Recall:
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Definition: A function f : G1 → G2 is well-defined if
x = y =⇒ f (x) = f (y ).
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Definition: An isomorphism from a group (G , ∗) to a group (Ḡ , )
is a one-to-one and onto function φ : G → Ḡ that preserves the group
action. That is,
φ is 1-1, φ is onto, and φ(a ∗ b) = φ(a) φ(b).
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If an isomorphism from G to Ḡ (any isomorphism) exists, then we say
that G and Ḡ are isomorphic and denote this by G ≈ Ḡ .
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As an example of an isomorphism, we constructed the canonical
isomorphism from a cyclic group < a > of order 5 and Z5 :
φ :< a >→ Z5 by φ(ak ) = k
Math 321-Abstracti (Sklensky)
In-Class Work
mod 5.
October 8, 2010
1 / 11
In Class Work
1. Find an isomorphism from the group U(8) to U(12).
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
2 / 11
In Class Work
1. Find an isomorphism from the group U(8) to U(12).
Hint: Sometimes, it’s easiest just to define a function
element-by-element.
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
2 / 11
In Class Work
1. Find an isomorphism from the group U(8) to U(12).
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
Hint: The even integers are of course 2Z = {2k|k ∈ Z}. Exploit the
obvious connection between Z and 2Z.
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
2 / 11
In Class Work
1. Find an isomorphism from the group U(8) to U(12).
Hint: Sometimes, it’s easiest just to define a function
element-by-element.
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
Hint: The even integers are of course 2Z = {2k|k ∈ Z}. Exploit the
obvious connection between Z and 2Z.
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
2 / 11
Hint:
1. Find an isomorphism from the group U(8) to U(12).
Look at the Cayley tables for each:
U(8)
· mod 8 1 3
1
1 3
3 1
3
5
5 7
7 5
7
5
5
7
1
3
7
7
5
3
1
U(12)
· mod 12 1
5
1
1
5
5
1
5
7
7 11
11 7
11
7
7
11
1
5
11
11
7
5
1
Each group consists of an identity and 3 elements of order 2.
While we don’t know much about isomorphisms yet, we know that
they are a way of showing that two groups are essentially the same,
so as a first guess, we’d choose to send each non-identity element in
U(8) to a different non-identity element of U(12).
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
3 / 11
Hint:
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
We know:
2Z = {2k|k ∈ Z}.
How is every element in 2Z related to an element in Z?
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
4 / 11
Solutions
1. Find an isomorphism from the group U(8) to U(12).
Look at the Cayley tables for each:
U(8)
· mod 8 1 3
1
1 3
3 1
3
5
5 7
7 5
7
5
5
7
1
3
7
7
5
3
1
U(12)
· mod 12 1
5
1
1
5
5
1
5
7
7 11
11 7
11
7
7
11
1
5
11
11
7
5
1
Each group consists of an identity and 3 elements of order 2.
While we don’t know much about isomorphisms yet, we know that
they are a way of showing that two groups are essentially the same,
so as a first guess, we’d choose to send each non-identity element in
U(8) to a different non-identity element of U(12).
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
5 / 11
Solutions
1. (continued)
Define f : U(8) → U(12) by
f (1) = 1
f (3) = 5
f (5) = 7
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f is clearly well-defined, 1-1, and onto
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Does f preserve the group operation?
f (7) = 11.
We need to show that f (n · m mod 8) = f (n) · f (m) mod 12.
Again, this may be easier to see element-by-element.
n, m
n, 1
a, a
3,5
3,7
5,7
f (n · m mod 8)
f (n mod 8) = f (n)
f (a2 mod 8) = f (1) = 1
f (7) = 11
f (5) = 7
f (3) = 5
f (n) · f (m) mod 12
f (n) · 1 mod 12 = f (n)
f (a)2 mod 12 = 1
5 · 7 mod 12 = 11
5 · 11 mod 12 = 7
7 · 11 mod 12 = 5
Thus f does indeed preserve the group operation
Hence f is indeed an isomorphism, so U(8) ≈ U(12).
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
6 / 11
Solutions
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
We know:
2Z = {2k|k ∈ Z}.
How is every element in 2Z related to an element in Z?
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
7 / 11
Solutions
2. Find an isomorphism from the group of integers under addition to the
group of even integers under addition.
We know:
2Z = {2k|k ∈ Z}.
This gives us a correlation between the elements of Z and 2Z. Let’s
use it to define what we hope will turn out to be an isomorphism.
Let φ : Z → 2Z by φ(k) = 2k.
0.1 Is φ well-defined?
If n = m, then φ(n) = 2n = 2m = φ(m).
0.2 Is φ one-to-one?
Suppose φ(k) = φ(m) for some m, n ∈ Z. Then 2k = 2m, and so
k = m.
0.3 Is φ onto?
Let y ∈ 2Z. Then there exists an n ∈ Z such that y = 2n. φ(n) = y ,
and so φ is onto.
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
8 / 11
Solutions
2. (continued))
(d) Does φ preserve theJ
group operation?
NTS φ(x ∗ y ) = φ(x) φ(y ). Since the group operation of both Z and
2Z are simple addition, we need to show that φ(x + y ) = φ(x) + φ(y ).
Let x, y ∈ Z. Then φ(x + y ) = 2(x + y ) = 2x + 2y = φ(x) + φ(y ).
Therefore φ does preserve the group operation.
Thus φ is an isomorphism.
Notice what this tells us!
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
9 / 11
Thm 6.2: Properties of Isomorphisms Acting on
Elements
Suppose that φ : G → Ḡ is an isomorphism. Then
1. φ carries the identity of G to the identity of Ḡ .
2. For every integer n and for every group elt a ∈ G , φ(an ) = [φ(a)]n .
3. For any elements a, b ∈ G , ab = ba ⇔ φ(a)φ(b) = φ(b)φ(a).
4. G =< a > if and only if Ḡ =< φ(a) > .
5. |a| = |φ(a)| for all a in G . (That is, isomorphisms preserve order).
6. For a fixed k ∈ Z and a fixed b ∈ G , the eqn x k = b has the same
number of solutions in G as does the eqn x k = φ(b) in Ḡ .
7. If G is finite, then G and Ḡ have exactly the same number of
elements of every order.
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
10 / 11
Thm 6.3: Properties of Isomorphisms Acting on
Groups
Suppose that φ : G → Ḡ is an isomorphism. Then
1. φ−1 is an isomorphism from Ḡ to G .
2. G is Abelian if and only if Ḡ is Abelian.
3. G is cyclic if and only if Ḡ is cyclic.
4. If K is a subgroup of G , then φ(K ) = {φ(k)|k ∈ K } is a subgroup of
Ḡ .
Math 321-Abstracti (Sklensky)
In-Class Work
October 8, 2010
11 / 11