Transcript Document
Math 344 Winter 07
Group Theory Part 2: Subgroups and
Isomorphism
Recall our Definition of Group
Definition: A group (G, •) is a set, G, together with an
operation, •, satisfying the following axioms:
1. (Associative)
(a•b)•c = a•(b•c), for all a, b, c G
2.
(Identity)
There exists an element e G such that a•e = e•a = a
for all a G
3.
(Inverses)
For each a G there is an element a-1 G such that
a•a-1 = a-1•a = e.
We used this definition in proofs a couple of
different ways
• We proved/disproved that a given set and
operation formed a group.
Example: Prove or disprove {0, 1} forms a group
under multiplication.
• We proved various things (both specific and
general) about groups
Example: Prove that a group G is commutative if
and only if (ab)2 = a2b2 a, b G
Some Theorems about Groups
• Theorem: The identity element of a
group is unique. (See Day 5 proof)
• Theorem: Each element of a group has a
unique inverse. (See Day 8 proof)
Subgroups
We defined a subgroup to be a subset of a group that is a
group itself under the same operation.
•
•
We found all the subgroups of D6, D8 and Z.
On a homework, you proved that the center of a group is a
subgroup (the center is the set of elements that commute with
every element of the group.)
We also proved two handy theorems about subgroups:
Theorem: Let G be a group and H a subset of G. Then H is a subgroup
iff 1) H is nonempty, 2) H is closed under the operation defined
on G, and 3) each element of H has its inverse in H.
Theorem: Let G be a group and H a nonempty finite subset of G. If H
is a closed under the operation defined on G then H is a subgroup
of G.
All about Functions: Everywhere Defined
•
By definition, all functions are everywhere defined and
well defined.
Definition: Suppose f: S T. f is everywhere-defined
if s S, t T such that f(s) = t.
This means each element of the domain gets sent
somewhere. If a “function” were not everywhere
defined, you would get a picture like this:
a
1
b
c
2
Example: Suppose I tried to make a function from the
integers to the rational numbers using the formula f(n) =
1/n. What if anything would go wrong?
All about Functions: Well Defined
•
By definition, all functions are everywhere-defined
and well-defined
Definition: Suppose f: S T. f is well-defined if
s1 = s2 f(s1) = f(s2) for all s1, s2 S.
This means each element of the domain gets Each
element of S goes to a unique element in T.
If a “function” were not well-defined, you would get
a picture like this:
a
b
1
2
3
Example: Suppose I tried to make a function from the
set of rotations of a triangle to the integers using the
formula f(Rn) = n. What if anything would go wrong?
All about Functions: Onto
•
Some functions are onto.
Definition: Let f: S T be a function. f is onto if
t T, s S such that f(s) = t.
This means each element of the co-domain (or
target) gets hit. In other words, the range is the
whole co-domain.
If a function were onto, you would get a picture like
this:
a
b
c
1
2
To prove a function is Onto
Definition: Let f: S T be a function. f is onto if t T,
s S such that f(s) = t.
This means each element of the co-domain (or target) gets hit.
In other words, the range is the whole co-domain.
So to prove a function is onto:
Take an arbitrary element of the codomain and show that there
is an element in the domain that maps to it.
Proof: Let t T
(Magic whereby you find a domain element, s, that
works)
Then f(s) = t. QED
All about Functions: One-to-One
•
Some functions are 1-1.
Definition: Let f: S T be a function. f is 1-1 if
f(s1) = f(s2) s1 = s2 for all s1, s2 S.
This means no two elements of the domain go to the
same element in the codomain.
If a function were 1-1, you would get a picture like
this:
a
b
1
2
3
To prove a function is One-to-One
Definition: Let f: S T be a function. f is 1-1 if f(s1) =
f(s2) s1 = s2 for all s1, s2 S.
This means no two elements of the domain go to the same
element in the codomain.
So to prove a function is 1-1, you suppose that two inputs give
the same output and show that these inputs must be equal.
Proof: Suppose f(s1) = f(s2)
(Magic happens until you get…)
Then s1 = s2. QED
Isomorphism
Informal: Two groups are isomorphic if they are
essentially the same.
Definition: Let (G, ) and (H, ) be groups. G is
isomorphic to H if there exists a bijective
function :GH such that a, b G,
(ab) = (a)(b).
• Because the function is bijective, we know the
groups are the same size.
• Because of the equation (ab) = (a)(b),
we know that the operation works the same in
each group.
A little lemma about isomorphisms
Lemma: Suppose (G, ) and (H, ) are groups and : GH is
an isomorphism.
Then (xn) = (x)n x G and n N .
Proof:Let x G.
(Will Proceed by Mathematical Induction)
Base case: Note that (x1) = (x) = (x)1 So the statement is true for
n=1.
Inductive step: Assume that (xk) = (x)k.
Then (xk+1) = (xkx) = (xk)(x) = (x)k (x) = (x)k+1. So if the
statement is true for n = k, it is true for n = k+1.
Therefore by the Principle of Mathematical Induction, the statement is
true for all n n N if we don’t count 0. Does it work for 0?
Why?
Two Theorems about Isomorphic Groups
Theorems: Suppose G and H are isomorphic groups then,
1. If G is abelian, then H is abelian.
2. If G is cyclic, then H is cyclic.
Proof of 1: Suppose : GH is an isomorphism. And suppose G is
abelian. Will show H is abelian.
Let c, d H. Since is onto, there exist a, b G such that (a)= c and
(b)= d. Since G is commutative, ab = ba. Thus since is an
isomorphism,
cd = (a) (b) = (ab) = (ba) = (b)(a)= dc. H is abelian.
Proof of 2: Suppose : GH is an isomorphism. And suppose there
exists x G such that G = {xn n Z}.
Will show that H = {(x)nn Z} (This will show H is cyclic).
Let h H. Since is onto, there exist gG such that (g)= h. Then
there exists nZ such that g = xn. Then h = (g) = (xn) = (x)n
This shows that H {(x)nn Z}. But since H is a group (closed) it is
clear that H {(x)nn Z}. So H = {(x)nn Z}.
Therefore, H is cyclic.
Two Theorems about Isomorphisms
Theorems: Suppose (G, ) and (H, ) are groups and : GH is
an isomorphism. Let a G. Then,
1. (a)-1 = (a-1) and 2. a = (a).
Proof of 1: Let eG be the identity of G. Note (eG) is the identity of H
(proved in class Day 10). Since is an isomorphism,
(a)(a-1) = (aa-1) = (eG) = eH (the identity in H) and
(a-1) (a) = (a-1 a) = (eG) = eH . So the inverse of (a) is (a-1)
In other words, (a)-1 = (a-1).
Proof of 2: Suppose a= n. Then (a)n = (an) = (eG) = eH.
Now we need to show n is the smallest integer with this property.
Suppose that k n and (a)k = eH. But since (a)k = (ak) and
(eG) = eH, this means (ak) = (eG). So, since is 1-1, ak = eG.
This cannot be! (Becauseais n and kn.)
We have a contradiction. Thus n is the smallest integer such that
(a)n = eH. Therefore, (a) = a.
Why do we like these theorems?
Because they make it easy to prove groups are not
isomorphic! And they give a way to check whether we
need to go to the trouble of creating an isomorphism.
When asked to prove/disprove whether two groups are
isomorphic, check to see if they have the same
properties:
• Commutative? (if one is, they both need to be)
• Cyclic? (if one is, they both need to be)
• Same number of self inverses?
• Same number of elements of elements of order 2, 3,
etc…
If the two groups are different in any of these ways, they
are NOT isomorphic. You prove they are not
isomorphic simply by showing that they are different in
the way you claim they are.
Why do we like these theorems? (Part 2)
If it looks like the groups probably are isomorphic, these
theorems can help when you try to create your
isomorphism.
Make sure to:
• Send the identity to the identity
• Send self-inverses to self-inverses
• (cyclic groups) Send generators to generators.
In the case of small finite groups: You do NOT need make
a formula for your isomorphism. Simply use arrows to
match up the elements then use the tables to test to see
if it works.
In the case of infinite groups: You WILL need to make a
formula for your isomorphism. You can use the points
above to make sure you are on the right track. Be
careful to make sure you are actually defining a
function (well-defined and everywhere-defined).
And that is where we are so far!