Unit 10 Assignment
August 4, 2022 | Author: Anonymous | Category: N/A
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Unit 10
Group Homomorphisms Submitted by: Shruti Pathak 5078 Sahil Gupta 5079 Hariom Sarswat 5104
Insight View :10.1 Homomorphism :Definition and Examples. 10.2 Kernel of Homomorphism 10.3 Properties of Homomorphism 10.4 Isomorphism theorems 10.5 Solved Exercises
HOMOMORPHISM
Homo+ Morphe Morphe {Greek words words } }
Like
Form
If we drop the basic idea of one - one ,onto from the isomorphic mapping &
{ : → is one- one, onto ′
= . } then the remaining mapping is called Homomorphism.
The concept of Homomorphism was introduced by CAMILLE JORDAN in 1870.
Definition: Homomorphism
from a group ,∙ to a group ′,∗ is a mapping from to ' that preserves the group operation ,that is, , ∙ = ∗ ,∀ , ∈ .
A homomorphism
NOTE: It is to be be noted that the binary operation operation on the left hand hand side is is that of , whereas on the right hand side is that of
′. ,
,
.
∙
*
∙ =∗
Example 1: Any isomorphism is homomorphism that is also onto and one- one. Example 2: Consider two groups and with identity elements and ' , respectively. Define
′
: → = , ∀ ∈ . . ′
Then
′
is a homomorphism. It is called trivial homomorphism.
Example 3: Consider any group
,∗. Define
: → = , , ∀ ∈ . Then
is a homomorphism. It is called identity homomorphism , as every element maps to itself.
Before going onto further examples let us define another important concept i.e. kernel of a homomorphism.
10.2 Kernel 10.2 Kernel of Homomorphism In a homomorphism, all those elements that are mapped to the identity element are of special importance.
Definition:
′ ′ ′ . . ∈ : =′} .
If is a homomorphism of a group to a group , then the set of all those elements of which are mapped by onto the identity is called the kernel of of the homomorphism .The kernel of is denoted by
Theorem(10.a):
′ ′
Let and be any two groups and let and of into , then (i)
′ be their respective identities. If is a homomorphism homomorphism
−=) =′ [ x]− ∈
(ii) (
Proof:
∈ , ∈′. ⋅ ′ = = = ⋅ , and therefore by using left cancellation law we have ′ = = ′. (ii) Since for any ∈ , − = we get (). ( −) = ( −) = = ′
(i) We know that for
−
−
Similarly in Hence by the definition of gives −
=,[ x] ′ we ⋅ obtain = the′ result
− = [ ]−. Note: The Kernel of isomorphism is identity. Let us give some examples of homomorphism:
Example 4: Consider the group
therefore
ℝ∗,∙ .The mapping : ℝ∗ → ℝ∗ defined by = || is a homomorphism . = ⇒ =
is well defined . . = |..| =||.|| =.
Consider,
is homomorphism. = ∈ ℝ∗ = 1} = ∈ ℝ∗: || = 1} = 1,1 1,1}}.
= 2, ℝ.Prove that the mapping : → ℝ∗,∙ defined by = is a homomorphism with kernel 2, ℝ.
Example 5: Let
Sol. Let
, , ∈ . Then
=det =det =det =det det det ∗= . Hence is a homomorphism . The identity element of ℝ is 1. = ∈ : = 1} =∈: = 1} = 2, ℝ. = ℤ, and = ℤ , for some 1. Define ∶ → by = []. Then is a homomorphism.
Example 6 :. Let
Sol. Since operation in both groups is addi additi tion, on,
= .. = [] [] = [ ] =
Claim:
[] [] = [ ] holds by definition of addition in ℤ).
(where equality
Note : All isomorphism are homomorphism (although not all homomorphism are isomorphism).
Isomorphism are homomorphism by definition. Kernels of isomorphism are the identity (or else it would not be one-to-one), and images are the entire target group (or else it would not be onto).
Example 7:
φ: ℤ , → ℤ ,⨁ , ⨁ defined by = , ℎ ℎ = . Prove that is
group homomorphism .
= ⇒ = ⇒ = Let , ∈ℤ be s.t. = & = = = ( ) = ⊕ = ⊕ therefore is a homomorphism. Ker= ∈ ℤ| = 0} Sol : is well defined as if
i.e.
= ∈ ℤ| = 0} = ℤ (set of all multiple of n) Ker=
= ℝ[] , group of polynomials with real coefficients under operation addition. Define [ ] : ℝ → ℝ[] given by = ′ is a group homomorphism. Example 8:
Sol : Since derivative of a polynomial in is again a polynomial in
= ∈ ℝ[] is well defined. = = =
is homomorphism.
= ∈ | = 0} = ∈ | = 0 }
then,
= Set of all constant polynomials.
= [],the greatest integer less than or equal to , is not a Homomorphism , since [1/21/2]≠[1/2] [1/2].
Example 9: The mapping from the group of real numbers under addition to itself given by
Types of group homomorphism Monomorphism A group homomorphism that is is injective injective (or one-to-one one-to-one)) i.e. injective homomorphism
Epimorphism A group homomorphism that is is surjective surjective (or onto onto). ). i.e. i.e. surjective surjective homomorphism
Isomorphism A group homomorphism that is is bijective bijective;; i.e., injective and surjective. Its inverse is also a group homomorphism. i.e. bijective i.e. bijective homomorphism homomorphism
Endomorphism A homomorphism, h: G → G; the domain and codomain are the same. Also called an endomorphism of G. i.e. homomorphism of a group to itself.
Automorphism An endomorphism that is bijective , and hence an isomorphism from a group G onto itself. It is denoted by Aut (G (G). i.e. isomorphism of a group with itself.
Example 10: The mapping
∶ ℝ, → ℝ+ ,·
→ is an isomorphism, and − = .
Example 11: The mapping
∶ ℝ → ℝ ,, → , is a surjective homomorphism and = 0, ,0 ,0 ∶ ∈ }.
.
Example 12: The mapping
∶ ℤ ℤ,⊕ → ℤ,⊕ → 3 Show that is a homomorphism and find −, where = 0,6}. Sol: = 3 =33 = ∴ is a homomorphism. now, = ∈ ℤ| = 0} = ∈ ℤ|3 = 0} = 0 0,4,4,8,8}}
∵∴ | || = 3 is 3 1 mapping ∵ 2 = 6 ] [∵ = , ℎ ℎ − = ∴ −6 = 2 = 2 2,6,6,1,10} 0} Next , Now,
= 0,6} =6⊆ℤ = − 0 = 0 0,4,4,8,8}} −6 = 2,66,10 ,10}}
∴ − = 0,2,4,6,8,10} = 2 ⊆ = ℤ. Example 13: For every
∈ ℚ , the mapping ∶ ℚ, → ℚ, → ℚ,.
is an automorphism of
(10.b) In 1770, L. Euler proved that every positive integer can be written as sum of the four . square can be written in form Q. Prove that no integer equal to [i.e. sum of three squares ] Sol: Let To show: Let, if possible
7 8
= 7 8
≠ , ,, ,, ∈ ℤ = , , ,, ∈ ℤ ⇒ 8 = 8 8 8 8 7 = 8 8 8 8 8 But, square of every even integer is 0 4 8 , and square of every odd integer = = 1 8 and no combination of 0,1 0,1 & 4 will result in 7. Hence, ≠
Example 14: Example of a function that is not a homomorphism.
: ℝ ⇒ ℝ defined by =[] , the greatest integer function is not a homomorphism Take
= 0.6 , = 0.4
= [ ] = 1 = [] [] = 0 0 = 0 ≠
10.3 Properties of Homomorphism: Recall: A function
Let
∶ → ′ ′ is a homomorphism if . = ∗ ∀ , ∈ .
be a homomo homomorphism rphism from a group group to a group ′ and be an element of .
Then,
(1) (2) (3) (4) (5)
ℎ ℎ ′. = [] . || ,ℎ , ℎ || ||. . − = ′, ′, ℎ ℎ = ∈ : = } =
PROOF: (1) Since
= . ,
we have
= . = . Also, ′ being the identity element element of of ′ ,we have = Thus, .. = ′′ By the left cancellation law in ' , we get, = ′
′
(2) If
0 , then
= [] , as is homomorphism. If 0 , then = for some ∈ ℤ, 0. So, = − = − = − = []− = [] Hence,
= 0,
Also if Hence,
= [] = = , = [] , ∀ .
then
∈ || = ⇒ = Since, is homomorphism
(3) Let
= =
using (i)
=′ = ′
Thus,
property prope rty (1)
| divides ||.
Hence, |
(4) The kernel
of a homomorphism ∶ → ′′ is a subgroup of . Proof . Let = ∈ | = ′} ′} Clearly, ∈ as = ′ Therefore, ≠ Let , ∈ be arbitrary arbitrary , ⇒ = = Now , − = . − = . ()− =′.′− = ′ ∴ − ∈ Hence , by One-Step Subgroup Test Test ≤ . (5) First, we prove that Let then, therefore therefore Thus We now prove that
Now
− ⊆ = ∈ −. = = . ′ = ()− = − = − − ∈ − ⊆ .
⊆ − . Let ∈ , then = , for some ∈ = = = . ′ = = ′ Thus , ∈ − so that ⊆ −
Hence ,we get
− = .
(10.c)Few other important properties of subgroup under Homomorphism . Let
homomorphism rphism from group group to group ′ and is a subgroup of , then be a homomo (a) is a subgroup of . (b) is Abelian subgroup of ⇒ is Abelian . (c) = is cyclic then = is cyclic. (d) is normal subgroup of then is normal subgroup of (e) if | | = , then is 1 mapping from ′. (f) if is finite ..| ..||| = , then then is finite and divides ||. (g) if is subgroup of ′, then − is a subgroup of . (h) − ) is a normal subgroup of if is a normal subgroup of ′. (i) is normal subgroup of . (j) If is onto and = }, }, then is an isomorphism from ′.
Proof:
∵ = ⇒ ≠
(a)
, [∵ =∈]
, , ∈ be be arbitrary
Let
there exist
and
⇒
− == −= = − ∈ − =∈,.. , Hence , by one step subgroup test ≤ ..
Now,
= ℎ:ℎ ∈} Let , ∈ be arbitrary arbitrary
(b
Consider,
. = . [ as is operation preserving ] = . [ is abelian] =
.
∀ , ∈ Therefore, is abelian. This holds
(c)
= = :∈}. Then,
Hence
= : ∈ } = () : ∈ } = is cyclic and is generated by .
be a normal subgroup of . Let ℎ ∈ and ∈ be be arbitrary Then , ℎ()− = ℎ (− )
(d)Let
1
1
−1 =
as is a homomorphism ( ℎ ) ∈ , is a normal subgroup of Hence,
is a normal subgroup of .
(e) We know that
= , then − = Since | | = , ⇒ |− | = | | = | | =
if
⇒every element has pre-images. Hence, is an 1 mapping . ≤ and | | = . then the restriction of of to is a homomorphism of . [ :→ s.t. ℎ = ℎ∀ ℎ ∈ ] Let | |= then by above theorem (e) , is an 1 mapping ,so | || = || |÷÷ . therefore , || = || ||.. (f) Suppose
Hence
| | divides ||.
is a subgroup of ′ , then − = ∈ : : ∈ } Since =′ ∈ ⇒ − ′ ∈ − so that − is non empty.
(g) if
1
Let
, ∈ −.
∈ , ⇒ − ∈, since is a subgroup f ⇒ ( − ) ∈ ⇒ − ∈ −, by definition of − .
then
2
1
2
1
1
2
1
Hence
− is a subgroup of . 1
(h) by (g) ,
−1
is a subgroup of
.
∈ ∈ − be arbitrary, arbitrar y, Then ∈ and ∈ ⇒ ()− ∈ , since is normal subgroup of '. ⇒ (− ) ∈ , as is a homomorphism ⇒− ∈ − . ⇒ − is a normal subgroup of .
Let
′
1
1
1
1
1
= } ⇒ is 1 1 mapping from and is onto and operation preserving as
(j) well.
Hence,
is an isomorphism.
Fig 10.1: Pictorial representation of properties.
:40 → 40 be homomorphism with Kernel = 1,99,17,3 ,17,33}. 3}. if if 1 11 = 11, Let us find all the element which map to 11. −11= 11 =11⊙ 1 1,11 ,11 ⊙ 9 9,11 ,11 ⊙ 1 17,11 7,11 ⊙ 33 } = 11, 11,19 19,2,27,7,3} 3} = 3,1 3,11,1 1,19,9,27 27}} Thus, is 4 1 map. Example 15: Let
Example 16: There is no homomorphism from Since, if there is a homomorphism
32
ℤ ℤ ⊕ ℤ. 32
32
4
4
from ℤ ℤ ⊕ ℤ, then ℤ = ℤ ⊕ ℤ
ℤ is cyclic whereas ℤ ⊕ ℤ is not cyclic .Therefore, there is no homomorphism possible from ℤ ℤ ⊕ ℤ
but
32
4
4
32
Theorem(10.d): Kernels are Normal
be a group homomorphism from ′ . Then is a normal subgroup of . Proof: Let ∶ → be a homomorphism Claim: ⊲ G
Let
Let
arbitrary ∈ and ∈ be arbitrary
= (− ) = ()− since − = ′ () ′
Then
1
1
Now ,
is homomorphism
1
()− = ′ Hence, − ∈ , so so that is a normal subgroup of . 1
=
Theorem(10.e): Normal subgroups are are Kernels
Every normal subgroup of a group is a kernel of a homomorphism of subgroup is the Kernel of mapping (from
. In particular , a normal
→ /. homomorphism from from Proof : Define : → = {this mapping is called the natural homomorphism /. /.}} is well defined: Let = ⇒ = ⇒ = ∴ is homomorphism: =
== ∴ is a homomorphism & is clearly onto. Now
= ∈ | = = } = ∈ | = } } = ∈ | ∈ } = ∩ =N [∵ ⊲ ⇒ ⊆ ]
Note:(1) We We can always define define a natural natural homomorphism homomorphism from a group group onto any of its quotient quotient group. group. (2) If
:→′ be an onto homomorphism , then G' is called the homomorphic image of G.
Example 17 : Let then
: → ′ be a homomorphism. Let ∈ be s.t. || = , | | =
| = , 1 1 | | = ⇒ = ⇒ = ⇒ ( ) = ∴|
Proof: (i)
(ii) Let
= ⇒ | | = ||
Let
= ⇒ − = ⇒ | −| = 1 ⇒ | −| = 1 ⇒ − = ⇒ = ∴ 1 1.
Conversely, Let
1 1
||= ⇒ ( ) = ⇒ = ⇒ = ⇒| ∴ = ∵ |
Example 18 : If
⊴ , then for all , , ∈ , , = . .
Proof . Since
is a normal subgroup of , we have = − ⇒ = .
[ − = ]
Example 19 : Let be a group and
a normal subgroup of . Let ∶ → / / = ∈ . Then for all ,ℎ , ℎ ∈ , , ℎ ℎ = ℎ ℎ = ℎ ℎ = ℎ ℎ.. We will see that every homomorphism can be considered to be of this form.
∶ → ,, then for all , ∈ , , = if and only if = .. − = − − ∈ . Then the cosets = . . Then ′ = Proof . Suppose and − are the same, and Example 20 : If is the kernel of a homomorphism
⇒ = − ⇒ = = .. − ∈ , − = ′ ′ = − ′ = − = − =
Conversely, Then
Then
10.4: ISOMORPHISM THEOREM : In this section we will deal with the group G, normal subgroup N of G and quotient group group G/N , their their interconnection interconnection and Cayley Cayley table and relation relation between between number number of Homomorphic images and number of quotients.
First Isomorphism Theorem: The first group isomorphism theorem, also known as the fundamental homomorphism theorem, states that if be a homomo homomorphism rphism from a group group to a group ' . Then
/ / ≈ . i.e. the mapping from / ,given by → ,is an isomorphism.
Proof: Let
= .For ∈ , ∈ , ∈ /
Define
:/→ by = . We will prove the following
2. is one- one 3. is onto
1. is well defined
4.
is homomorphism. Step 1: To prove that is well defined. Let ,ℎ ∈ . . = ℎ .. ⇒ ℎ− ∈ ⇒ (ℎ− ) = ⇒ ( ℎ − ) = ⇒ (ℎ)− = 1
1
′
1
′
1
Hence,
′
⇒ = ℎ = ℎ ℎ ,so is well defined.
is one -one. Let ,ℎ , ℎ ∈ / / .... = ℎ ℎ then = ℎ ⇒ (ℎ− ) = , as is homomorphism ⇒ ℎ− ∈ ⇒ = ℎ Hence, is one -one. Step3: To prove is onto. Let ∈ . Since is onto ∃ ∈ .. = Then ∈ / / and = = Hence is onto. Step 4: To prove is homomorphism. Let ,ℎ ∈/
Step 2 :.To prove that
1
′
1
ℎ ℎ = ℎ ℎ
ℎ = ℎ (as is homomorphism) = ℎ
=
is homomorphism.
Hence,
Hence
is a homomorphism from / onto , which is one- one also , so ≈ .. / / ≈ .
Diagrammatically this theorem can be represented as
/
Then
=
Result: Each quotient of quotient group thereof.
is a homomorphic image and each homomorphic image is isomorphic to some
Second Isomorphism Theorem : If
is a subgroup of a group and is a normal subgroup , then
/∩≈/ . Proof: Since is normal subgroup of , therefore = so that is a subgroup of . Moreover, = ⊆ So that is a normal subgroup of . Thus / is defined. Define :→/ by = Then is a homomorphism since it is operation preserving .Moreover, for if, ∈ / / then ∈
is onto
for some
∈ , ∈ ⇒⇒ ==, = =
∈ ⇒ = So, is onto.
Thus ,
By first isomorphism theorem
/ ≈ / / / . Let us now find
(1)
. ∈ ⇔ = , as is the identity element of the /.
⇔ = ⇔ ∈ ⇔ ∈ ∩ , as the mapping is defined on so ∈ . = ∩ . from (1) & (2) we get /∩≈/ . Hence
(2)
Third Isomorphism Theorem:
and are normal subgroup of a group and ⊆ , Then / / / // / ≈ / /.. Proof : Since ⊲ and ⊲ , so /,/ / are defined. Define : :/ / → / / =
If
we will prove that
is well defined 2. is onto 3. is homomorphism 4. =/ 1.
Step 1. Let Then
, ∈ / / . . = − ∈ ⊆ 1
⇒ =
⇒ = .
Hence, Step 2. Let Hence, Step 3. Let
is well defined.
∈ /. /. Then ∈ / , so that = is onto. , , ∈ / /.. = = = =
s homomorphism Step 4: ∈ ⇔ = / / ⇔ = ⇔ ∈ ⇔ ∈ / / Hence, = / / Hence,
Since the Kernel of Homomorphism is a normal subgroup ,
/ is a normal subgroup of / .
By the first isomorphism theorem ,
Therefore,
/ ≈ /. / /// ≈ /.
Freshman's 's Theorem. This is also known as Freshman
N/C Theorem: Let be a subgroup of a group . Then normalizer of
in is
. Consider the − = } and the centralizer of in is = ∈ |ℎ − | = ℎ ∀ ℎ ∈ = } ∈ mapping from to given by → , where is the inner automorphism of induced by [that is, ℎ =ℎ− ∀ ℎ ∈ ]. This mapping is a homomorphism with kernel . So, by fundamental theorem of homomorphism , / is isomorphic to a subgroup of .
Example 21 : The mapping
is a surjective homomorphism with and thus, by first isomorphism theorem
∶ ℤ ℤ, → ℤ, → 3 = 0 0,3,3,6,6,9,9}} = 3ℤ ℤ/3ℤ and ℤ are isomorphic.
1 isomorphism theorem
Example 22 : To illustrate Consider
: →
Then
′
= , } = ,,, }
= , , } , → as Define : = = = = = = = = Then
is desired isomorphism.
homomorphism from Example 23 : If is a homomorphism
ℤ onto a group of order 5, Determine .
5
Sol: Let be a group of order .By first isomorphism theorem ,
/ ≈ , where =
Comparing the Order, we get
being a subgroup of cyclic group ℤ , it is cyclic ||=6 = , where ||=6 Since, ℤ has unique subgroup of order 6 [by fundamental theorem of cyclic group ] So, = 5 ℤ⨁ℤ onto ℤ. : ℤ⨁ℤ → ℤ.
Example 24: Prove that there is no Homomorphism from Sol: Let there exist an onto homomorphism By first isomorphism theorem ,
ℤ⨁ℤ/ ≈ ℤ.
where
=
Comparing the order , we get
Since
||=2 ℤ. is a cyclic group of order 88, so ℤ⨁ℤ/ is a cyclic group of order 8. 4 = 0 ∀ , ∈ ℤ⨁ℤ ⇒ 4(, ) = ∀ , ∈ ℤ⨁ℤ ⇒ no element of ℤ⨁ℤ is of order 8. ℤ⨁ℤ/ is a cyclic group of order 8 8. But
which is contradiction to
Hence, no such homomorphism exist.
: ℂ∗ → ℂ∗ = ∵ . = .. = . = .
Example 25: Consider , Given by ,
∴ is a homomorphism. ker = ∈ ℂ∗| = 1} = ∈ ℂ∗| = 1} = 1, 1,,} ∴ | || = 4
is 4 1 mapping
So,
ℂ∗ that are mapped to 2. ∈ ℂ∗ .. .. = 2 = 2 i.e. 4 complex roots of 2
We will find all elements of i.e. we find
We know
( √ 2) = 2 ∴ −2 = √2 = √ 2 , , √ 2 , , √ 2 , √ 2}
SOLVED EXERCISE : be the group of non zero real numbers under under multiplication multiplication,, and let be a positive ℝ∗ be integer. Show that the mapping that takes to is a homomorphism from ℝ∗ to ℝ∗ and determine the kernel. Which values of yield an isomorphism?
Problem 1: Let
Sol:
: ∗ → ∗ 0 & € → . = . . = . =. ⇒ is a group homomorphism. = ∈ ∗| = 1}
For odd values of
= 1 1}} And for even values of = ±1} ⇒ n isomorphism if is odd.
is a homomorphism from to and is a homomorphism from to , show that is a homomorphism from to . How are and related? If and are onto and is finite, describe [ [ : : ] in terms of || || and ||. Sol: .Clearly, : → is well defined. Problem 2: If
We need to prove that
preserve operation. We know that and preserve operation, Now, = = = = ∴ is a homomorphism from to Take any ∈ .. This means = , and then , = () = = This implies that ⊆ . The mappings are onto implies that = =. Consequently, mapping is onto
, .., = .
Now using First Isomorphism Isomorphism Theorem Theorem we conclude conclude that Analogously,
|| = ||| | = |||| .
|| = || . || = | || ||
Now we can can compute compute the desired desired index:
|| || = [ [ ∶ ] ] = = || || || || ⨁ to G given by ,ℎ → is a homomorphism . What is the kernel? This mapping is called the ⨁ . ,ℎ ℎ , ℎ Sol : (, ℎ , ℎ ) = , ℎℎ = = ,
Problem 3: Prove that the mapping from
= , ℎ| | ℎ ∈ } ℤ⨁ ℤ/ , 0 × 0, is isomorphic to ℤ ⨁ ℤ . , , = ,, . Sol: We define a mapping :ℤ ⊕ ℤ → ℤ ⊕ ℤ with
Problem 4: Prove that
Kernal of mapping
, ) = 0,0.This translates to two simultaneous is found in solving equation (,
equation:
= 0 = 0 Solutions to first equation form a set
, and to second . Then the Kernel of is:
= , , | ∈ , ∈ } = ,0 ,0 × ,0 ,0 Now we apply apply First Isomorphism Isomorphism Theorem to mapping
, which exactly yields the desired isomorphism.
ℤ⨁ ℤ/ , 0 × 0, ≈ ℤ ⨁ ℤ
(
Problem 5: Prove that
⊕ / ⊕} ≈.
Sol: We define a mapping
:⊕ → with (, ) =
Kernel of mapping
is found solving the equation (, ) = .This translates to equation = .
Then the Kernel of
is ⊕ } }..
Now we apply apply First Isomorphism Isomorphism Theorem to mapping
, which exactly yields the desired isomorphism.
is a homomorphism from ℤ ℤ and = 0,10 0,10,, 20}. If 23 = 9, determine all elements that map to 9. Problem 6: Suppose that Sol:
: ℤ → ℤ = 0,10,20} If
23 = 9 Then −9 9 = 23 = 23 23,, 13 13,, 3} [∵ = , ℎ ℎ − = ] ℤ ⊕ ℤ onto ℤ ⊕ ℤ. Sol: Assume on the contrary that there is such a homomorphism ∶ ℤ ⊕ ℤ → ℤ ⊕ ℤ . Problem 7: Prove that there is no homomorphism from
Then applying the First Isomorphism Theorem to homomorphism
because ofof ℤ ⊕ ℤ = ℤ ⊕
ℤ ( is onto) yields the following: ℤ ⊕ ℤ / ≈ ℤ ℤ ⊕ ℤ = ℤ ⊕ ℤ This further implies that
⊕ℤ | = = 2, which we will use now. || = |ℤ|ℤ⊕ℤ |
are now: 8,0 8,1 0,1, as these are the only elements of order 2 in ⊕ .
The only possibilities for
in any case
|1,1,00 | = 8 1166 8
Since there are no elements of order nor
16 in ℤ ⊕ ℤ, we conclude it can not be isomorphic
ℤ ⊕ ℤ/ , which is a contradiction.
to
Problem 8:. Suppose that there is a homomorphism one. Determine
from ℤ to some group and that is not one-to-
.
is a sub group of cyclic group ℤ. Since the mapping is not one-one, it has to be , where = | |
Sol: We know that
But we know that order of sub group of cyclic group divides the order of the group, i. e
.|17. This
= 1 7 (since 17 is prime and ≠ 1, which means = ℤ. Hence, maps every element of ℤ to identity.
implies
is a homomorphism from ℤ onto a group of order 5, determine the kernel of . Sol: Since is onto, we know that | ℤ | = 5 . Then first isomorphism theorem implies Problem 9: If
|| = || = = 6. | But is a sub group of ℤ and there is only one sub group of order 6 in ℤ . that
This implies
= 5
Problem 10: Suppose that
is a homomorphism from ℤ to a group of order 24.
a. Determine the possible homomorphic images. b. For each image in part a, determine the corresponding kernel of
.
Sol: Part a: First of all note that homomorphic image of cyclic group is cyclic as well. We use corollary of first isomorphic theorem which says that |
| divides both || and |′|.
ℤ that divides |ℤ| = 36 and 24, which implies that it divides 36,24 = 12. The cyclic group of order which divides 12 are following:
In this case we have
ℤ ℤ ℤ
ℤℤ
ℤ And these are or possible homomorphic images.
Part B :From first isomorphism theorem we know that
| | || = ̅|| || , so we can easily determine the
order of kernel. But in cyclic groups, there is only one sub group of given order, so the kernels are of the following order:
= 1 = 2 = 3 = 4 = 6 = 12
Problem 11: Determine all homomorphisms from
ℤ to itself.
Sol: Any homomorphism acting on cyclic group is uniquely determined on its effect on any generator, so we just have to see where we have to send element
There are options
1 ∈ ℤ .
0,1,…….1. If 1 = then we can easily compute = 1 =
Problem 12: Suppose that is a homomorphism from
7 = 7 find all elements of 30 that map to 7.
30 to 30 and that = 1, 1, 11}. 11}. IfIf
of properties of homomorphism homomorphis m [ = ′ ′,, ℎ ℎ − = =use Sol: = 5 ]to conclude that ∈ We: just }statement − 7 = 7 = 71,11 1,11}} = 7,17 7,17}}
is a homomorphism from 40 to 40 and that = 1,, 9,17 1 9, 17,, 33 33}.}. If 11 = 11, find all elements of 40 that map to 11. − = = ′ ′, , ℎ ℎ Sol: We just use statement 5 of properties of homomorphism[ ∈ : = } = ] to conclude that 11 = 111,9,17,33} = 11,19,27,3} −11 = 11 Problem 13: Suppose that
: ℤ ⨁ ℤ → ℤ given by , ⟶ is a homomorphism. − What is the kernel of ? Describe the set 3 .(that is, all elements that map to 3). Sol: To verify that :ℤ⊕ℤ →ℤ defined like this is homomorphism, we need to make sure that it Problem 14: Prove that the mapping
preservess operation. preserve
Here is the calculation:
φ(a,a,bb , ) = , = = = , , The kernel of is obtained from equation , =0, which translatesto = 0, . . .. = . This implies that = , | ∈ ℤ } One element that maps to 3 is easy to spot it is 3,0 − = ∈ : = (1) Use statement 5 of properties of homomorphism [ = ′ ′, , ℎ ℎ ] we get−that set of all elements that map to 3 is: } = 3 = 3,0 = 3,| ∈ }} Problem 15: Let
= ∈ ∗ ||| = 1}. Prove that ∗/ is isomorphic to ℝ+ , the group of positive
real numbers under multiplication.
Sol:
: ∗ → ℝ∗defined by = || = || = ||. || =
i.e.
is a homomorphism ∗
∗ ∗ is onto ⇒ = ≈
ℝ+
Problem 16: Prove that the mapping
= ℤ ∈ ||||| = 1 }
→ from ℂ∗ ℂ∗ is a homomorphism. What is the kernel?
, ∈ ∗ and because of commutativity of multiplication of complex numbers, we get directly: = The kernel is given with equation = 1 has 6 solutions 1 √ 3 1 √ 3 1 √ 3 1 √ 3 = 1,1, 1,1, 2 2 , 2 2 , 2 2 , 2 2 } Sol: Take any
Isomorp hism Theorem) If is a subgroup of Problem 17: (Second Isomorphism prove that that
/ / ∩ is isomorphic to /.
and is a normal subgroup of ,
Sol: See prove of isomorphism theorem [ second isomorphism theorem in text ]
are normal subgroup of a group and ⊆ ,
Problem 18: (Third Isomorphism Theorem) If and Then
// // / / ≈ / /..
Sol: See prove of isomorphism theorem [ Third isomorphism theorem in text ]
images of Problem 19: Determine all homomorphic images
Sol: Let
(up to isomorphism). isomorphism).
: → be a homomorphism
By first isomorphism theorem
≈
∵ | | | divides 8 ∴ | | | = 1,1,2,2,44 8 ⇒ | | = 1,1,2,2,44,8,8 Case 1: | | = 1 ⇒ = Case 2: | | = 2 ⇒ = , } ∵ , , } is the only subgroup of order 2 .Hence = , , , } Every element of is of order 2 therefore ≈ ℤ ⨁ ℤ Case 3:
| | = 4 ⇒ = 2 therefore ≈ ℤ
| | = 8 ⇒ = } = Hence, all the homomorphic images of = ℤ , ℤ ⨁ ℤ , , . Case 4:
(up to isomorphism). isomorphism). Sol: : → is an homomorphism, then ≈
images of Problem 20: Determine all homomorphic images
so the possible order of images of
|| = 8
By first isomorphism theorem
1, 1,2,2,4, 4,88 is same as=the factor group /
Problem 21: Prove that Sol:
Define
is well defined: Let
is homomorphism: And
} ≈ .
: → as = , ∀ ∈ , , ∈ be arbitrary, such that = ⇒ = = = . = . = . is clearly onto ∴ is onto homomorphism
By first isomorphism theorem
≈}
(1)
= ∈ | = } = becomess ∴.1 become ≈} , , is the homomorphic image of a group . What can we
each prime Problem 22: Suppose that for each say about
||? Give an example of such a group.. p. Sol: If is finite, |ℤ | = will divide | |, for all prime p. Since this is not possible, we conclude that || = ∞ . One example is
= , because Image of mapping = is ≈ .
Problem 23: Prove that Sol: Define
} ≈ .
: → as =
, ∀ ∈
is well defined: Let
is homomorphism: And
, , ∈ be arbitrary, such that = ⇒ = . = . = . is clearly onto ∴ is onto homomorphism
By first isomorphism theorem
≈
(1)
= ∈ | = } = ∈ | = } = } becomess ∴.1 become ≈ }
Problem 24: Prove that the mapping from ∗ is isomorphic to ∗ .
ℂ /1,1}
ℂ
ℂ∗ ℂ∗ given by = is a homomorphism and that
: ℂ∗ → ℂ∗
Sol:
→ = ()(). = 1,1 1,1}} =
φ is onto on ℂ∗ By first isomorphism theorem
∗
ℂ ≈ ℂ∗. ⇒ ,− }
Problem 25: Let be an Abelian group. Determine all homomorphisms from
. Since | | must divide 6 we have that | | = 1,2, 2,33, , 6. Sol: Let be a homomorphism homomorphism from
.
maps every element to 0. If | | = 2 , then n is even and maps the even permutations to 0, and the odd permutations to /2.
In the first case
Case 3:
|| = 3 cannot occur because it implies that is a normal subgroup of order 2 whereas has no normal subgroup of order 2. Case 4: | | = 6 cannot occur because it implies that is an isomorphism from a non-Abelian group to an Abelian group.
Problem 26: Prove that
ℤ ≈ ℤ .
: ℤ → ℤ = ,
Sol: Define as Then
where
=
is an onto homomorphism = = = ⊕
Then, by first isomorphism theorem
ℤ
≈ ℤ = ℤ
[as is onto]
= ∈ ℤ| = 0} = ∈ ℤ| = 0} = ∈ ℤ| |} = = ℤ {multiple of } Hence,
Problem 27: Prove that every group of order
has an element of
77 is cyclic.
77. By Lagrange’s Theorem every nonidentity of has order 7,11 7, 11,, 77. order 77, then is cyclic.
Sol: Let be a group of order If
ℤ ≈ℤ
7 11
So, we may assume that all non-identity elements of have order . Not all non-identity elements of can have order because[we know in a finite group, the number of elements of order is a multiple of ], the n number umber of such elements is a multiple multiple of . Similarly not all nonidentity elements of can have order , the number of such elements is a multiple multiple of 6.
Φ
11
10
7
| || | = 11
| | = 7
=
So, must have elements and such that and . Let . Then is the only subgroup of of order for if is another one then − is also a But is a subset of and only has elements. Because for every in , − subgroup of of order , we must have .
121.
11 77 | || | = ||| |||| |/| /| ∩ | = 11 • 11/1 11/1 = 11 = So, = . Since Since has prime order, is cyclic and therefore Abelian. This implies that contains . So, 11 divides || and || divides This implies 77 that = or = . If = , then | | || = 77. If = , then | / /| | = 7. But by the “/” ℎ / is isomorphic to a subgroup of ≈ ≈ 11 . Since 11 11 = 10, we have a contradiction. Thus is a cyclic.
Problem 28: Find all homomorphisms from Sol: Let,
ℤ ℤ.
: ℤ → ℤ be a homomorphism ∵ ∈ ℤ can be written as = 1 1 1 ⋯ 1 ∴ = 1 1
∴ can be determined using the value of 1. ∵ 1 ∈ ℤ and order of an element divides order of the group
∴ |1| 30 |1| 12 Also, |1| 12,30 i.e. ∴ possible value of |1| are 1,2,3,6 1 ℤ.
Now, is the only element of order in
0
∵ |1| ||]
[
ℤ 10 and 20 are two element of order 3 in ℤ and, 5 and 25 are two element of order 6 in ℤ ∴ 1 can take value 0,5,10,15,20,25 So, there are six homomorphism from ℤ to ℤ ,given by : ℤ → ℤ as = 0 : ℤ → ℤ as = 5 : ℤ → ℤ as =10 : ℤ → ℤ as =15 : ℤ → ℤ as =20 : ℤ → ℤ as =25 15 is the only element of order 2 in
Problem 29: Prove that an infinite cyclic group is isomorphic to
ℤ , , .
= be an infinite cyclic group (⇒ || is not finite ..∄ any positive integer . . . = ) Define: : → ℤ as ( ) = , ∈ ℤ Claim: is well defined, 1-1, onto and operation preserving Let = ⇒ = [for if ≠ then − = ⇒ || = | |is finite ] ⇒ ( ) = ( ) ∴ is well defined Sol: Let
One-One: Let
() = ( ) ⇒ = ⇒ =
so,
is one one.
Onto: For any s. t.
∈ ℤ , ∈ ( ) = ∴ is onto.
Operation Preserving: Consider
( . ) = (+ ) = = ( ( ) ( ) Therefore, is operation preserving. Hence , ≈ℤ,
Remark: Since subgroup of an infinite cyclic group is infinite cyclic group.
Let
= & ≤ then = ⇒ ≈ ℤ, ℤ, ≈ ⇒ ≈
Thus , we note that subgroup of an infinite cyclic group is isomorphic to group itself .
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