The document discusses group isomorphism and homomorphism, defining isomorphism as a one-to-one mapping between groups that preserves group operations. It outlines the steps to prove isomorphism between two groups, detailing the concepts of kernel, types of homomorphisms, and provides examples showing whether specific mappings are homomorphisms or isomorphisms. Additionally, it introduces Cayley's theorem and Cayley tables to represent group operations and structures.

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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