Definition and examples – Sub groups – Homomorphism – Cosets – Normal Subgroups. examples,theorems,Basic Definitions – Paths, Reachability Connectedness – Matrix Representation of graphs – Trees.Introduction – alphabet, words, languages – regular expressions, regular languages - Finite state Automata – Grammars – Gödel Numbers.

1. M.VIJAYALAKSHMI
ASSISTANT PROFESSOR
DEPARTMENT OFMATHEMATICS
SRI SARADA NIKETAN COLLEGE OF SCIENCE
FOR WOMEN ,KARUR-5
MATHEMATICAL FOUNDATIONS FOR
COMPUTER APPLICATIONS –UNIT -3

2. CONTENT
 Sub groups
 Cosets
 Normal subgroups.

3. subgroup
 A nonempty subset H of the group G is a
subgroup of G if H is a group under the binary
operation (*) of G. We use the notation H ≤ G to
indicate that H is a subgroup of G. Also, if H is
a proper subgroup then it is denoted by H < G.
 For a subset H of group G, H is a subgroup of G
if,
 H ≠ φ
 if a, k H then ak H
∈ ∈
 if a H then a
∈ -1
H
∈

4. Examples of Subgroup
Some examples of subgroups are listed as
follows:
Integer under Addition (Z, +): The set of even
integers is a subgroup of the group of all
integers under addition.
Modular Arithmetic: In modular arithmetic,
the set of integers modulo n forms a group,
and the set of integers that are multiples of a
divisor d of n forms a subgroup.

5. Properties of Subgroups
Some of the common properties of subgroups are:
 G is a subgroup of itself and {e} is also a subgroup of
G, these are called trivial subgroups.
 Subgroup will have all the properties of a group.
 A subgroup H of the group G is a normal
subgroup if g-1
Hg = H for all g G.
∈
 If H < K and K < G, then H < G
(subgrouptransitivity).
 If H and K are subgroups of a group G then
H ∩ K is also a subgroup.
 If H and K are subgroups of a group G then H K
∪
may or may not be a subgroup.

6. COSETS
Definition:
LEFT COSET
Let H be a subgroup of a group (G,o). If 𝑎
then the subset aoH of G defined by aoH
∈ 𝐺
= { aoh : } is called a left coset of H in
ℎ ∈ 𝐻
G determined by element .
𝑎 ∈ 𝐺
RIGHT COSET
The subset Hoa of G defined by Hoa = { hoa :
} is called a right coset of H in G
ℎ ∈ 𝐻
determined by element .
𝑎 ∈ 𝐺

7. Properties of Cosets:
Theorem: 1
If G is an abelian group and then aH = Ha.
𝑎 ∈ 𝐺
Proof: Let . Then x = ha for some .
𝑥 ∈ 𝐻𝑎 ℎ ∈ 𝐻
As . Again, and G is
ℎ ∈ 𝐻 ⇒ ℎ ∈ 𝐺 𝑎 ∈ 𝐺
abelian, ha = ah x = ah for some .
⇒ ℎ ∈ 𝐻 ⇒ 𝑥
. Thus, Ha Similarly, if .
∈ 𝑎 𝐻 ⊆ 𝑎𝐻 𝑥 ∈ 𝑎𝐻
Then x = ah for some . As
ℎ ∈ 𝐻 ℎ ∈ 𝐻 ⇒ ℎ
. Again, and G is abelian, we have, ah
∈ 𝐺 𝑎 ∈ 𝐺
= ha x = ha for some G
⇒ ℎ ∈
⇒ 𝑥 ∈ 𝐻𝑎.Thus, aH . Hence, a H = Ha.
⊆ 𝐻𝑎

8. Theorem:2
If H is a subgroup of G and a, b , then (i)
∈ 𝐺
Ha = H if and only if a (ii) aH = H if and
∈ 𝐻
only if a (iii) Ha = Hb if only if −1
∈ 𝐻 𝑎𝑏
(iv) aH = bH if and only if −1 .
∈ 𝐻 𝑎 𝑏 ∈ 𝐻
Proof:
(i) Firstly, suppose Ha = H. As H is subgroup of
G, so Thus,
𝑒 ∈ 𝐻 𝑒𝑎 ∈ 𝐻𝑎 ⇒ 𝑎 ∈ 𝐻𝑎 ⇒ 𝑎
. Hence, Ha = H . Conversely,
∈ 𝐻 ⇒𝑎 ∈ 𝐻
suppose . To prove Ha = H.
𝑎 ∈ 𝐻

9. Let = . Now,
𝑥 ∈ 𝐻𝑎 ⇒ 𝑥 ℎ𝑎 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 ℎ ∈ 𝐻
, . This shows
ℎ 𝑎 ∈ 𝐻 ⇒ℎ𝑎 ∈ 𝐻 ⇒𝑥 ∈ 𝐻
that . …..……
𝑥 ∈ 𝐻𝑎 ⇒ 𝑥 ∈ 𝐻 ⇒ 𝐻𝑎 ⊆ 𝐻
(eq. 1) Now, take .
𝑥 ∈ 𝐻
−
𝐺𝑖𝑣𝑒𝑛 𝑎 ∈ 𝐻 ⇒𝑥𝑎 1 . ( −1 )
∈ 𝐻 ⇒ 𝑥𝑎 𝑎
( −1 ) .
∈ 𝐻𝑎 ⇒𝑥 𝑎 𝑎 ∈ 𝐻𝑎 ⇒𝑥 𝑒 ∈ 𝐻𝑎 ⇒𝑥
This proves that if
∈ 𝐻𝑎 𝑥 ∈ 𝐻 ⇒𝑥 ∈ 𝐻𝑎
. ……..(eq. 2) From eq. (1) & (2),
⇒𝐻 ⊆ 𝐻𝑎
we have Ha = H. (ii) Proof is similar to (i).

10. (iii) Firstly, suppose Ha = Hb. Now , as H is a
𝑒 ∈ 𝐻
subgroup of G , ,
⇒ 𝑒𝑎 ∈ 𝐻𝑎 𝑎 ∈ 𝐻𝑎 ⇒ 𝑎 ∈ 𝐻𝑏
since Ha = Hb = −1 =
⇒ 𝑎 ℎ𝑏 𝑓𝑜𝑟 ℎ ∈ 𝐻 ⇒ 𝑎𝑏
( ) −1 = ( −1 ) = = Thus, −1
ℎ𝑏 𝑏 ℎ 𝑏𝑏 ℎ 𝑒 ℎ 𝑎𝑏
.
∈ 𝐻
Conversely,
suppose −1 . Therefore, −1 =
𝑎𝑏 ∈ 𝐻 𝑎𝑏 ℎ 𝑓𝑜𝑟
. ( −1 ) = ( −1 ) =
𝑠𝑜𝑚𝑒 ℎ ∈ 𝐻 ⇒ 𝑎𝑏 𝑏 ℎ𝑏 ⇒ 𝑎 𝑏 𝑏
= . Thus, Ha = H (hb) = (Hh)b = Hb.
ℎ 𝑏 ⇒ 𝑎 ℎ𝑏
(iv) Proof is similar to (iii).

11.  Theorem: 2
 Prove that any two right(left) cosets of a subgroup are
either disjoint or identical.
 Proof:
 Let H be a subgroup of a group G and , . Let
𝑎 𝑏 ∈ 𝐺
Ha and Hb be two right cosets of H in G. We have to
show that either Ha = Hb or ∩ = . Case 1:
𝑯𝒂 𝑯𝒃 ∅
If ∩ = , then nothing to prove. Case 2:
𝐻𝑎 𝐻𝑏 ∅
Let ∩ ≠ . We have to show that Ha = Hb.
𝐻𝑎 𝐻𝑏 ∅
Since ∩ ≠ , so there exist atleast one
𝐻𝑎 𝐻𝑏 ∅
element ∩ &
𝑥 ∈ 𝐻 𝑎 𝐻 𝑏 ⇒ 𝑥 ∈ 𝐻 𝑎 𝑥 ∈ 𝐻 𝑏
= 1 1 & = 2
⇒𝑥 ℎ 𝑎 𝑓 𝑜𝑟 𝑠𝑜𝑚𝑒 ℎ ∈ 𝐻 𝑥 ℎ 𝑏 𝑓 𝑜𝑟
2 h
𝑠𝑜𝑚𝑒 ℎ ∈

12.  Thus, 1 = 2 1 1 ( 1 ) = 1
ℎ 𝑎 ℎ 𝑏 ⇒ ℎ − ℎ 𝑎 ℎ
1 ( 2 ) ( 1 1 1) = ( 1
− ℎ 𝑏 ⇒ ℎ − ℎ 𝑎 ℎ
1 2) = 3 , 3 =
− ℎ 𝑏 ⇒ 𝑒𝑎 ℎ 𝑏 𝑤ℎ𝑒𝑟𝑒 ℎ
( 1 1 2) = 3
ℎ − ℎ ∈ 𝐻 ⇒ 𝑎 ℎ 𝑏 ⇒ 𝐻𝑎
= ( 3 ) = ( 3 ) =
𝐻 ℎ 𝑏 𝐻ℎ 𝑏 𝐻𝑏 𝑠𝑖𝑛𝑐𝑒
3 = . Hence, Ha = Hb.Thus, if
𝐻ℎ 𝐻 𝐻𝑎
≠ , then Ha = Hb. So, either Ha
∩ 𝐻𝑏 ∅
= Hb or =
𝐻𝑎 ∩ 𝐻𝑏 ∅

13. Normal Subgroups
A subgroup H of G is called a normal subgroup of G if
every left coset of H in G is equal to the corresponding
right coset of h in G. i.e., aH = Ha, for all .
𝑎 ∈ 𝐺
Note
(i) If (G,+) is an additive group and H is called normal
subgroup of G iff a + H = H + a for all .
𝑎 ∈ 𝐺
(ii) (ii) If G is an Abelian group then every subgroup H
of G is a normal subgroup.
(iii) (iii) The subgroups {e} and G of any group G are
always normal subgroups of G. These are called
trivial normal subgroups.

14. Theorem: 1
A subgroup H of G is a normal subgroup of G if and only if 𝑔ℎ𝑔
−1 , .
∈ 𝐻 ∀ ℎ ∈ 𝐻 𝑔 ∈ 𝐺
Proof:
Firstly, suppose H is a normal subgroup of G.
Therefore, = . Let , .
𝑔𝐻 𝐻𝑔 ∀ 𝑔 ∈ 𝐺 ℎ ∈ 𝐻 𝑔 ∈ 𝐺
Then = .
𝑔ℎ ∈ 𝑔𝐻 𝐻𝑔 ⇒ 𝑔𝐻 ∈ 𝐻𝑔
This implies that = 1 for some 1 −1 = 1
𝑔ℎ ℎ 𝑔 ℎ ∈ 𝐻 ⇒ 𝑔ℎ𝑔 ℎ ∈ 𝐻
−1 .
⇒ 𝑔ℎ𝑔 ∈ 𝐻
Conversely,
suppose H is a subgroup of G such that −1 ,
𝑔ℎ𝑔 ∈ 𝐻 ∀ ℎ ∈ 𝐻 𝑔
.
∈ 𝐺
We have to show that H is a normal subgroup, i.e., =
𝑎 𝐻 𝐻𝑎 ∀ 𝑎 ∈
G

15. Let .
𝑎 ∈ 𝐺
Then by given condition −1 .
𝑎ℎ𝑎 ∈ 𝐻 ∀ ℎ ∈ 𝐻
Suppose . Then = ( −1 )
𝑎ℎ ∈ 𝑎𝐻 𝑎𝐻 𝑎𝐻𝑎 𝑎 ∈ 𝐻𝑎 ⇒
… . (1)
𝑎ℎ ∈ 𝐻𝑎 ⇒ 𝑎𝐻 ⊂ 𝐻𝑎
Again, let = −1 .
𝑏 𝑎 ∈ 𝐺
Then by given condition −1 .
𝑏ℎ𝑏 ∈ 𝐻
But −1 = −1 ( −1 ) −1 = −1 .
𝑏ℎ𝑏 𝑎 ℎ 𝑎 𝑎 ℎ𝑎 ∈ 𝐻
Let .
ℎ𝑎 ∈ 𝐻𝑎
Then = ( −1 ) = ( −1 )
ℎ𝑎 𝑎𝑎 ℎ𝑎 𝑎 𝑎 ℎ𝑎 ∈ 𝑎 𝐻 ⇒ ℎ𝑎 ∈
…….(2) From (1) and (2),
𝑎 𝐻 ⇒ 𝐻𝑎 ⊂ 𝑎 𝐻
we get = PROF Hence, H is a normal
𝑎𝐻 𝐻𝑎 ∀ 𝑎 ∈ 𝐺
subgroup of G

16. Theorem: 2
Let H be subgroup of a group G.
Then the following are equivalent:
( ) −1 , ∀𝑔∈ 𝐺, ℎ∈ 𝐻.
∈∀ ∈∈
( ) −1 = , . ( ) = .
𝑖𝑖 𝑔𝐻𝑔 𝐻 ∀ 𝑔 ∈ 𝐺 𝑖𝑖𝑖 𝑔𝐻 𝐻𝑔 ∀ 𝑔 ∈ 𝐺
Proof:
( ) ( ) Given −1 𝐻 , ∀ 𝑔 ∈ 𝐺, ℎ∈ 𝐻 .
⇒ ∈ ∀ ∈ ∈
Let −1 = 1 1 , −1 =
𝑔ℎ𝑔 ℎ ∀ ℎ ∈ 𝐻 ⇒ 𝑔𝐻𝑔 𝐻 ∀ 𝑔
𝐺.
∈
( ) ( ) Given −1 = ,
𝒊𝒊 ⇒ 𝒊𝒊𝒊 𝑔𝐻𝑔 𝐻 ∀ 𝑔 ∈ 𝐺 ⇒
( −1 ) = , G
𝑔𝐻𝑔 𝑔 𝐻𝑔 ∀ 𝑔 ∈

17. ⇒ 𝑔𝐻( −1 ) = ,
𝑔 𝑔 𝐻𝑔 ∀ 𝑔 ∈ 𝐺 ⇒ 𝑔𝐻𝑒
= , = ,
𝐻𝑔 ∀ 𝑔 ∈ 𝐺 ⇒ 𝑔𝐻 𝐻𝑔 ∀ 𝑔 ∈ 𝐺
( ) ( ) = ,
𝑖𝑖𝑖 ⇒ 𝑖 𝐺𝑖𝑣𝑒𝑛 𝑔𝐻 𝐻𝑔 ∀ 𝑔 ∈ 𝐺 ⇒ 𝑔
= 1 , 1 −1 = 1
ℎ ℎ 𝑔 ∀ ℎ ℎ ∈ 𝐻 ⇒ 𝑔 ℎ𝑔 ℎ ∈
−1 .
𝐻 ⇒ 𝑔 ℎ𝑔 ∈ 𝐻
Hence, the theorem.

18. THANK YOU