# Isomorphism is an equivalence relation on groups

1. Mar 12, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
Prove that isomorphism is an equivalence relation on groups.

2. Relevant equations

Need to prove reflexivity, symmetry, and transitivity for equivalence relationship to be upheld.

**We will use ≅ to define isomorphic to**
3. The attempt at a solution

Let G, H, and K be groups.
We want to show that G≅G, G≅H, and if G≅H H≅K then G≅K.

Reflexivity: Define f:G->G by f(x)=x. we show f is an isomorphism. For x,y ∈ G if f(x) = f(y), then by definition of f we see that x=y. f is thus one to one. f(x)=x so f is onto. f(xy) = xy = f(x)f(y).
There f is an isomorphism and G≅G.

Symmetric: We assume G≅H. Then there is an isomorphism f:G->H, that is a bijection and homomorphism.
Claim: f^(-1) from H to G is an isomorphism.
Since f is a bijection, so is f^(-1).
let h, h' ∈H and g,g' ∈G. f^(-1)(h*h') = f^(-1)(h)f^(-1)(h').
Since f^(-1) is a bijection, there exists g,g'∈G such that f(g) = h and f(g') = h'.
f(gg') = f(g)f(g') = hh' => f^(-1)(hh') = gg'.
f^(-1)(h)f^(-1)(h') = gg'.
Thus, f^(-1)(hh') = f^(-1)(h)f^(-1)(h') as required. Therefore H≅G.

Transitive: Assume G≅H and H≅K, then there are isomorphisms g:G->H and f->K.
Claim: (g°f):G->K is an isomorphism.
The composition of two bijections is also a bijection.
For any h,h' ∈G. (g°f)(hh') = g(f(hh')) = g(f(h)f(h')) = g(f(h))g(f(h')) = (g°f)(h)(g°f)(h')
Hence, G≅K.

Thus, isomorphisms of groups is an Equivalence relationship.

Question: I am most concerned about the symmetric part. Did I do this part right?

How can I arbitrarily define isomorphisms like f:G->G by f(x)=x?

Thank you.

2. Mar 12, 2016

### PeroK

It looks okay to me. I've underlined one thing that I think is superfluous. On the last question, you can say "let I or f be the identity mapping".

3. Mar 12, 2016

### RJLiberator

Is the underlined part due to me using an asterisk there when it is not needed?

4. Mar 12, 2016

### PeroK

The bit I underlined can all be deleted. It's not necessary. Nothing to do with the asterisk.

5. Mar 12, 2016

### RJLiberator

I see. Reading it through, that makes sense. Thanks.

6. Mar 12, 2016

### geoffrey159

Yes if $f$ is a group isomorphism, so is its inverse. However, you don't explicitly assign internal laws in your notations, and it can be bothering for the reader.

7. Mar 12, 2016

### RJLiberator

I am unfamiliar with this term 'internal laws'

8. Mar 12, 2016

### geoffrey159

When you describe a group, you have to specify a pair $(S,\star)$, where $S$ is a set, and $\star : S\times S \to S$ is the composition law ( or internal law ).

9. Mar 12, 2016

### RJLiberator

Ahh, I get it. So you are saying, perhaps I should assign internal laws to my proofs for easier reading/ a more clear proof. Such as (S,*).

10. Mar 12, 2016

### geoffrey159

Yes, just for easier reading, not to criticize you