

Hence identity exists.

Let a Є H, then a^m-1 Є H.

i.e., a^m-1 = a^m * a^-1 Є H

i.e., e * a^-1 Є H

i.e., a^-1 Є H.

⇒ inverse exists.

Since every element of H is G, associative property is true in H. Hence (H, *) is a group and so H is a subgroup of G.

Theorem 3.

The kernal of a homomorphism g from a group <G, *> to <H, ∆> is a subgroup of <G, *>.

Proof: Since g (e_G) = е_H, e_G Є ker (g)

Also, if a, b Є ker (g),

i.e., g (a) = g(b) = e_H, then

g (a*b) = g(a) ∆ g(b) = e_H ∆ e_H = е_H

so that a*b Є ker (g).

Finally, if a Є ker (g), then g (a^-1) = [g (a)^-1] ̄1 = е_H^-1 = e_H.

Hence a^-1 Є ker (g) and ker (g) is a subgroup of <G, *>.

Theorem 4.

Every cyclic group is abelian.     [A.U. M/J 2013, N/D 2013]

Solution: Let (G, *) be a cyclic group generated by an element a Є G.

(i.e.,) G = (a)

Then for any two elements x, y Є G

We have x = aⁿ, y = a^m, where m, n are integer.

Therefore   x*y = aⁿ * a^m = a^n+m

= a^m+n = a^m * aⁿ

= y * x

Thus, (G, *) is abelian.

Problems based on sub group

Example 1. Is the union of two subgroups of a group, a subgroup of G? Justify your answer.

Solution: The union of two subgroups of a group need not be a subgroup of G.

Let the group (Z, +)

Let H = 3Z = {0, ±3, ±6, ...}

Let K = 2Z = {0, ±2, ±4, ...}

⇒ H and K are subgroups of (Z, +).

⇒ 3 Є 3Z Є 3Z U 2Z = HUK

⇒ 2 Є 2Z Є 2Z U 3Z = HUK

But    3+2 = 5 € 2Z U 3Z

HUK is not a subgroup of (Z, +)

Example 2. The identity element of a subgroup is same as that of the group.    [A.U N/D 2012]

Solution: Let H be the subgroup of the group G and e and e' be the identity elements of G and H respectively.

Now if a Є H, then a Є G and ae = a, because e is the identity element of G.

Again a Є H, then ae' = a since e' is the identity element of H.

Thus ae = ae' which gives e = e'

Example 3. If H and K are subgroup of G, prove that HUK is a subgroup of G if and only if either H ≤ K or K ≤ H. [A.U N/D 2014]

Solution: Given H and K are two subgroups of G and H≤K or K≤H.

If H≤K then H≤K = K which is a subgroup of G.

If K≤H then H≤K = H which is a subgroup of G.

Conversely suppose K¢ H and H ¢ K.҂

Then there exists a Є H and a ҂ K and there exists a b Є K and b ҂ H.

Now a, b Є HUK. Because HUK is a subgroup, it follows that a * b Є H U K. Hence a * b Є H or a * b Є K.

Case (i): If a * b Є H

Then a^-1 *(a*b) Є H

That is b Є H which is a contradiction.

Case (ii) If a * b Є K

Then a * b * b^-1 Є K

i.e., a Є K which is a contradiction.

Thus either H≤K or K≤H

Example 4. Prove that the intersection of two subgroups of a group is a subgroup of G.         [A.U M/J 2013, N/D 2013, N/D 2014]

Solution: Given H and K are subgroups of G.

Let a, b Є H∩K ⇒ a, b Є H and a, b Є K

⇒ a*b^-1 Є H and a*b^-1 Є K (as H and K are subgroups)

⇒ a*b^-1 Є H∩K.

Thus H∩K is a subgroup of G.

Example 5. Show that the set of all elements a of a group (G, *) such that a*x = x*a for every x Є G is a subgroup of G.    [A.U N/D 2010]

Solution : Let H = {a Є G | ax = xa, xEG}

As ey = ye = y, ∀ y Є G, e Є G, H is non empty.

Let x and z in H

Then  xy = yx and zy = yz for all y Є G

(xz) y = x (yz) ⇒ (yx) z = y (xz), ∀ y ЄG

x z Є H, ∀ x, z Є H

X Є H ↔ xy = yx,    ∀ y Є G

↔ x^-1 (xy) x^-1 = x^-1 (yx)x^-1, ∀ y Є G

↔ (x^-1 x) (y x^-1) = (x^-1 y) (x x^-1)

↔ yx^-1 = x^-1y

↔ x-1 Є H

H is a subgroup.

Example 6. If 'a' is a generator of a cyclic group G, then show that 'a^-1 is also a generator of G.      [A.U M/J 2012]

Solution: Let G = (a) be a cyclic generated by 'a'

If x Є G, then x = aⁿ for some n Є Z

x = aⁿ = (a^-1)^-n, (- n Є Z)

a^-1 is also a generator of G.

Example 7. Find all the subgroups of (Z₉, +₉)       [A.U M/J 2014]

Solution: Z₉ = {0, 1, 2, 3, 4, 5, 6, 7, 8}

The operation is addition modulo 9.

Consider the subsets

H₁ = {0, 2, 4, 6, 8}

H₂ = {0, 3, 6}

H₃ = {0, 4, 8}

H₄ = {0, 5}

The improper subgroups of (Z₉, +₉) are [{0}, +₉] and [Z₉, +₉]

The operation tables shows that

H₁, H₂, H₃ and H₄ are closed for +₉

The possible proper subgroups of (Z₉, +₉) are (H₁, +₉),

(H₂, +₉), (H₃, +₉) and (H₄, +₉)

Example 8. Any cyclic group of order n is isomorphic to the additive group of residue classes of integers modulo n.

Proof :

Let G {a, a²,...,aⁿ = e} be a cyclic group of order n generated by a.

We know that (Z_n, +_n) is the additive group of residue classes modulo n.

⇒ Z_n = {[1], [2], ..., [n] = [0]}

Let f: G→ Z_n defined by f (a^r) = [r] for all a^r Є G.

For all [r] Є Z_n, there exists a a^r Є G such that ƒ (a^r) = [r]

⇒ f is onto.

For r ≠ s, [r] ≠ [s] and hence ƒ (a^r) ≠ f (a^s)

⇒ f is one-to-one.

For all a^r, a^s Є G, ƒ (a^r .a^s) = ƒ (a^r+s) = [r+s] = [r] + [s]

= f (a^r) + _n ƒ (a^s)

⇒ f is a homomorphism. Hence (G, .) is isomorphic to (Z_n, +_n)

Example 9. Prove that every finite group of order "n" is isomorphic to a permutation group of degree n.           [A.U M/J 2013]

(OR)

State and prove Cayley's theorem on permutation groups.       [MCA, Nov, 93, May 95]

Proof: Let G be the given group and A (G) be the group of all permutations of the set G.

For any a Є G, define a map f: G→G such that f(x) = ax.

f_a is well defined :

Let x = y ⇒ ax = ay ⇒ f_a (x) = f_a (y). Thus f_a is well defined.

f_a is 1-1:

Again f_a (x) = f_a (y) ax ⇒ x = y. Thus f_a is 1-1.

f_a is onto: For any y Є G, f_a (a^-1y) = a a^-1 y

= y Є G

Thus we find a preimage a^-1 y for any "y" in G. Thus f_a is onto.

Hence f_a is permutation. (i.e.,) f_a Є A (G).

Let K be the set of all such permutations. We can show that K is a subgroup of A (G). Since e Є G, f_a Є K. Thus K is non-empty.

Let f_a, f_b Є K.

Then (f_a o f_a^-1)(x) = f_a (a^-1 x)

= aa^-1 x

= ex

= ƒ_e (x)

Thus the inverse of f_a is f_a^-1

(f_a o f_b) (x) = f_a (f_b (x))

= f_b (bx)

= abx

= f_ab (x)

f_a o f_b = f_ab Є K.

Thus K is a subgroup of A (G).

Next we will show that G is isomorphic to K.

Define a map: x : G → K such that x (a) = f_a

X is well defined :

For a, b Є G, a = b ↔ ax = bx

↔ f_a (x) = f_b (x)

↔ f_a = f_b

↔ x (a) = x (b)

x is one-one and onto.

x is a homomorphism :

x (ab) = f_ab = f_a o f_b = x (a). x (b)

Thus is a homomorphism and hence an isomorphism which proves the theorem.

Example 10. Show that every cyclic group of order n is isomorphic to (Z_n +_n)

Solution: Let (G, o) be a cyclic group of order n.

The element of G are {a, a², a³, ..., aⁿ = e}.

The elements of Z_n are {[0], [1], [2], ..., [n-1]}.

Define

f: D → Z_n by

f(e) = [0] and f (aⁱ) = [i] for i < n where f is one-one and onto.

Then ƒ (aⁱ a^j) = f (a^i+j) = [i+j]

= [i] + _n[j]

= f (aⁱ) + _nf (a^j)

Hence ƒ is an isomorphism.

Example 11. Define : Symmetric group, Dihedral group. Show that if (G, *) is a cyclic group, then every sub group of (G, *) must be cyclic.

(OR)                    [MCA, May 93, M.U]

Show that every subgroup of a cyclic group is cyclic.

Solution: Let (G, *) be a cyclic group generated by "a", and let H be a subgroup of G. If H contains the identity element alone, then trivially H is cyclic and H = (e). Suppose that H ≠ (e). Since H≤G, any element of H is of the form a^k for some integer K. Let

"m" be the smallest positive integer such that a^m Є H. We shall show that H is a cyclic group generated by a^m. If a^k Є H, then by division algorithm we can write

K = qm +r, where 0 ≤r<m.

Hence a^k = a^qm+r = a^qm, a^r = (a^m)^q. a^r and from this it follows that a^r = (a^m)^-q. a^k. Since a^m, a^k Є H, we have a^r Є H which is a contradiction that "m" is so small such that a^m Є H. Hence we have r = 0.

(i.e.,) K = qm ⇒ a^K = e^qm = (a^m)^q

Thus every element of H is a power of a^m. Hence H is a cyclic group generated by a^m.

Example 12. Show that U₉ is a cyclic group. What are all its generators?

Solution: We have U₉ = {1, 2, 4, 5, 7, 8}

It is easy to verify that U₉ = <2>, since

2¹ = 2.2² = 4, 2³ = 8, 2⁴ = 7, 2⁵ = 5, 2⁶ = 1

Thus U₉ is a cyclic group. We have o (U₉) = 6.

The positive integers less than 6 and prime to 6 are 1, 5.

Hence all the generators of U₉ are 2 and 5,

since 2¹ = 2 and 2⁵ = 5.

Example 13. Show that U₈ is not a cyclic group.

Solution: We have U₈ = {1, 3, 5, 7}

Here 3² = 1, 3³ = 3, 3⁴ = 1, 3⁵ = 3, etc.

Thus 3 cannot be a generator of U₈.

Now 5² = 1, 5³ = 5, 5⁴ = 1, 5⁵ = 5, etc.

Thus 5 cannot be a generator of U₈.

Again 7² = 1, 7³ = 7, 7⁴ = 1, 7⁵ = 7, etc.

Thus 7 cannot be a generator of U₈.

Clearly, 1 is also not a generator of U₈.

Hence U₈ is not a cyclic group.

Example 14. Show that U₁₇ is a cyclic group. What are all its generators?

Solution: U₁₇ = {1, 2, …, 16} and o (U₁₇) = 16.

All the positive integers less than 16 and prime to 16 are 1, 3, 5, 7, 9, 11, 13, 15.

It can be verified that U₁₇ = <3> i.e., every element of U₁₇ is a power of 3. All the generators of U17 are

3¹ = 3, 3³ = 10, 3⁵ = 5, 3⁷ = 11, 3⁹ = 14, 3¹¹ = 7, 3¹³ = 12, 3¹⁵ = 6.

Hence all the generators of U₁₇ are

3, 5, 6, 7, 10, 11, 12, 14

Example 15. Show that group homomorphism preserves, identity, inverse and subgroup.   [A.U. M/J 2006]

Solution :

(i) Group homomorphism preserves identity.

Let g: (G, *) → (H, ∆) be a group homomorphism.

As e_G * e_G = e_G, we have

g(e_G) = g(e_G * e_G)

= g (e_G) ∆ g (e_G)

⇒ g(e_G) is an idempotent element in the group (H, ∆).

But only idempotent of a group is the identity element, g (e_G) = e_H

g (e_G) = e_H

(ii) Group homomorphism preserves inverse

Since a * a^-1 = e_G = a^-1*a we have

g(a*a ^-1) = g(e_G) = g(a^-1*a)

⇒ g(a) ∆ (a^-1) = е_H = g(a^-1) ∆ g(a)

⇒ g (a^-1) is the inverse of g (a)

g (a^-1) = [g (a)]^-1

(iii) Group homomorphism

Let S be a subgroup of (G, *)

To show that g (S) = {x Є H/x = g(a) for some a Є G}

is a subgroup of (H, ∆)

(i) As e_G Є S, g (e_G) еH Є g(s)

(ii) For each x Єg (s), Ǝ      a Є s such that g (a) = x

Since s is a sub group of G,

for each a Є s, a^-1 Є s

g (a^-1) = [g (a)]^-1 Є g(s)

⇒ x^-1 Є g(s)

(iii) For x, y Є g(s), Ǝ       a, b Є s

Such that g (a) = x and g (b) = y

As s is a subgroup, a * b Є s

⇒ g (a*b) = g(a) ∆ g (b)

= x ∆ y Є g(s)

g (s) is a subgroup of H.