Proving a function is an isomorphism

[RJLiberator]

Homework Statement

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.

Homework 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)

The Attempt at a Solution

[/B]
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 ?

 

[Samy_A]

Homework Statement

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.

Homework 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)

The Attempt at a Solution

[/B]
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 ?

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.

 

[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.

 

[Samy_A]

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.

Correct.

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.

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.

Now,
Let x = g^(-1)*y
g*x = y
so φ is onto.

Again, what exactly is g?

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

 

[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

 

[Samy_A]

Could you just say:

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

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).

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

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)).

 

[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).

 

[Samy_A]

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

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.

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

So how do you prove that φ is surjective (onto)?

 

[RJLiberator]

So how do you prove that φ is surjective (onto)?

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.

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.

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.

 

[Samy_A]

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.

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 y^m = e for some integer m (e is the unit element of G). Is m even or odd?

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.

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.

 

[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...

 

[Samy_A]

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 ?

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?

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...

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

 

[RJLiberator]

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

Ah... yes...

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

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

 

[Samy_A]

It would be odd as there are no elements of order 2.

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

 

[RJLiberator]

y^(m+1) = y?

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

 

[Samy_A]

y^(m+1) = y?

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

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

 

[RJLiberator]

wooooohoooo. What beauty.