isomorphism

of 3

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 = 𝑥2𝑦2 = 𝑥2𝑦2 = 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}.

of 3

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 𝑎𝑏+𝑐 = 𝑎𝑏 (𝑎𝑐) that the addition of exponents corresponds to multiplication of two quantities.

Thus we try defining 𝜑: 𝑅 → 𝑅+ by 𝜑 𝑥 = 𝑒𝑥 for 𝑥 ∈ 𝑅. Note that 𝑒𝑥 > 0 for all 𝑥 ∈ 𝑅 so indeed 𝜑 𝑥 ∈ 𝑅+. Step 2. If 𝜑 𝑥 = 𝜑(𝑦), then 𝑒𝑥 = 𝑒𝑦 . 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 = 𝑒𝑥+𝑦 = 𝑒𝑥 ∙ 𝑒𝑦 = 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.

of 3

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 = 0.5𝑟 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 𝜑𝑎 : 𝐺 → 𝐺′ be defined by 𝜑𝑎 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>.

of 3

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