Math 344 Winter 07Group 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

• (Identity)

There exists an element e G such that a•e = e•a = a for all a G

• (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 give 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 HW5 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.)

• On HW 6 you proved that the centralizer of a group element is a subgroup.

• 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: abc12

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: ab123

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:

abc12

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: ab312

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 GroupsTheorems: 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 IsomorphismsTheorems: 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) = (e) = eH (where eH is the identity in H.) and

(a-1)(a) = (a-1a) = (e) = eH . So the inverse of (a) is (a-1) In other words, (a)-1 = (a-1).

Proof of 2: Let eG be the identity of G. Suppose a= n. Then an = eG. So (a)n = (an) = (eG) = eH the identity in H.

Now we need to show n is the smallest integer with this property. Suppose that kn and (a)k = (eG) = eH.

Then since (a)k = (ak), (ak) = (eG) But, then since is 1-1, ak = eG. This cannot be! (Becauseais n and kn.)

We have a contradiction, so n is the smallest integer n 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!