Algebra: Section - 1 (Groups) | Mathematics Optional Notes for UPSC PDF Download

 Page 1


Edurev123 
ALGEBRA 
SECTION -
 
1	
 
GROUPS
 
PRELIMINARIES
 
We briefly recall some notations and definitions from set Theory.
 
?? ?B
 
: Union of ?? 
and ?? 
?? n?? 
: Intersection of ?? 
and ?? 
?? -?? ={?? ??? :?? ??? }
 
?? =
 
empty set
 
?? ×?? 
(Cartesian) product of ordered pairs {(?? ,?? ),?? ??? ,?? ??? 
In particular?? ×?? ={(?? ?? ,?? ?? ):?? ?? ,?? ?? ˜?? ,?? =1,2,…,?? ,  ?? =1,2,…,?? }
 
The subset
 
?? ?? ??? ×?? 
defined by
 
?? ?? ={(?? ,?? ):?? ??? }
 
is called the diagonal of
 
?? ×?? 
 
Consider a subset ?? ??? ×?? having the properties: 
(1)   1
°
(?? ,?? )??? ,? a in ?? , i.e., the diagonal ?? ?? ??? 
        2
°
(?? ,?? )??? ?(?? ,?? )??? 
        3
°
(?? ,?? ) and (?? ,?? ) in ?? ?(?? ,?? )??? 
After fixing such a subset of ?? ×?? , let us say that ?? is equivalent ???? ?? , in notation ?? =?? 
or ?? -?? (?? ) if and only if (???,?? )=?? 
Then the conditions 1
°
,2
°
,3
0
 above can be alternatively written as 
(2) 1
°
a=a 
2
°
?? =?? ??? =?? 
3
°
?? =?? and ?? =?? ,??? =?? 
When we have fixed ?? , and are nol considering any other such subset we omit ?? from 
the notation we shall say that the subset ?? ??? ×?? in (1) defines the equivalence 
relation (2) among the elements of ?? . 
Page 2


Edurev123 
ALGEBRA 
SECTION -
 
1	
 
GROUPS
 
PRELIMINARIES
 
We briefly recall some notations and definitions from set Theory.
 
?? ?B
 
: Union of ?? 
and ?? 
?? n?? 
: Intersection of ?? 
and ?? 
?? -?? ={?? ??? :?? ??? }
 
?? =
 
empty set
 
?? ×?? 
(Cartesian) product of ordered pairs {(?? ,?? ),?? ??? ,?? ??? 
In particular?? ×?? ={(?? ?? ,?? ?? ):?? ?? ,?? ?? ˜?? ,?? =1,2,…,?? ,  ?? =1,2,…,?? }
 
The subset
 
?? ?? ??? ×?? 
defined by
 
?? ?? ={(?? ,?? ):?? ??? }
 
is called the diagonal of
 
?? ×?? 
 
Consider a subset ?? ??? ×?? having the properties: 
(1)   1
°
(?? ,?? )??? ,? a in ?? , i.e., the diagonal ?? ?? ??? 
        2
°
(?? ,?? )??? ?(?? ,?? )??? 
        3
°
(?? ,?? ) and (?? ,?? ) in ?? ?(?? ,?? )??? 
After fixing such a subset of ?? ×?? , let us say that ?? is equivalent ???? ?? , in notation ?? =?? 
or ?? -?? (?? ) if and only if (???,?? )=?? 
Then the conditions 1
°
,2
°
,3
0
 above can be alternatively written as 
(2) 1
°
a=a 
2
°
?? =?? ??? =?? 
3
°
?? =?? and ?? =?? ,??? =?? 
When we have fixed ?? , and are nol considering any other such subset we omit ?? from 
the notation we shall say that the subset ?? ??? ×?? in (1) defines the equivalence 
relation (2) among the elements of ?? . 
Conversely, if ? an equivalence relation = (or ~) on ?? 1
 the corresponding in subset ?? ?
?? ×?? is defined by 
(3) ?? ={(?? ,?? ):?? =?? } 
DEFINITION: If ?? is a set ; io = is an equivalence relation ?? on ?? 1
, then, for given ?? ??? ?? 
[?? }
R
:={?? ??? ,?? :?? (?? )} 
(We omit ?? in the notation). [a] as the equivalence class corresponding to ?? . Since ?? =?? 
[?? ]
p 
 must always contain ?? , and 
?? ?[?? ]
?? ??? =?? (?? ) 
Basic fact 
(*) Two equivalence classes must either be disjoint or must completely coincide. 
Proof 
Let [a] and [b] be NOT disjoint. Let ?? 0
?[?? ]n[?? ], then 
(4) ?? 0
=?? and ?? 0
=?? 
We assert  [a]=[b] 
In fact, let ?? ?[?? ], then ?? =?? , from (4) 
?? =?? =?? ?? ??? , i.e., ?? =?? ,?? ?[?? ] 
whence 
{?? ]?[?? ] 
Similarly, [?? ]?[?? ]. This proves (*) 
consequently, we can state  
Any equivalence relation on a set A yields us a partition of A into distinct equivalence 
classes. 
Given two sets ?? and ?? , we say that ?? :?? ??? is a map. 
If corresponding to every ?? ??? , there exists a well-defined element ?? ??? , called the 
image of a' under the map ?? . Denoted by ?? (?? ) the subset of ?? consisting of the images, 
under ?? ,  if the elements of ?? 
?? (?? )??? 
We write 
Page 3


Edurev123 
ALGEBRA 
SECTION -
 
1	
 
GROUPS
 
PRELIMINARIES
 
We briefly recall some notations and definitions from set Theory.
 
?? ?B
 
: Union of ?? 
and ?? 
?? n?? 
: Intersection of ?? 
and ?? 
?? -?? ={?? ??? :?? ??? }
 
?? =
 
empty set
 
?? ×?? 
(Cartesian) product of ordered pairs {(?? ,?? ),?? ??? ,?? ??? 
In particular?? ×?? ={(?? ?? ,?? ?? ):?? ?? ,?? ?? ˜?? ,?? =1,2,…,?? ,  ?? =1,2,…,?? }
 
The subset
 
?? ?? ??? ×?? 
defined by
 
?? ?? ={(?? ,?? ):?? ??? }
 
is called the diagonal of
 
?? ×?? 
 
Consider a subset ?? ??? ×?? having the properties: 
(1)   1
°
(?? ,?? )??? ,? a in ?? , i.e., the diagonal ?? ?? ??? 
        2
°
(?? ,?? )??? ?(?? ,?? )??? 
        3
°
(?? ,?? ) and (?? ,?? ) in ?? ?(?? ,?? )??? 
After fixing such a subset of ?? ×?? , let us say that ?? is equivalent ???? ?? , in notation ?? =?? 
or ?? -?? (?? ) if and only if (???,?? )=?? 
Then the conditions 1
°
,2
°
,3
0
 above can be alternatively written as 
(2) 1
°
a=a 
2
°
?? =?? ??? =?? 
3
°
?? =?? and ?? =?? ,??? =?? 
When we have fixed ?? , and are nol considering any other such subset we omit ?? from 
the notation we shall say that the subset ?? ??? ×?? in (1) defines the equivalence 
relation (2) among the elements of ?? . 
Conversely, if ? an equivalence relation = (or ~) on ?? 1
 the corresponding in subset ?? ?
?? ×?? is defined by 
(3) ?? ={(?? ,?? ):?? =?? } 
DEFINITION: If ?? is a set ; io = is an equivalence relation ?? on ?? 1
, then, for given ?? ??? ?? 
[?? }
R
:={?? ??? ,?? :?? (?? )} 
(We omit ?? in the notation). [a] as the equivalence class corresponding to ?? . Since ?? =?? 
[?? ]
p 
 must always contain ?? , and 
?? ?[?? ]
?? ??? =?? (?? ) 
Basic fact 
(*) Two equivalence classes must either be disjoint or must completely coincide. 
Proof 
Let [a] and [b] be NOT disjoint. Let ?? 0
?[?? ]n[?? ], then 
(4) ?? 0
=?? and ?? 0
=?? 
We assert  [a]=[b] 
In fact, let ?? ?[?? ], then ?? =?? , from (4) 
?? =?? =?? ?? ??? , i.e., ?? =?? ,?? ?[?? ] 
whence 
{?? ]?[?? ] 
Similarly, [?? ]?[?? ]. This proves (*) 
consequently, we can state  
Any equivalence relation on a set A yields us a partition of A into distinct equivalence 
classes. 
Given two sets ?? and ?? , we say that ?? :?? ??? is a map. 
If corresponding to every ?? ??? , there exists a well-defined element ?? ??? , called the 
image of a' under the map ?? . Denoted by ?? (?? ) the subset of ?? consisting of the images, 
under ?? ,  if the elements of ?? 
?? (?? )??? 
We write 
= domain (?? ),?? =co -domain (?? ),?? (?? )=lm (?? ) Range ( f) 
DEFINITION: Given: a map f:A?B. 
 
1
0
?? is called a (one - one) map, or injective if  (5) ?? (?? 1
)=(?? 2
)??? 1
=?? 2
, 
2
*
 is called an ONTO map or surjective if  ?? (?? )=?? . 
3
°
f is called bijective, if f is both injective and surjective. 
If f
i
:A?B(?? =1,2) are two maps, we say: 
?? 1
=?? 2
??? 1
(?? )=?? 2
(?? ) for every a in A.  
If we have maps ?? and ?? such that 
A? B??
2
C 
Wo denotes by ?? . ?? the composition map 
(6) A?C 
Then 1
°
f,g surjective ? g.f surjective  
(7) 
2
°
f.?? injectivs =?? .?? injective 
 
Proof of ?? °
 
Put h=?? .?? . To prove 1
0
 we must show ?? = on h, tet ?? ??? . since ?? is surjective  ??? in ?? 
with ?? =?? (?? ) , since ?? is surjective, ??? in ?? with ?? =?? (??) . Then 
?? =?? (?? )=?? (?? (?? ))?? (?? ,?? )??? )=h(?? ) 
Hence h is surjective. 
 
Proof of ?? ?? :  
Suppose h(?? )=h(?? 2
) o ( 1 a ?? (?? (?? 1
))=?? (?? (?? 2
))· ?? -6 
Page 4


Edurev123 
ALGEBRA 
SECTION -
 
1	
 
GROUPS
 
PRELIMINARIES
 
We briefly recall some notations and definitions from set Theory.
 
?? ?B
 
: Union of ?? 
and ?? 
?? n?? 
: Intersection of ?? 
and ?? 
?? -?? ={?? ??? :?? ??? }
 
?? =
 
empty set
 
?? ×?? 
(Cartesian) product of ordered pairs {(?? ,?? ),?? ??? ,?? ??? 
In particular?? ×?? ={(?? ?? ,?? ?? ):?? ?? ,?? ?? ˜?? ,?? =1,2,…,?? ,  ?? =1,2,…,?? }
 
The subset
 
?? ?? ??? ×?? 
defined by
 
?? ?? ={(?? ,?? ):?? ??? }
 
is called the diagonal of
 
?? ×?? 
 
Consider a subset ?? ??? ×?? having the properties: 
(1)   1
°
(?? ,?? )??? ,? a in ?? , i.e., the diagonal ?? ?? ??? 
        2
°
(?? ,?? )??? ?(?? ,?? )??? 
        3
°
(?? ,?? ) and (?? ,?? ) in ?? ?(?? ,?? )??? 
After fixing such a subset of ?? ×?? , let us say that ?? is equivalent ???? ?? , in notation ?? =?? 
or ?? -?? (?? ) if and only if (???,?? )=?? 
Then the conditions 1
°
,2
°
,3
0
 above can be alternatively written as 
(2) 1
°
a=a 
2
°
?? =?? ??? =?? 
3
°
?? =?? and ?? =?? ,??? =?? 
When we have fixed ?? , and are nol considering any other such subset we omit ?? from 
the notation we shall say that the subset ?? ??? ×?? in (1) defines the equivalence 
relation (2) among the elements of ?? . 
Conversely, if ? an equivalence relation = (or ~) on ?? 1
 the corresponding in subset ?? ?
?? ×?? is defined by 
(3) ?? ={(?? ,?? ):?? =?? } 
DEFINITION: If ?? is a set ; io = is an equivalence relation ?? on ?? 1
, then, for given ?? ??? ?? 
[?? }
R
:={?? ??? ,?? :?? (?? )} 
(We omit ?? in the notation). [a] as the equivalence class corresponding to ?? . Since ?? =?? 
[?? ]
p 
 must always contain ?? , and 
?? ?[?? ]
?? ??? =?? (?? ) 
Basic fact 
(*) Two equivalence classes must either be disjoint or must completely coincide. 
Proof 
Let [a] and [b] be NOT disjoint. Let ?? 0
?[?? ]n[?? ], then 
(4) ?? 0
=?? and ?? 0
=?? 
We assert  [a]=[b] 
In fact, let ?? ?[?? ], then ?? =?? , from (4) 
?? =?? =?? ?? ??? , i.e., ?? =?? ,?? ?[?? ] 
whence 
{?? ]?[?? ] 
Similarly, [?? ]?[?? ]. This proves (*) 
consequently, we can state  
Any equivalence relation on a set A yields us a partition of A into distinct equivalence 
classes. 
Given two sets ?? and ?? , we say that ?? :?? ??? is a map. 
If corresponding to every ?? ??? , there exists a well-defined element ?? ??? , called the 
image of a' under the map ?? . Denoted by ?? (?? ) the subset of ?? consisting of the images, 
under ?? ,  if the elements of ?? 
?? (?? )??? 
We write 
= domain (?? ),?? =co -domain (?? ),?? (?? )=lm (?? ) Range ( f) 
DEFINITION: Given: a map f:A?B. 
 
1
0
?? is called a (one - one) map, or injective if  (5) ?? (?? 1
)=(?? 2
)??? 1
=?? 2
, 
2
*
 is called an ONTO map or surjective if  ?? (?? )=?? . 
3
°
f is called bijective, if f is both injective and surjective. 
If f
i
:A?B(?? =1,2) are two maps, we say: 
?? 1
=?? 2
??? 1
(?? )=?? 2
(?? ) for every a in A.  
If we have maps ?? and ?? such that 
A? B??
2
C 
Wo denotes by ?? . ?? the composition map 
(6) A?C 
Then 1
°
f,g surjective ? g.f surjective  
(7) 
2
°
f.?? injectivs =?? .?? injective 
 
Proof of ?? °
 
Put h=?? .?? . To prove 1
0
 we must show ?? = on h, tet ?? ??? . since ?? is surjective  ??? in ?? 
with ?? =?? (?? ) , since ?? is surjective, ??? in ?? with ?? =?? (??) . Then 
?? =?? (?? )=?? (?? (?? ))?? (?? ,?? )??? )=h(?? ) 
Hence h is surjective. 
 
Proof of ?? ?? :  
Suppose h(?? )=h(?? 2
) o ( 1 a ?? (?? (?? 1
))=?? (?? (?? 2
))· ?? -6 
Since ?? is injective, this gives ?? (?? 1
)=?? (?? 2
) . Since ?? is injective, we get ?? 1
=?? 2
. 
Consequently, h is injective.  
 
Example 1 
Prove that a map ???? ? is, injective if and only if ?? can be left-cancelled in the sense that 
f.g=?? ·h?g?h  
?? is surjective if and only if it can be right-cancelled in the sense that 
?? .?? =?? .?? ??? =?? 
(a) INJECTIVE: 
(i) injective ? left cancellation law holds 
Suppose ?? injective, and f.g. = f.h. Then, for every ?? in ?? 
 
 (?? .?? )(?? )=(?? .h)(?? )
?? (?? (?? ))=?? (h(?? ))
 
since ?? is injective, we get ?? (?? )=h(?? ) , since this holds for every ?? in ?? ,?? =h, 
 i.e. ?? .?? =?? h??? =h.  
(ii) Conversely left cancellation law holds ??? injective. Suppose, on the contrary, ?? is 
not injective: ??? 1
 and ?? 2
 in ?? ,
 ?? 1
??? 2
, such that 
(8) ?? (?? 1
)=?? (?? 2
'
) 
Choose the maps g an??  h:X?X 
such that  ?? (?? )={?? 1
}             h(?? )={?? 2
} 
i.e., ?? (?? )=?? 1
,h(?? )=x
2
        … (9) 
for all ?? ??? . 
Page 5


Edurev123 
ALGEBRA 
SECTION -
 
1	
 
GROUPS
 
PRELIMINARIES
 
We briefly recall some notations and definitions from set Theory.
 
?? ?B
 
: Union of ?? 
and ?? 
?? n?? 
: Intersection of ?? 
and ?? 
?? -?? ={?? ??? :?? ??? }
 
?? =
 
empty set
 
?? ×?? 
(Cartesian) product of ordered pairs {(?? ,?? ),?? ??? ,?? ??? 
In particular?? ×?? ={(?? ?? ,?? ?? ):?? ?? ,?? ?? ˜?? ,?? =1,2,…,?? ,  ?? =1,2,…,?? }
 
The subset
 
?? ?? ??? ×?? 
defined by
 
?? ?? ={(?? ,?? ):?? ??? }
 
is called the diagonal of
 
?? ×?? 
 
Consider a subset ?? ??? ×?? having the properties: 
(1)   1
°
(?? ,?? )??? ,? a in ?? , i.e., the diagonal ?? ?? ??? 
        2
°
(?? ,?? )??? ?(?? ,?? )??? 
        3
°
(?? ,?? ) and (?? ,?? ) in ?? ?(?? ,?? )??? 
After fixing such a subset of ?? ×?? , let us say that ?? is equivalent ???? ?? , in notation ?? =?? 
or ?? -?? (?? ) if and only if (???,?? )=?? 
Then the conditions 1
°
,2
°
,3
0
 above can be alternatively written as 
(2) 1
°
a=a 
2
°
?? =?? ??? =?? 
3
°
?? =?? and ?? =?? ,??? =?? 
When we have fixed ?? , and are nol considering any other such subset we omit ?? from 
the notation we shall say that the subset ?? ??? ×?? in (1) defines the equivalence 
relation (2) among the elements of ?? . 
Conversely, if ? an equivalence relation = (or ~) on ?? 1
 the corresponding in subset ?? ?
?? ×?? is defined by 
(3) ?? ={(?? ,?? ):?? =?? } 
DEFINITION: If ?? is a set ; io = is an equivalence relation ?? on ?? 1
, then, for given ?? ??? ?? 
[?? }
R
:={?? ??? ,?? :?? (?? )} 
(We omit ?? in the notation). [a] as the equivalence class corresponding to ?? . Since ?? =?? 
[?? ]
p 
 must always contain ?? , and 
?? ?[?? ]
?? ??? =?? (?? ) 
Basic fact 
(*) Two equivalence classes must either be disjoint or must completely coincide. 
Proof 
Let [a] and [b] be NOT disjoint. Let ?? 0
?[?? ]n[?? ], then 
(4) ?? 0
=?? and ?? 0
=?? 
We assert  [a]=[b] 
In fact, let ?? ?[?? ], then ?? =?? , from (4) 
?? =?? =?? ?? ??? , i.e., ?? =?? ,?? ?[?? ] 
whence 
{?? ]?[?? ] 
Similarly, [?? ]?[?? ]. This proves (*) 
consequently, we can state  
Any equivalence relation on a set A yields us a partition of A into distinct equivalence 
classes. 
Given two sets ?? and ?? , we say that ?? :?? ??? is a map. 
If corresponding to every ?? ??? , there exists a well-defined element ?? ??? , called the 
image of a' under the map ?? . Denoted by ?? (?? ) the subset of ?? consisting of the images, 
under ?? ,  if the elements of ?? 
?? (?? )??? 
We write 
= domain (?? ),?? =co -domain (?? ),?? (?? )=lm (?? ) Range ( f) 
DEFINITION: Given: a map f:A?B. 
 
1
0
?? is called a (one - one) map, or injective if  (5) ?? (?? 1
)=(?? 2
)??? 1
=?? 2
, 
2
*
 is called an ONTO map or surjective if  ?? (?? )=?? . 
3
°
f is called bijective, if f is both injective and surjective. 
If f
i
:A?B(?? =1,2) are two maps, we say: 
?? 1
=?? 2
??? 1
(?? )=?? 2
(?? ) for every a in A.  
If we have maps ?? and ?? such that 
A? B??
2
C 
Wo denotes by ?? . ?? the composition map 
(6) A?C 
Then 1
°
f,g surjective ? g.f surjective  
(7) 
2
°
f.?? injectivs =?? .?? injective 
 
Proof of ?? °
 
Put h=?? .?? . To prove 1
0
 we must show ?? = on h, tet ?? ??? . since ?? is surjective  ??? in ?? 
with ?? =?? (?? ) , since ?? is surjective, ??? in ?? with ?? =?? (??) . Then 
?? =?? (?? )=?? (?? (?? ))?? (?? ,?? )??? )=h(?? ) 
Hence h is surjective. 
 
Proof of ?? ?? :  
Suppose h(?? )=h(?? 2
) o ( 1 a ?? (?? (?? 1
))=?? (?? (?? 2
))· ?? -6 
Since ?? is injective, this gives ?? (?? 1
)=?? (?? 2
) . Since ?? is injective, we get ?? 1
=?? 2
. 
Consequently, h is injective.  
 
Example 1 
Prove that a map ???? ? is, injective if and only if ?? can be left-cancelled in the sense that 
f.g=?? ·h?g?h  
?? is surjective if and only if it can be right-cancelled in the sense that 
?? .?? =?? .?? ??? =?? 
(a) INJECTIVE: 
(i) injective ? left cancellation law holds 
Suppose ?? injective, and f.g. = f.h. Then, for every ?? in ?? 
 
 (?? .?? )(?? )=(?? .h)(?? )
?? (?? (?? ))=?? (h(?? ))
 
since ?? is injective, we get ?? (?? )=h(?? ) , since this holds for every ?? in ?? ,?? =h, 
 i.e. ?? .?? =?? h??? =h.  
(ii) Conversely left cancellation law holds ??? injective. Suppose, on the contrary, ?? is 
not injective: ??? 1
 and ?? 2
 in ?? ,
 ?? 1
??? 2
, such that 
(8) ?? (?? 1
)=?? (?? 2
'
) 
Choose the maps g an??  h:X?X 
such that  ?? (?? )={?? 1
}             h(?? )={?? 2
} 
i.e., ?? (?? )=?? 1
,h(?? )=x
2
        … (9) 
for all ?? ??? . 
 
Then  (?? .?? )(?? )=?? (?? (?? ))=?? (?? 1
) 
(?? ,h)(?? )=?? (h)(?? ))=?? (?? 2
) 
By (8) 
(?? ,?? )(?? )=?? (?? 1
)=?? (?? 2
)·(?? ,h)(?? ) for ail ?? in ?? 
i.e., ??   
?? .h
???????
?? .?? ?? , ?? .g=?? .h 
But herein, ?? cannot be cancelled otherwise one would get ?? =h, g(?? )=h(?? ) for all ?? 
contradicting (9). (In fact, since ?? 1
??? 2
,?? (?? )?h(?? ) for every ?? in ?? .) 
(b) SURJECTIVE 
(i) f Surjective ? right cancellation holds 
Suppose ?? is surjective, and ?? .?? =?? .?? 
Let ?? be any element in  ?? . Since ?? is surjective, ??? n ?? such that ?? (?? )=?? 
Now 
 (?? .?? )(?? )=?h.?? )(?? (?? (?? ))=h(?? (?? ))
 
?? (?? )=h(?? ) 
Since y was arbitrary this gives ?? =h, 
i.e., g. ?? =h.?? ??? =h