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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 ∈ G : x = hk, h ∈ H, k ∈ 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 ∈ Z : x = 2i + 4j = 2(i + 2j)} = 2Z.
Problem-13.1: Let
H = { 8 a + 14b : a, b ∈ 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) ∈ 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. ¯a
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 Herstein’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 ¯7 = 11 17 = 19 13 = 11 ¯9(¯9)
− 1
Theorem 0.5. If H and K are finite subgroups of G, then
o(HK) =
o(H)o(K)
o(H ∩ K
(2) gN g
− 1
= N for every g ∈ 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 ∈ G, aN = N a.
Proof..
Let N is normal. Let g ∈ G, by normality of N we get gng
− 1
∈ N
for every n ∈ N. So we get gN g
− 1
⊂ N. Now we show N ⊂ gN g
− 1
.
Let n ∈ N , again the normality of N implies that g
− 1
n(g
− 1
)
− 1
= k ∈
N. So we get n = gkg
− 1
∈ gN g
− 1
which implies that gN g
− 1
⊂ N.
Let g ∈ G and n ∈ N. Given that gN g
− 1
= N which directly
implies that gng
− 1
∈ N.
Suppose left coset multiplication is well defined. I want to show N
is normal G. Let g ∈ G and h ∈ N. I will show ghg
− 1
∈ 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
− 1
N both side we get,
(gh)N g
− 1
N = gN g
− 1
H
(ghg
− 1
)N = eN = N.
Hence ghg
− 1
∈ N for every g ∈ G, n ∈ 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 ∈ N. Now
a 2
b 2
= a 1
nb 1
k.
Since N is normal so b
− 1
1
nb 1
∈ N , let b
− 1
1
nb 1
= l ∈ 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 ⇔ aN = bN for a, b ∈ 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
− 1
N is the inverse of aH.
aHa
− 1
= aa
− 1
N − eN = N = a
− 1
N aN
Remark 0.9. It is abelian if G is abelian.
Let a ∈ G and an ∈ aN , since ana
− 1
∈ N implies that an ∈ N a.
So we get aN ⊂ N a, similarly we can prove N a ⊂ aN. Hence
aN = N a.
Exercise-13.1: Let G be a group and H be a subgroup G and fix g ∈ G.
(1) Prove that gHg
− 1
is a subgroup of G of the same order as H.
