FinalalloMorphism

Let 16 , %) and (H 0) ,
be
groups.
A homomorphism # :
-H is a function
that satisfies

& (x
o y) =
P(x)ob(y)fx ,
y
+ 6

EX :
0 (1230 :
, x) >
- (1 ,+ )

X - In X

· (xy) =

(nxy = 1nx +
my = bX +
by
Let p :
/6 ) :
+ (H 0) ,
be a
group homomorphism
Then 1) $(eb) en = +

2) P(g)) =
24(g)3

Proof : 1) & (ex =

ple-ed = dedo P(ed)

24(20)3"d(es) = (e)
=

2) b(g) =
P(g)
Proof : g
.

g" =
ex
P
(g ) EP(g)3

I
=

P(g g ) :
=
p(eb)

↓ (g) od (g) =
eH
 6LnIIR) IR" EIR(303)
1)
>
- =

A- >
def(A) (nonzero)

det AB = det A det B

sign

2) SneE113 ,
de sgns
Sgnd =
sgnW sght

3) 6-H ,
gees Itrivial homomorphism

4) 6 abelian
,
6e 6

geg" / for some nEIN)

homomorphism d 6-H is isomorphism if

:
A an
it is
bijective
&
position Let d : Get be an
isomorphism

1) If
ge
6 , ordd(g) =

ord(g)
2) If K is a
subgroup of 6 with (k) =
then P(K) is a
subgroup of H with
P(k) = n

3) G is abelian E His abelian

4) Gis EXH cyclic

cyclic
is

5) &" :
H -G is an
isomorphism
 Direct product
If (A0) and (B )
, .
are
groups then AXB
is a
group under
(a .. b . ) ·
(a2 , bi) = La , 092 , bobe)

Groups with K elements

2/22x2/2 and /42 2

4
(a , b) + (a , b) =

(2a2b) [1]phasorar
=
(02 , 02)

element has at most

Every order 2

Groups of order 6

2/62 and Sz are not isomorphic

↓ ↓

abelian not a belian

Theorem
Let6 be with elements
a
cyclic group n
then to In 2
G is isomorphic
(bE2(n2)

Isomorphic means there exists an isomorphism

between .
them
Proof : % :
/nz - 6 G = <
g
>

[an] -ga
↑: is well defined

taln [b]n # a-b is divisible

by
=

= a -b = K forsore KEY

ga-bne
ga gb sop([a]n) P([b]n)
=
=

↓ is surjective :

let + 6 there exists OxMCU-1S .

y
t
. .

y
=
gm
y = q(cm)n)
since (z/n21 =
16) = n c + 10

↓ is
injective

isomorphism :

+ b
P[a + b(n
P
=

is
ga =
gag" =
P([a]n)P([b]n)
so an
isomorphism
 Chinese Remainder Theores

Let mine N tuen 21mn & 2ImIXIve

iff gcd(m , n) = 1

Proof :
Suppose that g(d(m n) , /
>

Then km (min) is
m n
= ,

Let (a , b) - 2 1m2 x * /n 2 24 XXo

Ekm (min) 3(a b) ([0]m [0]n) (12a 12b) =

= ,
, ,

(On 0x),

Then 2/m I x *In * has no element of

order mn so is not
cyclic .
So not isomorphic
to 21 mn 2

Ff 1
gcd(m n) =

([13m . [1]n) has order on n

(HW) and hence 2/MIXZ/NI is

cyclic
so 21mn ** [Im2x * 194
 23/10/24
·
Do Exercise 1 of HW6 and P1z(Q8 question
in midterm

Lagrange's theorem format's little theorem

·

Cyclic groups
·

won't have
symmetric groupe any questions
·

no need bijection induction ,

,

-
encledean

cycli groups
-

lagrange
-

applications
-

formatis and ealer theoress

If p is a
prime and
p = 3mod 4

Show that X has solution

= -1 modp no

s t
solution 3 + 4k 4- IN
p
=
,

that X- = -1modp
Suppose
If p divices X then X2
, = 0 modp
Then p does not divide X

Since is (x ,p) 1
p prive , g(d =
 *"
Hence FLT 1
by
=
, X
p
- 1
modp
(x2) = 1
modp
Since - Elmodp
+N
pl
+ 2
( -

= /
+2
modp
=
-

1 = /mudp

2 =
omodp
Contradiction since P33
 Homomorphism
:

Let 6 andH be
groups . A homomorphism is a

function : Get such that (xy)

=
p(x) b(y)

Let G be and NAE

a
group
Then 6/N is a
group.
The map #n : G- 6/N is a
group homomorphisms
gN
+
g

#n (g 92) ..
=

(g gzN)
, (gzN) =
in (g )N(g2)
,

Ex : N = e0396 .
6/N =

Eges] ge 63 ,

=
5593 ge 63 ,

61e3 = C Since
#
= 10 50

- 61ze0

g - Eg3
=
g5203
what about :

616 = 213
6 - 616
g
+
g6
=
e66 =
6
 Kernel
Defi
Let 6 :
6-I be a
homomorphism
The Kernel knd E 6
b(g) en3
:
= =
=
g
Ex : det GLn(IR) >
- IR" =
IRIS03

Kn det =2 At GLn(IR) :
det A =
13

Ex :
sgn
·

Sn-5113

knsgn
=
EdeSn :
Sgn(0) 13 = :
An

EX : 6+ 6/N (Nab)
ge gN knTn =

Egt6
:

gN =
N3 =
N

Properties Kernel of a
group of homomorphism ?
Prop Let 66 ·

6- be a
group homomorphism
Then Ker -6

ProofKerTed(eb) =
e

2) xy- Kerb P(x) b(y)

Let . Then =
en and =
en

Hence P(xy) =

q(x)d(y) =
en
-
+
H =
2H

Hence
Xye Kerd
3) x = kerd .
P(x-) = P(x)" =
en" =
en

so t Kerd

We
prove that Kerd is normal in C

Let
geG and Xe Kerd
↓ (gxg ) b(g)P(x)P(g)
-
=

= (g) e + P(g ) +

=
d(g)d(g)
=
d(gg")
= d(es)
=
CH

Hence
gxg" >
-

Ker
Hence Kerd16
Prop If : Get is a
group of homomorphism
Then o is C Kerd
injective
=
Eec3

&of Assume o is
injective . We show Kerd =
Se03
ext Kerb ples)
since =
en

If Xt Kerd ,
Hence &(x) =
CH =
P(es)
since o is injective X =
ec

Hence Kerd =
Ees3

# suppose Kerd =
Eeu3 We'll .
show o is
injective
let G be such that
x ,
y
d(x)
d(y) =

d(x)(d(y)) " =
eH

P(x)d(y 1) em
-

d(xy -) = e
+

xy = kerp =
Ee63
so
xy ep so Hence o is
injective
y
= x= .
 The first Isomorphism Theorem

Let 6 and H be and 1 6- :

groups a
group
homomorphism
Then
6/erd = $(6)

Proof :
Define Gard :
>
- &(6)

gkerd >
-
P(g)
#
well-defined
is

Suppose g Kerd
,
=

gz Kerd
>
- gz"g ,
t kerd
=> d(gz" g) =
et

> ↓(gz")f(g ) en
-
=
.

=> P(g ) P(gz)" en

=
.

=> P(g ) .
=
P(qz)
⑦ is a
homomorphism
↑ (g Kerd) (gzkerd) = (g
,
.
g2 kerd)
=
p (g 92) ,

=
p(g ) +(2) .

= (g Kerd) (g2 kerP)

,
#
injectie4)
is

= e
+

= P(g) =
en

>
-

ge Kard
E gkerd =
kerd

Ker =

Ekerd3 =
es/kerd
so ot is
injective

↑ is surjective
let
ye
$(6) , then y =
&(x)
for some X+ 6 and hence

↑ (xkerd) =
b(x) y =