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: (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 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: 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: 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:

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:

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 :GH such that a, b  G, (ab) = (a)(b). Because the function is bijective, we know the groups are the same size. Because of the equation (ab) = (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 : GH 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) = (xkx) = (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, If G is abelian, then H is abelian. If G is cyclic, then H is cyclic. Proof of 1: Suppose : GH 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 : GH is an isomorphism. And suppose there exists x G such that G = {xn n Z}. Will show that H = {(x)nn Z} (This will show H is cyclic). Let h H. Since  is onto, there exist gG such that (g)= h. Then there exists nZ such that g = xn. Then h = (g) = (xn) = (x)n This shows that H  {(x)nn Z}. But since H is a group (closed) it is clear that H  {(x)nn Z}. So H = {(x)nn Z}. Therefore, H is cyclic.

Two Theorems about Isomorphisms Theorems: Suppose (G, ) and (H, ) are groups and : GH 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) = (aa-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! (Becauseais n and kn.) 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!