Basic proof for Homomorphism of Abelian Groups

[dabokey]

Homework Statement

Let f : G → H be a homomorphism of Abelian groups.

1. Show that f (0) = 0.
2. Show that f (−x) = −f (x) for each x ∈ G.

Homework Equations

The Attempt at a Solution

My background in topology / group theory is next to nothing.

1. Show that f(0) = 0. My attempt is as follows:

f(x+y) = f(x) + f(y)

Let y = 0

f(x+0) = f(x) + f(0)
f(x) -f(x) = f(x) -f(x) + f(0)
0 = 0 + f(0)
0 = f(0)

Is this a valid proof?

2. I tried the following:

f(-x) = f(-1 * x) = f(-1)*f(x) = -1 * f(x) = -f(x)

I'm fairly certain that for a homomorphism, f(1) = 1. Is that correct? I'm guessing that f(-1) = -1, though I'm not sure whether this is correct.

Can somebody please provide me with some comments or suggestions?

Thank you!

 

[NoMoreExams]

Well f(n) = n right? Since let's take f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2.

You can check that f(1) = 1 since f(1) = f(1*1) = f(1)*f(1) therefore 1 = f(1)

I guess I'm not remembering my algebra right what's the operation for G and H. In one case we seem to be dealing with addition, the other multiplication or does it not matter?

 

[dabokey]

I'm not sure that f(n) = n. That isn't given in the problem statement.

Verifying that f(1) = 1 isn't difficult, but I do appreciate your proof.

For a homomorphism, f(x+y) = f(x) + f(y) because f maps G → H

How do I show that f(-x) = -f(x)?

For that matter, is it true that f(-1) = -1?

Thank you!

 

[Hurkyl]

f(-1 * x)

This doesn't make sense. What is -1? What is *? In the problem at hand, there is no such thing as '1'.

 

[dabokey]

This doesn't make sense. What is -1? What is *? In the problem at hand, there is no such thing as '1'.

Hurkyl,

I am at the beginning of learning group theory on my own, so my questions may seem a bit basic.

Why doesn't f(-1 * x) make sense? Does f(-x) actually mean the f(inverse of x)?

With regards to '*' and '-1', I meant to factor '-x' into two separate parts so that I can somehow apply the axiom f(x+y) = f(x) + f(y). Based upon your reply, I believe that you're telling me that '*' and '-1' are not valid operations in this problem?

Why is there no such thing as '1'? I thought that '1' was the identity element.

Thank you for your help.

 

[NoMoreExams]

Hurkyl,

I'm not sure I understand your reply.

What doesn't f(-1 * x) make sense? I'm confused.

Why is there no such thing as 1?

I thought I proved that f(n) = n, it doesn't need to be given since

$$f(n) = f(\sum_{i=1}^{n}{1}) = \sum_{i=1}^{n}{f(1)} = \sum_{i=1}^{n}{1} = n$$

1 doesn't make sense because 1 is the identity element for multiplication. For hom f(e) = e. So if you have addition then e is 0 hence f(0) = 0.

That's why I asked which operation are you working under.

 

[radou]

First of all, if you have chosen additive notation for an Abelian group, you should be perfectly consistent with it.

OK, for the second one, what does f(-x) + f(x) equal? (Hint: f is a group homomorphism.)

 

[Hurkyl]

Why is there no such thing as '1'? I thought that '1' was the identity element.

The name of the identity element is whatever name you assign to it. You were previously calling the identity element '0'. Furthermore, you had been using '+' for the binary operation, and '-' for the inverse operation.

If you really meant to use '1' for the identity, and '*' for the binary operation, then nothing you wrote previously makes sense. (And the way you used '-' would be nonsensical)


There are structures called 'rings' that have two binary operations (usually written '+' and '*'), and the corresponding two identities (usually written '0' and '1'), and given the identities for rings, mixing them like that would make sense. But the question isn't about rings...

 

[dabokey]

I thought I proved that f(n) = n, it doesn't need to be given since

$$f(n) = f(\sum_{i=1}^{n}{1}) = \sum_{i=1}^{n}{f(1)} = \sum_{i=1}^{n}{1} = n$$

1 doesn't make sense because 1 is the identity element for multiplication. For hom f(e) = e. So if you have addition then e is 0 hence f(0) = 0.

That's why I asked which operation are you working under.

Ok, so I am working under the operation of addition because I'm given that f(x+y) = f(x) + f(y) for f: G-->H.

Thanks to your clarification, I now understand the difference in identity elements between the additional and multiplication operations.

Now, I'm working with addition and am trying to show that f(-x) = -f(x).

Does -x mean the inverse of x?

f(-x) = -f(x)
= 0 - f(x)
= f(0) - f(x)

But where do I go from there?

 

[NoMoreExams]

Ok, so I am working under the operation of addition because I'm given that f(x+y) = f(x) + f(y) for f: G-->H.

Thanks to your clarification, I now understand the difference in identity elements between the additional and multiplication operations.

Now, I'm working with addition and am trying to show that f(-x) = -f(x).

Does -x mean the inverse of x?

f(-x) = -f(x)
= 0 - f(x)
= f(0) - f(x)

But where do I go from there?

Wait why are you starting with what you are trying to prove? If you are doing a direct proof, start with f(-x) and through a series of logical connections, get to -f(x). Use the hint given above, those people seem to know more than I do :)

And yes, -x is the additive inverse of x.

So perhaps 0 = f(0) = f(x - x) = f(x) + f(-x) therefore f(x) = -f(x)

 

[dabokey]

The name of the identity element is whatever name you assign to it. You were previously calling the identity element '0'. Furthermore, you had been using '+' for the binary operation, and '-' for the inverse operation.

If you really meant to use '1' for the identity, and '*' for the binary operation, then nothing you wrote previously makes sense. (And the way you used '-' would be nonsensical)There are structures called 'rings' that have two binary operations (usually written '+' and '*'), and the corresponding two identities (usually written '0' and '1'), and given the identities for rings, mixing them like that would make sense. But the question isn't about rings...

Ok, I've heard of rings and your answer makes sense.

I believe that I've proved f(0) = 0 as follows:

f(x + y) = f(x) + f(y)

Take y = 0
f(x + 0) = f(x) + f(0)
f(x) = f(x) + f(0)
f(x) - f(x) = f(x) - f(x) + f(0)
0 = f(0), which is what I was trying to prove.

But to prove -f(x) = f(-x), I'm at a loss.

Hurkyl helped tremendously with this:

0 = f(0)
0 = f(x - x)
0 = f(x) + f(-x)
-f(x) = -f(x) + f(x) + f(-x)
-f(x) = 0 + f(-x)
-f(x) = f(-x), which is what I was trying to prove.

Do both of these look valid?

Thank you!

 

[dabokey]

Wait why are you starting with what you are trying to prove? If you are doing a direct proof, start with f(-x) and through a series of logical connections, get to -f(x). Use the hint given above, those people seem to know more than I do :)

And yes, -x is the additive inverse of x.

So perhaps 0 = f(0) = f(x - x) = f(x) + f(-x) therefore f(x) = -f(x)

Thank you, your proof seems so simple.

I honestly appreciate your help and feel humbled. :)

Regarding your question as to why I started with what I was trying to prove: The answer is that I'm new to proofs and am trying to learn how to do them. I've been out of school for about 4 years and took math up to linear algebra. I didn't have to do any proof based math in the pursuit of my degree.

 

[dabokey]

First of all, if you have chosen additive notation for an Abelian group, you should be perfectly consistent with it.

OK, for the second one, what does f(-x) + f(x) equal? (Hint: f is a group homomorphism.)

That's an excellent hint and I wish that I would have seen it a little earlier. I appreciate your help!