Normal Subgroups and Factor Groups · Normal Subgroups and Factor Groups INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 3 We have just learned that, in order to show a subgroup

Normal Subgroups and Factor Groups

INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 1


Subject: Mathematics

Course Developer: Harshdeep Singh

Department/ College: Assistant Professor,

Department of Mathematics,

Sri Venkateswara College,

University of Delhi



Introduction

If G is a group, and H is a subgroup of G, and g is an element of G, then

gH = {gh : h an element of H} is the left coset of H in G with respect to g, and

Hg = {hg : h an element of H} is the right coset of H in G with respect to g.

For example, consider a group G = S3 and the subgroup H = { (1), (12) }. Then for

g = (13) ∈ G, gH = (13)H = { (13), (123) }, and Hg = H(13) = { (13), (132) }.

In this chapter we will study the algebraic properties of the subgroups whose left

coset and right coset are same, and for those subgroups, their cosets form a group

called the quotient or factor group. Then we will study characteristic, commutator

subgroups. We will look into problems for better understanding of the text.

Definition: Normal Subgroup

A subgroup H of a group G, is called a normal subgroup if gH = Hg, for all g in G.

Notation: A subgroup H is normal subgroup of G is denoted by H ⊴ G.

The property gH = Hg, for g ∈ G, means that gh1 = h2g for some h1, h2 ∈ H.

Theorem 1: Normal Subgroup Test

A subgroup H of a group G, is normal in G, if and only if gHg-1 ⊆ H, for all g ∈ G.

Proof: (⇒) If H ⊴ G, then for any g ∈ G and h ∈ H, gh = h1g for some h1 ∈ H (by

definition); i.e., ghg-1 = h1. Hence, gHg-1 ⊆ H.

(⇐) If gHg-1 ⊆ H, for all g ∈ G, then for some a ∈ G, aHa-1 ⊆ H, or aH ⊆ Ha. Also, if g

= a-1 (as a-1 is also a member of G), then a-1Ha ⊆ H, or Ha ⊆ aH. Hence, aH = Ha, for

all a ∈ G.

Remarks:



We have just learned that, in order to show a subgroup H of G normal, we

have to show, either aH = Ha for all a ∈ G, or aHa-1 ⊆ H for all a ∈ G; and,

In order to show a subgroup H of G, not normal, i.e., H ⋬ G, we have to find

at least one element g ∈ G, and h ∈ H, such that ghg-1 ∉ H.

Example 1: Every subgroup of an Abelian group is normal (Abelian group: Group in

which xy = yx ∀ x, y ∈ G).

Solution: Let G be an Abelian group, and H be its subgroup. For a ∈ G and h ∈ H, ah

= ha (as h ∈ H ⊆ G, ⇒ h ∈ G). Hence, aha-1 = h, which implies, a-1Ha ⊆ H. Therefore,

H is normal subgroup of G.

Example 2: The center of a group is always normal subgroup of the group.

Solution: Let G be a group and Z(G) be the center of group G. We know that Z(G)

is a subgroup of G. In order to show Z(G) ⊴ G, we need to show gZ(G)g-1 ⊆ Z(G), ∀

g ∈ G. So, let x ∈ Z(G), or gxg-1∈ gZ(G)g-1. Since, gx = xg, as x ∈ Z(G), Therefore,

gxg-1 = x, and hence, gxg-1 ∈ Z(G). ⇒ gZ(G)g-1 ⊆ Z(G). Hence, Z(G) ⊴ G.

Example 3: Let Sn be the group of permutations on n elements {1, 2, 3, . . . , n}.

Let An be the subgroup of Sn, consisting of all even permutations. Then An ⊴ Sn.

Solution: For σ ∈ Sn, we have to show that, σAnσ-1 ⊆ An. Let h ∈ An. Then h is an even

permutation (by definition of An).

Case – 1: If σ is an even permutation, then σhσ−1 is also an even permutation for all

σ ∈ Sn (product of even permutations).

Case – 2: If σ is an odd permutation, then σh is also an odd permutation, also, σ-1 is

an odd permutation, which implies σhσ-1 is an even permutation (product of two odd

permutations is an even permutation).

Therefore, σhσ−1 ∈ An. Hence, σAnσ-1 ⊆ An. Therefore An ⊴ Sn.

Example 4: The subgroup of rotations in Dn is a normal subgroup of Dn.



Solution: Let subgroup of rotations be denoted by R. We need to show that gRg-1 ⊆

R, ∀ g ∈ Dn.

Case – 1: If g ∈ Dn, be rotation, then for any rotation r, grg-1 is also a rotation.

Case – 2: If g ∈ Dn, be reflection, then for any rotation r, grg-1 = r-1, hence is a

rotation. (as, frf-1 = r-1)

⇒ gRg-1 ⊆ R. Hence, subgroup of rotations ⊴ Dn.

Example 5: The subgroup SL(n, ℝ) of n x n matrices with determinant 1, is a

normal subgroup of GL(n, ℝ), the group of n x n matrices with non-zero

determinant.

Solution: Let A ∈ SL(n, ℝ), B ∈ GL(n, ℝ). Since, det(BAB-1) = det(B)det(A)det(B-1),

and det(B-1)= 1/ det(B). ⇒ det(BAB-1) = det(B)det(A)/det(B) = det(A) =1(as A ∈

SL(n, ℝ). Hence, B SL(n, ℝ) B-1 ⊆ SL(n, ℝ), ∀ B ∈ GL(n, ℝ). ⇒ SL(n, ℝ) ⊴ GL(n, ℝ).

Example 6: Let H = {(1), (12)} be a subgroup of S3. Then, H ⋬ S3.

Solution: Let (13) ∈ S3. Since (13)-1 = (13), and (12) ∈ H. Since, (13)(12) (13)-1 =

(13)(12)(13) = (23) ∉ H. Hence, H ⋬ S3.

Example 7: Let H = {[𝑎 𝑏0 𝑑

] | 𝑎, 𝑏, 𝑑 ∈ ℝ, 𝑎𝑑 ≠ 0}. Then, H ⋬ GL(2, ℝ).

Solution: Let A = [1 11 0

] ∈ GL(2, ℝ). Then A-1 = [0 11 −1

]. Let B = [1 10 1

] ∈ H. Since,

ABA-1 = [1 11 0

] [1 10 1

] [0 11 −1

] = [2 −11 0

] ∉ H. Hence, H ⋬ GL(2, ℝ).

Example 8: Let G = GL(2, ℝ), and let K be a subgroup of ℝ* i.e., (ℝ\{0},*). Then H

= { A ∈ G | det(A) ∈ K } ⊴ G.

Solution: Let B ∈ G, and A ∈ H. Then det(BAB-1) = det(B)det(A)/det(B) = det(A) ∈

K (as A ∈ H). Therefore, BAB-1 ∈ H, and hence, B H B-1⊆ H, which implies, H ⊴ G.

Example 9: If a subgroup H of G has index 2 in G, then H ⊴ G.



Solution: Subgroup H of G has index 2 in G means, there are exactly 2 distinct

left(or, right) cosets of H in G. Let x ∈ G.

Case – 1: If x ∈ H, then xH = H = Hx.

Case – 2: If x ∉ H, then xH ∩ H = 𝜙 , and. Also, Hx ∩ H = 𝜙. Since, there are exactly

2 distinct left(or, right) cosets of H in G, G = H ∪ Hx, and G = H ∪ xH. Hence, xH =

Hx.

Hence, for x ∈ G, xH = Hx. ⇒ H ⊴ G.

Example 10: Let H = { (1), (12)(34) } be a subgroup of A4. Then, H ⋬ A4.

Solution: Consider the even permutation (123) ∈ A4. Since, (123)(12)(34)(123)-1 =

(123)(12)(34)(132) = (14)(23) ∉ H. Hence, H ⋬ A4.

Factor Groups

Let N be a normal subgroup of a group (G,*). We define the set G/N to be the set of

all left (or, right) cosets of N in G, i.e., G/N = { aN : a ∈ G }. We define the binary

operation on G/N as follows:

For each aN and bN ∈ G/N, the binary operation of aN and bN is written as (aN)(bN),

which is equal to (a*b)N, where ‘*’ is the binary operation of the group G.

In view of this G/N is a group, which is defined and proved in the succeeding topics.

Definition: Factor Groups

For any group G, let N be a normal subgroup of G. Then, the set of all left (or right)

cosets of N in G, is also a group itself, known as factor group of G by N (or the

quotient group of G by N).

Notation: The factor group of G by H is denoted by G/H. And, G/H = { aH | a ∈ G }.

Theorem 2: Factor Groups



Let G be a group and let H be a normal subgroup of G. Then the set G/H = { aH | a

∈ G } is a group under the operation (aH)(bH) = abH.

Solution: Firstly, we have to show that the operation is well defined; for this, we

must show that the correspondence defined above from G/H x 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 (aH)(bH) = (a’H)(b’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 ∈ H, and therefore a’b’H = ah1bh2H =

ah1bH = ah1Hb = aHb = abH (using H is normal subgroup of G).

Now, eH = H is the identity; a-1H is the inverse of aH; and (aH)(bH)(cH) = (abH)cH

= (ab)(cH) = a(bc)H = aH(bc)H = aH(bH)(cH). This proves that G/H is a group.

Remark: The group G/H is known as Factor Group G by H.

Example 11: Since Z is an Abelian group, 4Z is normal subgroup of Z. Find the factor

group Z/4Z.

Solution: Z/4Z = { a + 4Z | a ∈ Z } = { 0 + 4Z, 1 +4Z, 2 +4Z, 3 +4Z } (using the

fact that, 0 + 4Z = 4 + 4Z = 8 + 4Z = . . ., as 0 – 4 ∈ 4Z; 1 + 4Z = 5 + 4Z = 9 +

4Z = . . .; so on).

Example 12: Let G = Z18, and let H = = { 0, 6, 12 }. Find G/H.

Solution: G/H = { 0 + H, 1 + H, 2 + H, 3 + H, 4 + H, 5 + H }, (as 0 + H = 6 + H

= 12 + H = . . .; and so on).

Example 13: Consider K = { R0, R180 } a subgroup of D4. Write the elements of D4/K.

Solution: Since D4 = { R0, R90, R180, R270, H, V, D1, D2 }, therefore, D4/K = { R0K,

R90K, R180K, R270K, HK, VK, D1K, D2K }. Since, R0K = K = R180K; R270K = R90K; HK =

VK; D1K = D2K, therefore, D4/K = { K, R90K, HK, D1K }.

Result used: aH = bH, if and only if, a-1b ∈ H.



The cayley table for this group D4/K is given as:

K R90K HK D1K

K K R90K HK D1K

R90K R90K K D1K HK

HK HK D1K K R90K

D1K D1K HK R90K K

Example 14: Let G = U(32), and let H = U16(32).

a) Find the order of G/H, and is it Abelian?

b) Which of three Abelian groups of order 8 is it isomorphic to:

Z8, Z4 ⊕ Z2, or Z2 ⊕ Z2 ⊕ Z2?

Solution: Since G = U(32) = { a ∈ Z32 | gcd(a,32) = 1 } = { 1, 3, 5, 7, 9, 11, 13,

15, 17, 19, 21, 23, 25, 27, 29, 31 }, and H = U16(32) = { a ∈ U(32) | a = 1mod(16)

} = {1, 17 }.

a) Order of G/H = (order of G)/(order of H). Hence, |G/H| = 16/2 = 8. Since G is

Abelian, so is G/H.

b) G/H = { 1H, 3H, 5H, 7H, 9H, 11H, 13H, 15H }(since, 17H = H, 19H = 3H, 21H

= 5H, 23H = 7H, 25H = 9H, 27H = 11H, 29H = 13H, 31H = 15H )

Result: |aH| = least n ∈ N | an ∈ H

Hence, |1H| = 1; |3H| = 4 (as, 32 = 9 ∉ H, 33 = 27 ∉ H, 34 = 81 = 17 ∈ H);

similarly, |7H| = |9H| = 2. Since, Z2 ⊕ Z2 ⊕ Z2 doesn’t have any element of

order 2, G/H cannot be isomorphic to it. Also, since, ∃ only 1 element of order

4 in Z8, and G/H has more than 1 element of order 4, hence G/H cannot be

isomorphic to it. Therefore, without loss of generality, G/H is isomorphic to Z4

⊕ Z2.

Example 15: Let G = U(32) and K = { 1, 15 }. Then, find |G/K| and, G/K is

isomorphic to Z8, Z4 ⊕ Z2, or Z2 ⊕ Z2 ⊕ Z2?



Solution: |G/K| = |G|/|K| = 16/2 = 8. Since, G is Abelian, G/K is also Abelian group.

Further, since, 3K ∈ G/K, and |3K| = 8, and since ∄ any element of order 8 in Z4 ⊕

Z2, and Z2 ⊕ Z2 ⊕ Z2, therefore G/H is isomorphic to Z8.

Example 16: Show that A4 has no subgroup of order 6.

Solution: If possible, let H be a subgroup of A4, such that |H| = 6. Since, |A4| = 12,

H has index 2 in G, as |G|/|H| = 2, hence H ⊴ G. Thus the factor group A4/H exists,

and |A4/H| = 2. Since, order of an element divides the order of group, for aH ∈ A4/H,

|aH| = 1, or 2. If |aH| = 1, aH = H, hence, a ∈ H. If |aH| = 2, then a2 ∈ H, for all a

∈ A4. Since there are nine different elements in { a2 | a ∈ A4 }, hence we have a

contradiction to order of the subgroup H. Hence, A4 has no subgroup of order 6.

Theorem 3: The G/Z(G) Theorem

Let G be a group and consider Z(G) to be the center of G. If G/Z(G) is cyclic, then G

is Abelian.

Proof: G/Z(G) is a cyclic factor group, so ∃ a generator of G/Z(G) say gZ(G). Let a,

b ∈ G, there exist integers p and q respectively such that

aZ(G) = (gZ(G))p = gpZ(G), and

bZ(G) =(gZ(G))q= gqZ(G)

Thus, a = gpy for some y ∈ Z(G) ( aZ(G) = gpZ(G) ⇒ a-1gp ∈ Z(G) ⇒ a-1gp= x for some

x in Z(G). ⇒ gp= ax ⇒ gpx-1 = a, x-1 = y in Z(G) ); Similarly b = gqt for some t in

Z(G). Therefore,

ab = (gpy)(gqt) = gp(ygq)t = gp(gqy)t = (gpgq)yt = (gpgq)ty = (gp+q)ty = (gq+p)ty =

(gqgp)ty = gq(gpt)y = gq(tgp)y = (gqt)(gpy) = ba.

Thus, G is Abelian.

Theorem 4: G/Z(G) ≈ Inn(G)

For any group G, G/Z(G) is isomorphic to Inn(G).



Remark: Inn(G) = { ϕg | ϕg: G → G , ϕg is an isomorphism, which is defined as,

ϕg(x) = g-1xg, for x ∈ G }

Proof: Define a mapping F: G/Z(G) → Inn(G) by F(gZ(G)) = ϕg . Firstly, we have to

show that F is a well-defined mapping. Consider gZ(G) = hZ(G) ⇒ g-1h ∈ Z(G)

⇒ g-1hx = x g-1h, ∀ x ∈ G. ⇒ hxh-1= gxg-1 ∀ x in G ⇒ ϕh(x) = ϕg(x) ∀ x in G ⇒ ϕh = ϕg

⇒ F(gZ(G)) = F(hZ(G)). Now, to show F injective, Consider F(gZ(G)) = F(hZ(G)) ⇒

ϕg = ϕh ⇒ ϕg(x) = ϕh(x) ∀ x in G ⇒ gxg-1 = hxh-1 ∀ x in G ⇒ x g-1h = g-1hx ∀ x in G

⇒ g-1h ∈ Z(G) ⇒ gZ(G) = hZ(G). F is onto because for any element ϕg ∈ Inn(G) ∃ an

element g ∈ G such that F(gZ(G)) = ϕg. F is an operation preserving map

(homomorphism) as F(gZ(G)hZ(G)) = F(g(Z(G)h)Z(G)) = F(g(hZ(G))Z(G)) =

F(ghZ(G)) = ϕgh = ϕg ϕh (by property of ϕ) = F(gZ(G))F(hZ(G)), Thus G/Z(G) is

isomorphic to Inn(G).

Example 17: Prove that Inn(D6) ≈ D3.

Solution: We know Inn(D6) ≈ D6/Z(D6); Since Z(D6) = { R0, R180 }, |Z(D6)| = 2,

hence, |Inn(D6)| = |D6/Z(D6)| = |D6|/|Z(D6)| = 12/2 = 6. Since, groups of order 6,

up to isomorphism are: D3, or Z6. Hence, Inn(D6) ≈ D3, or Z6. If Inn(D6) ≈ Z6, then,

Inn(D6) is cyclic, and thus, by G/Z(G) theorem, D6 is Abelian, which is a contradiction.

Hence, Inn(D6) ≈ D3.

Theorem 5: Existence of Element of Prime Order

Let G be a finite Abelian group and let p be a prime number that divides the order

of group G. Then G contain an element of order p.

Proof: Firstly, suppose that G has order 2 and only prime number which divide the

order of Group G is 2 itself; as group of order 2 has one element of order

1(identity) and other of order 2 so result holds trivially. Theorem will be proved by

using Second Principle of Mathematical Induction on order of group G, according to

which the theorem is assumed to be applicable for all those Abelian groups that

have their order less than the order of group G and use this assumption to prove

that theorem holds true for G also. Let x ∈ G, and |x| = n (G has finite order so



does x) and let n = mq where q is some prime number (any natural number can be

written as such), Thus, xm = y (say) has order q (as q is the least positive integer

such that xmq = xn = e, as n is the least), y ∈ G has order q, so if q = p then we are

done. So assume p ≠ q. Since every subgroup of an Abelian group is normal, we

can therefore construct factor group G’ = G/ ( is subgroup of Abelian group

G hence normal). Then G’ is Abelian as G is Abelian and p divides │G’│ since │G’│

= │G│/││ = │G│/q and p, q divides │G│ (subgroup order divides order of

group). So G’ is Abelian group with order less than G and p/│G’│, Thus by Second

Principle of Mathematical Induction, ∃ an element y of │G’│ of order p, Thus

(y)p = (identity in G’) ⇒ yp = ⇒ yp ∈ . So yp = e or yp has

order q (yp ∈ and has prime order). So if, yp = e ⇒ y has order p then this

will be the required element of order p, if yp has order q , so yq has order p thus yq

= y’ ∈ G will be the element of order p. Thus, existence of element of order p in G is

ensured.

Some problems from Seventh Edition,

Contemporary Abstract Algebra, by Joseph A. Gallian

Question 1. Prove that a factor group of an Abelian group is Abelian.

Solution: Recall: A group G is Abelian if ab = ba for all a, b ∈ G.

Suppose G is Abelian. Let G/H be any factor group of G. We have to show that G/H

is Abelian. Let aH, bH be any elements of G/H. Then (aH)(bH) = (ab)H by definition

of multiplication in factor groups. Then (ab)H = (ba)H since G is Abelian. (ba)H =

(bH)(aH) again definition of multiplication in factor group. Therefore (aH)(bH) =

(bH)(aH). Therefore G/H is Abelian.

Question 2. What is the order of the factor group (Z10 ⊕ U(10))/?

Solution: We know that |Z10 ⊕ U(10)| = |Z10||U(10)| = 10 x 4 = 40; and |(2, 9)|

= l.c.m.(|2| in Z10, |9| in U(10)) = l.c.m.(5, 2) = 10; Since, |(2, 9)| = ||,

|| = 10. Hence, |(Z10 ⊕ U(10))/| = 40/10 = 4.



Question 3. Show, by example, that in a factor group G/H it can happen that aH =

bH, but |a| ≠ |b|. (Do not use a = e or b = e.)

Solution: Let G = Z6, and H = {0, 3}; We know that Z6 is Abelian, and H is a subgroup

of it, therefore, H is normal subgroup of Z6. Hence, Z6 /H is a factor group. Let a =

1, and b = 4. Here,

aH = 1H = {1, 4}; bH = 4H = {4, 1}. Hence, aH = bH, but, |a| = |1| = 6, |b| = |4|

= 3; i.e., |a| ≠ |b|.

Question 4. The only subgroup of A4 of order 4. Why does this imply that this

subgroup must be normal in A4? Generalize this to arbitrary finite groups.

Solution: Let H be that only subgroup of A4 which has order 4. For g ∈ A4, gH is also

a subgroup of A4. Also, |gH| = |H| = 4. Since, ∃ a unique subgroup of order 4 of A4,

gH = H, similarly, Hg = H. Hence, gH = Hg for all g ∈ A4, which implies H is normal

in A4.

For generalization, let G be any group of finite order (say n), then for any m, s.t.,

m|n. If ∃ a unique subgroup of order m, then that subgroup is normal (using the fact

|aH| = |H|).

Question 5. Let p be a prime. Show that if H is a subgroup of a group of order 2p

that is not normal, then H has order 2.

Solution: Let G be a group, and |G| = 2p. Then possible orders of (non-trivial)

subgroups of G are: 2, or p. If the subgroup has order p, then [G:H] = |G|/|H| =

2p/p = 2, i.e., H has index 2 in G, then H is normal in G, which is a contradiction.

Hence, |H| = 2.

Question 6. Show that D13 is isomorphic to Inn(D13).

Solution: Result: For any prime p, Z(Dp) = { R0 }, i.e., |Z(Dp)| = 1.

We know G/Z(G) ≈ Inn(G), therefore, Inn(D13) ≈ D13/Z(D13). Since, Z(D13) is trivial

subgroup of D13, D13/Z(D13) = D13. Hence, Inn(D13) ≈ D13.



Question 7. If G is a group and |G : Z(G)| = 4, prove that G/Z(G) ≈ Z2 ⊕ Z2.

Solution: Since |G : Z(G)| = 4, G/Z(G) is isomorphic to either Z4, or Z2 ⊕ Z2. If

G/Z(G) ≈ Z4, then G/Z(G) is cyclic, and therefore, G is Abelian, which implies, Z(G)

= G, i.e., |G : Z(G)| = 1, which is a contradiction to hypothesis given. Hence, G/Z(G)

≈ Z2 ⊕ Z2.

Question 8. Suppose that G is a non-Abelian group of order p3, where p is a prime,

and Z(G) ≠ {e}. Prove that |Z(G)| = p.

Solution: Since, Z(G) ≠ {e}, |Z(G)| can be p, p2, or p3; Since, if |Z(G)| = p3 then

Z(G) = G, and G will be Abelian, which is a contradiction. So, Z(G) can have order

either p, or p2. If |Z(G)| = p2, then by G/Z(G) will have order p, hence G/Z(G) will

be cyclic, implying G to be Abelian, which is again a contradiction. Hence, |Z(G)| =

p.

Question 9. If |G| = pq, where p and q are primes that are not necessarily distinct,

prove that |Z(G)| = 1 or pq.

Solution: Let |G| = pq, then possibilities for |Z(G)| can be 1, p, q, or pq (using

Lagrange’s theorem). Our task is to eliminate the possibilities of |Z(G)| = p and q.

Suppose |Z(G)| = p, then G/Z(G) will have order q, which implies G/Z(G) is cyclic,

and hence, G is Abelian, implying Z(G) = G, or |Z(G)| = pq, which is a contradiction

in itself. Hence, |Z(G)| ≠ p. Similarly, |Z(G)| ≠ q. Hence, |Z(G)| can be 1 or pq only.

Question 10. Let G be an Abelian group and let H be the subgroup consisting of all

elements of G that have finite order. Prove that every nonidentity element in G/H has

infinite order.

Solution: If for aH ∈ G/H, where a ∉ H, i.e., aH ≠ H, and aH has finite order. Then

(aH)n = H, for some n ∈ ℕ. ⇒ anH = H for some n, ⇒ an ∈ H for some n ∈ ℕ. ⇒ (an)m

= e, for some m ∈ ℕ, since, H contains elements of finite order. Thus, anm = e, hence

|a| is finite, which is a contradiction. Hence, every nonidentity element in G/H has

infinite order.



Question 11. Determine all subgroups of ℝ* that have finite index.

Solution: Let H be a subgroup of G that has index n. ⇒ |G/H| = n, hence for any aH

∈ G/H, (aH)n = H, i.e., an ∈ H. Then (ℝ*)n = { xn | x ∈ ℝ*} ⊆ H. If n is odd, then (ℝ*)n

= ℝ*; if n is even, then (ℝ*)n = ℝ+. So, H = ℝ* or H = ℝ+.

Question 12. Let G = {±1, ±i, ±j, ±k}, where i2 = j2 = k2 = -1, -i = (-1)i, 12 = (-

1)2 = 1, ij = -ji = k, jk = -kj = i, and ki = -ik = j.

a) Construct the Cayley table for G.

b) Show that H = {1, -1} ⊴ G.

c) Construct the Cayley table for G/H. Is G/H isomorphic to Z4 or Z2 ⊕ Z2?

Solution: a) The cayley table for G is given as:

1 -1 i -i j -j k -k

1 1 -1 i -i j -j k -k

-1 -1 1 -i i -j j -k k

i i -i -1 1 k -k -j j

-i -i i 1 -1 -k k j -j

j j -j -k k -1 1 i -i

-j -j j k -k 1 -1 -i i

k k -k j -j -i i -1 1

-k -k k -j j i -i 1 -1

b) In order to show H = {1, -1} ⊴ G, we have to show gHg-1 ⊆ H, for all g ∈ G.

Let g ∈ G be any arbitrary element of G, then g1g-1 = gg-1 = 1 ∈ H. Also, g(-

1)g-1 = (-1)gg-1 = (-1)1 = -1 ∈ H. Hence, gHg-1 ⊆ H, and H = {1, -1} ⊴ G.

c) Here G/H = { H, iH, jH, kH } (as iH = (-i)H, jH = (-j)H, kH = (-k)H). Since,

(aH)2 = H for a = 1, i, j, or k, G/H is isomorphic to Z2 ⊕ Z2.

Cayley table for G/H is given as:

H iH jH kH



H H iH jH kH

iH iH H kH jH

jH jH kH H iH

kH kH jH iH H

Question 13. In D4, let K = { R0, D } and let L = { R0, D, D’, R180 }. Show that K ⊴

L ⊴ D4, but that K is not normal in D4.

Solution: Index of K in L is 2 ( |L|/|K| = 4/2 = 2 ). Therefore, K ⊴ L. Also, since,

index of L in D4 is 2 ( as |D4|/|L| = 2 ), L ⊴ D4.

Now, we have to show that K is not normal in D4. Since, R90DR90-1 = R90DR270 = D’ ∉

K, K is not normal in D4.

Question 14. Show that the intersection of two normal subgroups of G is a normal

subgroup of G.

Solution: Let H and K be normal subgroups of G. Let x ∈ H ∩ K. Then x ∈ H and x ∈

K. For any element g ∈ G, gxg−1 ∈ H (since H is normal) and gxg−1 ∈ K (since K is

normal). So gxg−1 ∈ H ∩ K. Thus H ∩ K is normal subgroup of G.

Question 15. Let N be a normal subgroup of G and let H be any subgroup of G.

Prove that NH is a subgroup of G. Give an example to show that NH need not be a

subgroup of G if neither N nor H is normal.

Solution: Suppose n1h and n2h2 ∈ NH. Then n1h1n2h2 = n1n’h1h2 ∈ NH (as h1n2h1-1 =

n’). Also, (n1h1)-1 = h1-1n1-1 = nh1-1 ∈ NH. By two-step subgroup test, NH is a subgroup

of G.

Take G = S3, H = and K = . Then H and K are subgroups of G (each

containing two elements) and HK = { (1), (12), (23), (132) } a set of size 4.

Therefore HK is not a subgroup of G (by Lagrange’s Theorem) since 4 does not divide

6.



Question 16. Suppose that a group G has a subgroup of order n. Prove that the

intersection of all subgroups of G of order n is a normal subgroup of G.

Solution: Let D = ∩ { S | S is a subgroup of G, and |S| = n }. Let g ∈ G.

Case – 1: If g-1Dg is a subset of each subgroup of G of order n, then g-1Dg is a subset

of the intersection of all subgroups of G of order n. Hence, g-1Dg ⊆ D for each g ∈ G,

and therefore D is normal in G.

Case – 2: If g-1Dg is not contained in a subgroup, say H, of G of order n for some g

∈ G, thus D is not contained in gHg-1, for if D is contained in gHg-1, then g-1Dg is

contained in H which is a contradiction. But |gHg-1| = n, and hence, D ⊆ gHg-1, a

contradiction. Thus, g-1Dg = D for each g ∈ G. Hence D is normal in G.

Question 17. Suppose that H is a normal subgroup of a finite group G. If G/H has

an element of order n, show that G has an element of order n. Show, by example,

that the assumption that G is finite is necessary.

Solution: Let |gH| = n. Then, gn ∈ H for some n ∈ N. Hence, (gn)t = e for some t ∈

N; Hence, |g| = nt and |gt| = n. Hence, G has an element of order n.

Consider G = Z, and H = 2Z, then 1 + 2Z ∈ G/H has order 2, but ∄ any element in Z

of order 2. Hence, for above result, the assumption that G is finite is necessary.

Automorphism of a group

Let G be a group. An automorphism of G is an isomorphism G → G. We write Aut(G)

for the set of all automorphisms of G.

Characteristic Subgroup

A subgroup N of a group G is called a characteristic subgroup if f(N) = N for all

automorphisms f of G.

Remark: It is sufficient to show f(N) ⊆ N, in order to prove a subgroup characteristic,

as f is bijective, and number of elements in f(N) =number of elements in N.



Question 18. Prove that every subgroup of cyclic group is characteristic.

Solution: Let G be a cyclic group. So, G = for some g in G. Now, let H be a

subgroup of G. So, H is cyclic as well, H = for some integer k. Next, any

automorphism f: G → G is of the form f(g) = gn for some integer n, where G =

= (automorphism takes generator to generator)

So, f(H) = (subgroup of cyclic group is cyclic) = .

Claim: = . (Then f(H) = H, which shows that H is characteristic.)

(i) ⊆ : This is true, because gnk = (gk)n.

(ii) ⊆ : Since = G, there exists an integer s such that gns = g.

Hence, gk = (gns)k = (gs)nk, and we are done.

Generating set of a group

Let G be a cyclic group, we say G is generated by { a }, i.e., G = , or if every

element of G can be written as an, for different values of n ∈ ℤ.

Now, if generating set for a group G = { a1, a2, a3, a4, . . . , am }, it means, every

element of G can be expressed as in the form of a1n1a2n2a3n3. . .amnm, for n1, n2, n3,

. . ., nm ∈ ℤ.

Question 19. Prove that the center of a group is characteristic.

Solution: Let G be a group, and Z(G) be the center of G. We have to show that

f(Z(G)) = Z(G) for every automorphism f of G. So, let f be some arbitrary

automorphism of G. Let b ∈ Z(G), Then, f(b) ∈ f(Z(G)). Let g ∈ G, then g = f(u) for

some u ∈ G (as, f is automorphism). Then, gf(b) = f(u)f(b) = f(ub) = f(bu) (as b ∈

Z(G)) = f(b)f(u) = f(b)g. Hence, f(Z(G)) ⊆ Z(G).

Next, Let h ∈ Z(G), ⊆ G. So h = f(y) for some y ∈ G. We want to show that y ∈ Z(G).

Consider yg for some g ∈ G. Since, g ∈ G, g = f(x) for some x ∈ G. So yg = f-1(h)f-

1(x) = f-1(hx) = f-1(xh) (as h ∈ Z(G)) = f-1(x)f-1(h) = gy. Hence, y ∈ Z(G) or h = f(y)

∈ f(Z(G)). Hence, Z(G) ⊆ f(Z(G)), for f an automorphism of G.



Hence, the center of a group is a characteristic subgroup.

Commutator Subgroup

Let G be a group, and G’ be a subgroup of G. Then G’ is called commutator subgroup

of G if G’ is generated by the set { x-1y-1xy | x, y ∈ G }.

Question 20. Prove that commutator subgroup G’ of a group G is characteristic

subgroup of G.

Solution: Let G be a group, G’ be commutator subgroup of G. Let G’ is generated by

the set { x-1y-1xy | x, y ∈ G }. Let f ∈ Aut(G). Then f(G’) is generated by the set {

f(x)-1f(y)-1f(x)f(y) | f(x), f(y) ∈ f(G) }. Since f is one-one, and onto homomorphism,

the set { f(x)-1f(y)-1f(x)f(y) | f(x), f(y) ∈ f(G) } = { x-1y-1xy | x, y ∈ G }. Hence, f(G’)

= G’, or G’ is characteristic subgroup of G.

Question 21. Let G be a group and let G’ be the subgroup of G generated by the set

S = {x−1y−1xy | x, y ∈ G}.

a) Prove that G’ is normal in G.

b) Prove that G/G’ is abelian.

c) If G/N is abelian, prove that G’ ≤ N.

Solution:

a) Let x, y, g ∈ G be given. Then g(x−1y−1xy)g−1 = g(x−1g−1gy−1g−1gxg−1gy)g−1 =

(gx−1g−1)(gy−1g−1)(gxg−1)(gyg−1) = (gxg−1)−1(gyg−1)−1(gxg−1)(gyg−1) ∈ G’.

Now let a ∈ G’ be given. Then a = a1a2 · · · an, where each ai = xi−1 yi−1xiyi , for

some xi , yi ∈ G. By the above argument, gaig−1 ∈ G’ for every i. Therefore,

gag−1 = g(a1a2 · · · an)g−1 = (ga1g−1)(ga2g−1)· · ·(gang−1) ∈ G’. Thus, G’ is

normal in G.

b) Let x, y ∈ G be given. Then (x−1y−1xy)G’ = G’ , since x-1y−1xy ∈ G’ . Therefore,

xy G’ = yx G’ (since, a-1b ∈ H, if and only if, aH = bH)

c) It suffices to show that x−1y−1xy ∈ N for every x, y ∈ G. Let x, y ∈ G be given.

Since G/N is abelian, (x−1y−1xy)N = N, which implies that x−1y−1xy ∈ N.



Question 22. Let H be a normal subgroup of a finite group G and let x ∈ G. If gcd(|x|,

|G/H|) = 1, show that x ∈ H.

Solution: Let |x| = m, and |G/H| = n; Since, xH ∈ G/H, (xH)n = H. Hence, xnH = H,

which implies, xn ∈ H. Since gcd(m, n) = 1, there exist integers t, and q s.t., mt +

nq = 1. Hence, xmt + nq = x1. ⇒ xmtxnq = x. ⇒ xnq = x (since |x| = m). Now, since xn ∈

H, and H is a subgroup of G, xnq ∈ H; Hence, x ∈ H.

References:

1. Dummit, David S., Foote, Richard M. (2004), Abstract Algebra (3rd edition)

2. Joseph A. Gallian (2013), Contemporary Abstract Algebra (8th edition)

Normal Subgroups and Factor Groups · Normal Subgroups and Factor Groups INSTITUTE OF LIFELONG LEARNING, DELHI UNIVERSITY 3 We have just learned that, in order to show a subgroup

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