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Normal subgroups and properties, Lecture notes of Mathematics

Lecture notes for groups and symmetry course , it tells about normal subgroups

Typology: Lecture notes

2018/2019

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INTRODUCTION TO GROUPS AND SYMMETRY (MTH 203)
Lecture-13, (30-08-2019)
Product of two subgroups and a counting principal. .
In this section we will generalize the notion of coset of a subgroup. Let
Gbe a group and H, K are subgroups of G. Their product is defined as,
HK ={xโˆˆG:x=hk, h โˆˆH, k โˆˆK}.
Remark 0.1. Similarly we can also define the product of two subsets of
G. The left coset aH of Hin Gis just a product of {a}with H.
Example-13.1: Let H= 2Zand K= 4Zare two subfrops of (Z,+).
With respect to addition operation we get,
HK = 2Z4Z={xโˆˆZ:x= 2i+ 4j= 2(i+ 2j)}= 2Z.
Problem-13.1: Let
H={8a+ 14b:a, b โˆˆZ}.
Is Ha subgroup of (Z,+)? If yes then write it in terms of mZ.
Home Work-13.1: Show that mZnZ= (m, n)Z.
Example-13.2: Let H={(1,2,3),(2,1,3)}and K={(1,2,3),(3,2,1)}
are subgroups of S3. In this case
HK ={(1,2,3)(1,2,3),(2,1,3)(1,2,3),(1,2,3)(3,2,1),(2,1,3)(3,2,1)}
HK ={(1,2,3),(2,1,3),(3,2,1),(2,3,1).
Also
KH ={(1,2,3)(1,2,3),(3,2,1)(1,2,3),(1,2,3)(2,1,3),(3,2,1)(2,1,3)}
KH ={(1,2,3),(3,2,1),(2,1,3),(3,1,2)}
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INTRODUCTION TO GROUPS AND SYMMETRY (MTH 203)

Lecture-13, (30-08-2019)

Product of two subgroups and a counting principal..

In this section we will generalize the notion of coset of a subgroup. Let

G be a group and H, K are subgroups of G. Their product is defined as,

HK = {x โˆˆ G : x = hk, h โˆˆ H, k โˆˆ K}.

Remark 0.1. Similarly we can also define the product of two subsets of

G. The left coset aH of H in G is just a product of {a} with H.

Example-13.1: Let H = 2Z and K = 4Z are two subfrops of (Z, +).

With respect to addition operation we get,

HK = 2Z 4 Z = {x โˆˆ Z : x = 2i + 4j = 2(i + 2j)} = 2Z.

Problem-13.1: Let

H = { 8 a + 14b : a, b โˆˆ Z}.

Is H a subgroup of (Z, +)? If yes then write it in terms of mZ.

Home Work-13.1: Show that mZnZ = (m, n)Z.

Example-13.2: Let H = {(1, 2 , 3), (2, 1 , 3)} and K = {(1, 2 , 3), (3, 2 , 1)}

are subgroups of S 3

. In this case

HK = {(1, 2 , 3)(1, 2 , 3), (2, 1 , 3)(1, 2 , 3), (1, 2 , 3)(3, 2 , 1), (2, 1 , 3)(3, 2 , 1)}

HK = {(1, 2 , 3), (2, 1 , 3), (3, 2 , 1), (2, 3 , 1).

Also

KH = {(1, 2 , 3)(1, 2 , 3), (3, 2 , 1)(1, 2 , 3), (1, 2 , 3)(2, 1 , 3), (3, 2 , 1)(2, 1 , 3)}

KH = {(1, 2 , 3), (3, 2 , 1), (2, 1 , 3), (3, 1 , 2)}

1

It is clear that HK 6 = KH and HK, KH both are not subgroups of S 3

(since (3, 2 , 1)(2, 3 , 1) = (1, 3 , 2) โˆˆ HK and (2, 1 , 3)(3, 1 , 2) = (1, 3 , 2) 6 =

KH.

Problem-13.2: Let

U

20

be the group with respect to the multiplication operation (i.e. ยฏa

b = ab).

H = {

and

K = {

are subgroups of U 20

. Compute HK and find out the number of element

HK. Is it a subgroup of U 20

Remark 0.2. From above examples it is clear that the product of two sub-

groups of a group may or may not be a subgroup.

The following Lemma gives a necessary and sufficient condition for prod-

uct to be a subgroup.

Lemma 0.3. HK is a subgroup of G if and only if HK = KH.

Proof. See Hersteinโ€™s Book Lemma 2. 5 .1. 

Corollary 0.4. If G is an abelian group then the product of two subgroups

is always a subgroup.

Example-13.3: Let

U

20

H = {

and

K = {

Check that ยฏ7 = 11 17 = 19 13 = 11 ยฏ9(ยฏ9)

โˆ’ 1

Theorem 0.5. If H and K are finite subgroups of G, then

o(HK) =

o(H)o(K)

o(H โˆฉ K

(2) gN g

โˆ’ 1

= N for every g โˆˆ G.

(3) Operation on the set of left cosets of N on G by aN bN = abN is well

defined. (i.e. if a 1

N = a 2

N and b 1

H = b 2

H, then a 1

b 1

N = a 2

b 2

N ).

(4) Operation on the set of left cosets of N on G by aN bN = abN gives

group structure on the set of left cosets of N on G.

(5) Operation on the set of right cosets of N on G by N aN b = N ab

is well defined (i.e. if N a 1

= N a 2

and N b 1

= N b 2

, then N a 1

b 1

N a 2

b 2

(6) Operation on the set of right cosets of N on G by N aN b = N ab gives

group structure on the set of right cosets of N on G.

(7) For any a โˆˆ G, aN = N a.

Proof..

Let N is normal. Let g โˆˆ G, by normality of N we get gng

โˆ’ 1

โˆˆ N

for every n โˆˆ N. So we get gN g

โˆ’ 1

โŠ‚ N. Now we show N โŠ‚ gN g

โˆ’ 1

.

Let n โˆˆ N , again the normality of N implies that g

โˆ’ 1

n(g

โˆ’ 1

)

โˆ’ 1

= k โˆˆ

N. So we get n = gkg

โˆ’ 1

โˆˆ gN g

โˆ’ 1

which implies that gN g

โˆ’ 1

โŠ‚ N.

Let g โˆˆ G and n โˆˆ N. Given that gN g

โˆ’ 1

= N which directly

implies that gng

โˆ’ 1

โˆˆ N.

Suppose left coset multiplication is well defined. I want to show N

is normal G. Let g โˆˆ G and h โˆˆ N. I will show ghg

โˆ’ 1

โˆˆ N. Since

hN = eN , and surely gN = gN , so (since left coset multiplication

is well-defined)

(gN )(hN ) = (gN )(eN )

(gh)N = gN

and by multiplying g

โˆ’ 1

N both side we get,

(gh)N g

โˆ’ 1

N = gN g

โˆ’ 1

H

(ghg

โˆ’ 1

)N = eN = N.

Hence ghg

โˆ’ 1

โˆˆ N for every g โˆˆ G, n โˆˆ N and thus N is normal.

Let N be normal in G.

Claim: The operation on the set of left cosets of N in G well

defined and gives a group structure on it.

(1) Well defined: Let a 1

N = a 2

N and b 1

N = b N

this implies

a 2

= a 1

n, b 2

= b 1

k

where n, k โˆˆ N. Now

a 2

b 2

= a 1

nb 1

k.

Since N is normal so b

โˆ’ 1

1

nb 1

โˆˆ N , let b

โˆ’ 1

1

nb 1

= l โˆˆ N. Now

a 2

b 2

= a 1

b 1

lk,

this implies that

a 2

b 2

N = a 1

b 1

N (why?).

(Remark: a = bN โ‡” aN = bN for a, b โˆˆ N )

(2) Associativity:

(aN bN )cN = (ab)N cN = (abc)N

and

aN (bN cN ) = aN (bc)N = (abc)N.

(3) Identity: N = eN is the identity.

aN.eN = a.eN = aN = eN.aN

(4) Inverse: For aN clearly a

โˆ’ 1

N is the inverse of aH.

aHa

โˆ’ 1

= aa

โˆ’ 1

N โˆ’ eN = N = a

โˆ’ 1

N aN

Remark 0.9. It is abelian if G is abelian.

Let a โˆˆ G and an โˆˆ aN , since ana

โˆ’ 1

โˆˆ N implies that an โˆˆ N a.

So we get aN โŠ‚ N a, similarly we can prove N a โŠ‚ aN. Hence

aN = N a.

Exercise-13.1: Let G be a group and H be a subgroup G and fix g โˆˆ G.

(1) Prove that gHg

โˆ’ 1

is a subgroup of G of the same order as H.