# Groups and Isomorphisms.

1. Oct 11, 2009

### cmj1988

Given a group G. J = {$$\phi$$: G -> G: $$\phi$$ is an isormophism}. Prove J is a group (not a subgroup!).

Well we know the operation is function composition. To demonstrate J a group we need to satisfy four properties:

(i) Identity: (I'm not sure what to do with this)
(ii) Inverses: Suppose a $$\phi$$ in J. We know that since $$\phi$$ is a one-to-one and onto function, we know that it has an inverse (demonstrated in class). Hence we know for every $$\phi$$ in J, there exists a $$\phi$$^-1.
(iii) Closure: Suppose a $$\phi1[\tex] and [tex]\phi2$$ such that $$\phi$$1(g)=g and $$\phi2$$(g)=g. Therefore $$\phi$$1($$\phi2$$(g))=g. This follows from what we proved about functions in class.
(iv) Associativity: This one is just moving a bunch of parentheses aroudn while composing and hoping (fingers crossed) that regardless of order of composition a g pops out.

My big problem is identity.

2. Oct 12, 2009

### Office_Shredder

Staff Emeritus
You might know of a map that's often called the identity map - f(g) = g for all g in G. This is not a coincidence.

I take issue with points ii and iii. For point ii, you need to show that its inverse is also a homomorphism, but maybe you did that in class also.

For point iii,. you've chosen both maps to be identity maps (they map g to g) not isomorphisms.

3. Oct 12, 2009

### cmj1988

How would I demonstrate closure? Since its an isomorphism from G-> G won't it just keep sending elements to itself?

4. Oct 12, 2009

### cmj1988

Also, demonstrating its a homomorphism. Doesn't that immediately follow since it is a function through composition, composing a function with its inverse is kosher.

5. Oct 12, 2009

### Office_Shredder

Staff Emeritus
An isomorphism does not send an element of G to itself, it sends elements of G to elements of G. For example the function
f(x)=-x defined on the integers (with + being your group operation) is an isomorphism, but f(x) never equals x unless x is zero

6. Oct 12, 2009

### cmj1988

Alright, so here is my new proof of closure:

Suppose a $$\phi$$1(g1)=g2 and $$\phi$$2(g2)=g3. Composing and such, it is still closed. I feel like I don't necessarily have to prove closure to demonstrate its a group just because binary operations are assumed to be closed.

Another thing is how do I demonstrate that homomorphism is preserved when demonstrating inverses and such. Since the operation is function composition, would we even have to worry about demonstrating that the action still works? Function composition is pretty versatile.

7. Oct 12, 2009

### Office_Shredder

Staff Emeritus
What? While it may be that you assumed that a group has a binary operation that maps to the group, and hence closure is not an axiom, you still need to show that composition works here. Otherwise, for example:

The set A of all rational numbers of the form n or 1/n, where n is a natural number, is a group under multiplication. It has inverses (1/n*n=1), identity (1) and is associative. The key element here is that I picked a binary operation which doesn't satisfy the definition of a binary operation that a group must have, i.e. multiplication here doesn't map AxA to A; for example 3*(1/2)

If $\phi :G \rightarrow G$ is an isomorphism, an algebraic inverse $\phi ^{-1}$ exists and is obviously bijective. What's left to prove is that $\phi ^{-1}$ is a homomorphism, namely that $\phi ^{-1}(ab) = \phi ^{-1}(a) \phi ^{-1}(b)$ You can write a and b in terms of $\phi$ since it is onto, and that should give you the result that you want