# Proving a function is an isomorphism

1. Mar 26, 2016

### RJLiberator

1. The problem statement, all variables and given/known data
Let G be a finite abelian group with no elements of order 2 Show that the function φ: G-> G defined as φ(g) = g^2 for all g ∈G, is an isomorphism.

2. Relevant equations
Abelian group means xy = yx for all x,y∈G

Isomorphic if there exists a bijection ϒ: G_1 -> G_2 such that for all x,y ∈ G, ϒ(xy) = ϒ(x)ϒ(y)

3. The attempt at a solution

We have a couple of main points.
-Abelian
-Definition of the function maps one element to that element squared.
-G is finite with no elements of order 2.

To prove that it is isomorphic, we use the definition, that there must exist a bijection from G1 to G1 such that for all xy we see ϒ(xy) = ϒ(x)ϒ(y)

I'm not sure where to go from here. Do I just say for some x, y in G we have ϒ(xy) = ϒ(yx) = (xy)^2 = (yx)^2 or some form of this ?

2. Mar 27, 2016

### Samy_A

Well, a group is obviously isomorphic with itself.
What you have to prove is that φ is a group isomorphism.

First thing you have to prove is that φ is a group homomorphism. You more or less did that, although I'm not sure why you used Y and not φ.

Once it is proved that φ is a group homomorphism, you have to prove it is a bijection.

Last edited: Mar 27, 2016
3. Mar 27, 2016

### RJLiberator

Thanks for the help.

How is this looking:

We seek to show that φ is a group isomorphism. First we show that φ is a group homomorphism.

φ(xy) = φ(yx) = (xy)^2 = (yx)^2
(xy)^2 = x^2y^2 = φ(x)φ(y)

Therefore φ(xy) = φ(x)φ(y) and we've proven that it is a group homomorphism.

To the prove that φ is a bijection we show the following:

Given a fixed g in G,
g*x = g*y

g^(-1) * g*x = g^(-1) * g*y
x = y.

so we see one to one correspondence.
Now,
Let x = g^(-1)*y
g*x = y
so φ is onto.

4. Mar 27, 2016

### Samy_A

Correct.
I'm not sure I understand what you do here. What is that fixed g in G?

You have to show that φ(x)=φ(y) implies x=y.
Again, what exactly is g?

You have to show that, given any y ∈ G, ∃ x ∈ G such that φ(x)=y.

5. Mar 27, 2016

### RJLiberator

Could you just say:

Assume φ(x) = φ(y)
then x^2 = y^2 and so x must equal y.

And then
if we let x = φ(-1)*y then φ(x) = φ(φ^(-1)y) = y

6. Mar 27, 2016

### Samy_A

You have to explain why x²=y² implies x=y. That is not true in general in a group (in ℤ, (-1)²=1², for example).
What is φ(-1)?
You have to show that φ is a bijection, so you can just assume φ-1 exists (assuming that's what you meant with φ(-1)).

7. Mar 27, 2016

### RJLiberator

x^2 = y^2 implies x = y as the group is abelian no elements of order 2.
Correct?

Sorry, yes, that is what I meant with φ^(-1).

8. Mar 27, 2016

### Samy_A

Yes, that is correct, but somewhat terse.
An examinator might expect a more detailed argument of how the group being Abelian and not having elements of order 2 implies that x²=y² means that x=y.
So how do you prove that φ is surjective (onto)?

9. Mar 27, 2016

### RJLiberator

I assume that φ^(-1) exists so we let x = φ^(-1)*y

We see φ(x) = φ(φ^(-1)*y) = y.
So φ is onto.

Absolutely. Hm.
I'm having trouble seeing it. By no elements of order 2, and since it is abelian, we know that it either has elements of order greater than 2 or just of order 1.

10. Mar 27, 2016

### Samy_A

You can't assume that φ-1 exists, that is part of what you have to prove.

You have to do it the "hard" way.
Take y ∈ G. You need to find an x ∈ G such that φ(x) = x² = y.

Hint: since G is finite, y as an order, ie ym = e for some integer m (e is the unit element of G). Is m even or odd?
Yes, but that doesn't help much, does it?
Hint: assume you have x² = y². Since x and y are elements of the group G, they have inverses, x-1, y-1.

Last edited: Mar 27, 2016
11. Mar 27, 2016

### RJLiberator

Take y ∈ G. You need to find an x ∈ G such that φ(x) = y.
Could you say that since y^m = e (as discussed), then φ(x) = y would be (φ(x))^m = e and so x = e ?

For the last part, since x and y are elements of the group G they have inverses x^(-1) and y^(-1).
So since we arrived at x^2 = y^2 we can then take x^(-1) on both sides and see
x^(-1)*x*x = x
and x^(-1) *y*y on the other side.

so x = x^(-1)*y*y

now take y^(-1) and see
y^(-1)*x = x^(-1)*y

Not sure if this helps tho...

12. Mar 27, 2016

### Samy_A

Not really. You don't have x yet.

Let's do it in steps. If m is the order of y, is m an even or an odd integer?

Way too complicated. Why not multiply both sides twice by y-1, and remember that the group is abelian?

13. Mar 27, 2016

### RJLiberator

Ah... yes....

It would be odd as there are no elements of order 2.
EDIT: wait, since the map takes an element in g and maps it to that element ^2 we see that it is even.
EDIT:EDIT My edit was wrong :P

14. Mar 27, 2016

### Samy_A

Correct.
So, ym = e, with m some odd integer.
Then what is ym+1? And can this lead to an x ∈ G satisfying φ(x) = x² = y?

15. Mar 27, 2016

### RJLiberator

y^(m+1) = y?

So then if some x = y^((m+1)/2) then x^2 would equal y?

16. Mar 27, 2016

### Samy_A

Yes, x = y(m+1)/2 satisfies φ(x) = y. That proves that φ is surjective.

17. Mar 27, 2016

### RJLiberator

wooooohoooo. What beauty.