This is a handout about Homomorphism, Isomorphism, Kernel and Cayley's Theorem. This is a group effort. #LoveforAbstractAlgebra

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ISOMORPHISM
GROUP ISOMORPHISM
Let G and G’ be groups with operations ∗and ∗ ′. An isomorphism 𝜑 from a group G to a group G’ is a
one – to – one mapping (or function) from G onto G’ that preserves the group operation. That is,
𝜑 𝑎 ∗ 𝑏 = 𝜑 𝑎 ∗ ′𝜑(𝑏) for all a, b in G.
If there is an isomorphism from G onto G’, we say that G and G’ are isomorphic and write G≅ G’.
Steps involved in proving that a group G is isomorphic to group G’:
1. “Mapping.” Define a candidate for isomorphism; that is, define a function 𝜑 from G to G’.
2. “1 – 1.” Prove that 𝜑 is one – to – one: that is, assume 𝜑 𝑎 = 𝜑(𝑏) and prove that 𝑎 = 𝑏.
3. “Onto.” Prove that 𝜑 is onto; that is, from any element g’ in G’, find an element g in G such
that 𝜑 𝑔 = 𝑔′.
4. “O.P.” Prove that 𝜑 is operation – preserving; that is, show that 𝜑 𝑎 ∗ 𝑏 = 𝜑 𝑎 ∗ ′𝜑(𝑏) for
all a andb in G.
Note: We can relate Steps 2 and 4 to Kernel and Homomorphism, respectively.
HOMOMORPHISM
Let G and G’ be groups with operations ∗and ∗ ′. A homomorphism of G into G’ is a mapping 𝜑 of G
intoG’ such that for every a and b in G,
𝜑 𝑎 ∗ 𝑏 = 𝜑 𝑎 ∗′
𝜑 𝑏 .
Examples: Determine whether the given map is a homomorphism.
1. Let 𝜑: 𝑅∗
→ 𝑅∗
under addition given by 𝜑 𝑥 = 𝑥2
.
Answer: 𝜑 𝑥 + 𝑦 = 𝑥 + 𝑦 2
= 𝑥2
+ 2𝑥𝑦 + 𝑦2
≠ 𝑥2
+ 𝑦2
= 𝜑 𝑥 + 𝜑(𝑦). Thus 𝜑 is NOT a
homomorphism.
2. Let 𝜑: 𝑅∗
→ 𝑅∗
under multiplication given by 𝜑 𝑥 = 𝑥2
.
Answer: 𝜑 𝑥𝑦 = 𝑥𝑦 2
= 𝑥2
𝑦2
= 𝑥2
𝑦2
= 𝜑 𝑥 ∙ 𝜑(𝑦). Thus 𝜑 is a homomorphism.
3. Let 𝜑: 𝑅∗
→ 𝑅∗
under multiplication given by 𝜑 𝑥 = −𝑥.
Answer: 𝜑 𝑥𝑦 = − 𝑥𝑦 = −𝑥𝑦 ≠ −𝑥 (−𝑦) = 𝜑 𝑥 ∙ 𝜑(𝑦). Thus 𝜑 is NOT
ahomomorphism.
Types of Homomorphisms:
1. Epimorphism – surjective homomorphism
2. Monomorphism – injective homomorphism
3. Isomorphism – bijective homomorphism
4. Automorphism – isomorphism, domain and codomain are the same group
5. Endomorphism – homomorphism, domain and codomain are the same group
KERNEL
Let 𝜑: 𝐺 → 𝐺′ be a homomorphism of groups. The subgroup 𝜑−1
𝑒′
= 𝑥 𝜖 𝐺 𝜑 𝑥 = 𝑒′ is the
kernel of 𝜑, denoted by Ker(𝜑).
Examples: Find the kernel of the following homomorphism:
1. Let 𝜑: 𝑅∗
→ 𝑅∗
under multiplication given by 𝜑 𝑥 = 𝑥2
.
Answer: The identity element of the set of images is 1 under the operation of multiplication.
If 𝑥2
= 1 then 𝑥 = −1 or 𝑥 = 1. Thus, Ker(𝜑)={−1,1}.

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2. Let 𝜑: 𝑍 → 𝑍 under addition given by 𝜑 𝑥 = 5𝑥.
Answer: The identity element of the set of images is 0 under the operation of addition. If 5x
= 0, then x = 0. Thus, Ker(𝜑)={0}.
COROLLARY:
A group homomorphism 𝜑: 𝐺 → 𝐺′ is one – to – one map if and only if Ker(𝜑) = {𝑒}.
In view of this corollary, we can modify our steps in showing that two groups are isomorphic.
To show 𝝋: 𝑮 → 𝑮′ is an isomorphism: (modified version)
1. Show 𝜑 is a homomorphism.
2. Show Ker(𝜑) = {𝑒}.
3. Show 𝜑 maps G onto G’.
Examples: (On showing isomorphism between G and G’)
A. Let us show that the binary structure <R, +> with operation the usual addition is isomorphic to
the structure <R+, ∙> where ∙ is the usual multiplication.
Step 1. We have to somehow convert an operation of addition to multiplication. Recall from
𝑎 𝑏+𝑐
= 𝑎 𝑏
(𝑎 𝑐
) that the addition of exponents corresponds to multiplication of two quantities.
Thus we try defining 𝜑: 𝑅 → 𝑅+
by 𝜑 𝑥 = 𝑒 𝑥
for 𝑥 ∈ 𝑅. Note that 𝑒 𝑥
> 0 for all 𝑥 ∈ 𝑅 so indeed
𝜑 𝑥 ∈ 𝑅+
.
Step 2. If 𝜑 𝑥 = 𝜑(𝑦), then 𝑒 𝑥
= 𝑒 𝑦
. Taking the natural logarithm, we see that 𝑥 = 𝑦, so 𝜑 is
indeed 1-1.
Step 3. If 𝑟 ∈ 𝑅+
, then ln⁡(𝑟) ∈ 𝑅 and 𝜑 ln 𝑟 = 𝑒ln 𝑟
= 𝑟. Thus 𝜑 is onto R+.
Step 4.For𝑥, 𝑦 ∈ 𝑅, we have𝜑 𝑥 + 𝑦 = 𝑒 𝑥+𝑦
= 𝑒 𝑥
∙ 𝑒 𝑦
= 𝜑 𝑥 ∙ 𝜑 𝑦 .
Thus, 𝜑 is an isomorphism.
B. Let 2𝑍 = 2𝑛 𝑛 ∈ 𝑍 , so that 2Z is the set of all even integers, positive, negative and zero. We
claim that <Z, +> is isomorphic to <2Z, +> where + is the usual addition.
Step 1. Define 𝜑: 𝑍 → 2𝑍 by 𝜑 𝑛 = 2𝑛 for 𝑛 ∈ 𝑍.
Step 2. If 𝜑 𝑚 = 𝜑(𝑛), then 2𝑚 = 2𝑛 so 𝑚 = 𝑛. Thus 𝜑 is 1-1.
Step 3. If 𝑛 ∈ 2𝑍, then n is even so 𝑛 = 2𝑚 for 𝑚 = 𝑛/2 ∈ 𝑍. Hence 𝜑 𝑚 = 2(𝑛/2) = 𝑛 so 𝜑 is onto
2Z.
Step 4. For𝑚, 𝑛 ∈ 𝑍, we have𝜑 𝑚 + 𝑛 = 2 𝑚 + 𝑛 = 2𝑚 + 2𝑛 = 𝜑 𝑚 + 𝜑(𝑛).
Thus, 𝜑 is an isomorphism.
C. Determine whether the given map 𝜑 is an isomorphism of the first binary structure with the
second.
1. <Z, +> with <Z, +> where 𝜑 𝑛 = 2𝑛 for 𝑛 ∈ 𝑍.
Answer: 𝜑 is 1-1, but NOT onto. Thus, 𝜑 is NOT an isomorphism.
2. <Z, +> with <Z, +> where 𝜑 𝑛 = 𝑛 + 1 for 𝑛 ∈ 𝑍.
Answer:𝜑 is 1-1, onto, but NOT operation preserving. Thus, 𝜑 is NOT an isomorphism.

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3. <R, ∙> with <R, ∙> where 𝜑 𝑥 = 𝑥3
for 𝑥 ∈ 𝑅.
Answer:𝜑 is 1-1, onto, and operation preserving. Thus, 𝜑 is an isomorphism.
4. <R, +> with <R+, ∙> where 𝜑 𝑟 = 0.5 𝑟
for 𝑟 ∈ 𝑅.
Answer:𝜑 is 1-1, onto, and operation preserving. Thus, 𝜑 is an isomorphism.
CAYLEY’S THEOREM
Every group is isomorphic to a group of permutations.
Example:
Let 𝐺 = {2,4,6,8}⊆Z10, where G forms a group with respect to multiplication modulo 10. Write out
the elements of a group of permutations that are isomorphic to G, and exhibit an isomorphism from
G to this group.
Solution:
Let 𝜑 𝑎 : 𝐺 → 𝐺′ be defined by 𝜑 𝑎 𝑥 = 𝑎𝑥 for each 𝑥 ∈ G. Then we have the following permutations:
𝜑2 =
𝜑2 2 = 4
𝜑2 4 = 8
𝜑2 6 = 2
𝜑2 8 = 6
𝜑4 =
𝜑4 2 = 8
𝜑4 4 = 6
𝜑4 6 = 4
𝜑4 8 = 2
𝜑6 =
𝜑6 2 = 2
𝜑6 4 = 4
𝜑6 6 = 6
𝜑6 8 = 8
𝜑8 =
𝜑8 2 = 6
𝜑8 4 = 2
𝜑8 6 = 8
𝜑8 8 = 4
Thus, the set 𝐺′
= {𝜑2, 𝜑4, 𝜑6, 𝜑8} is a group of permutations and the mapping 𝜑: 𝐺 → 𝐺′ is defined
by
𝜑:
𝜑 2 = 𝜑2
𝜑 4 = 𝜑4
𝜑 6 = 𝜑6
𝜑 8 = 𝜑8
is an isomorphism from G to G’.
CAYLEY TABLE
A Cayley table (or operation table) of a (finite) group is a table with rows and columns labelled by
the elements of the group and the entry 𝑔 ∗ ℎ in the row labelled g and column labelled h.
CAYLEY DIGRAPHS (DIRECTED GRAPHS)/CAYLEY DIAGRAMS
 For each generating set S of a finite group G, there is a directed graph representing the
group in terms of the generators of S.
 A digraph consists of a finite number of points, called vertices of the digraph, and some arcs
(each with a direction denoted by an arrowhead) joining vertices.
 In a digraph for a group G using the generator set S we have one vertex, represented by a
dot, for each element of G.
 Each generator in S is denoted by one type of arc.
Example:
At the right is a possible digraph for𝑍6 with𝑆 = {2, 3} using  for
<2> and --- for <3>.

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University of Santo Tomas
College of Education
España, Manila
MATH 115: Abstract Algebra
Performance Task 1: Oral Presentation
Written Report on
Isomorphism
Submitted by:
CAPIOSO, RIKKI JOI A.
CHUA, ERNESTO ERIC III A.
GONZALES, MA. IRENE G.
PARK, MIN YOUNG
4MM
Submitted to:
Assoc. Prof. JOEL L. ADAMOS
Course Facilitator
Date Submitted:
08 April 2015, Wednesday