You are Here: Home >< Maths

# One-to-one homomorphism watch

1. I have this theorem in my notes:

There is a homormophism with and if and only if

1.
2.
3.

Now this is what I want to prove:

If , , then f is one-to-one provided n>2. Hint: y should not be its own inverse.

Can anybody give me hint for this? I've been working on this for hours and I'm so confused.

2. (Original post by Artus)
I have this theorem in my notes:

There is a homormophism with and if and only if

1.
2.
3.
Firstly, this isn't really a theorem. All it says is that if is any function from the dihedral group to any other group then it is a homomorphism if and only if it is trivial on the relations of where the dihedral group is tacitly given by the presentation . This is just the definition of a homomorphism in the context of groups given by presentations.

(Original post by Artus)
Now this is what I want to prove:

If , , then f is one-to-one provided n>2. Hint: y should not be its own inverse.
[/latex]
Consider the restriction of to its image and then use the first isomorphism theorem.
3. Thanks a lot. I think I know how to prove it, but I just have 2 more questions:

1) , since n>2 and y cannot be its own inverse. But if we raise both sides of this equation to the power n, we get . I must be doing something wrong, but I just can't see it...

2)

(Original post by Mark85)
All it says is that if is any function from the dihedral group to any other group then it is a homomorphism if and only if it is trivial on the relations of where the dihedral group is tacitly given by the presentation .
I'm not sure what "trivial on the relations" means.
4. (Original post by Artus)
1) , since n>2 and y cannot be its own inverse. But if we raise both sides of this equation to the power n, we get . I must be doing something wrong, but I just can't see it.../
A couple of things here:
If there's no "" sign, then this cannot be an equation and thus cannot necessarily be treated as though it is one. Take the following, by your logic:
2≠-2 => 2^2≠(-2)^2
Which is, of course, nonsense because this isn't an equation so you can't carry out most of the usual operations and still expect the result to hold.

What you're saying here is that . This statement is logically equivalent to the statement [since (P=>Q) is identical to (not(Q)=>not(p))]. Why would you expect this equivalent statement to be true? As you should know, you can't "nth root" elements of groups in such equations so you should instinctively expect that claim to be false at first glance.

In this case, you're saying that "if two elements in a finite group are not the same thing, then they cannot have the same order" (note that your argument extends to any finite group with an element of order n [and beyond, in fact, but that's irrelevant] rather than simply D_(2n) because the only property you've used here is that y^n=e) or equivalently "if two elements of a finite group have the same order, then they must be the same". When said that way, alarm bells should be ringing as that can't possibly be true by countless counterexamples (C_3, Klein-4, Q_8 etc).

Broadly speaking, working with 'non-equalities' (for want of a better phrase) is often a bit useless/awkward as it doesn't usually give you a great deal of information in that form.

I'm not sure what "trivial on the relations" means.
The relations are the three equations given in Mark's post. To say f is trivial on the relations is to say that it carries over the way in which the elements are combined in D_(2n) to the group G.
5. Yeah Farhan's summed it up perfectly.
6. (Original post by Farhan.Hanif93)
A couple of things here:
If there's no "" sign, then this cannot be an equation and thus cannot necessarily be treated as though it is one. Take the following, by your logic:
2≠-2 => 2^2≠(-2)^2
Which is, of course, nonsense because this isn't an equation so you can't carry out most of the usual operations and still expect the result to hold.

What you're saying here is that . This statement is logically equivalent to the statement [since (P=>Q) is identical to (not(Q)=>not(p))]. Why would you expect this equivalent statement to be true? As you should know, you can't "nth root" elements of groups in such equations so you should instinctively expect that claim to be false at first glance.

In this case, you're saying that "if two elements in a finite group are not the same thing, then they cannot have the same order" (note that your argument extends to any finite group with an element of order n [and beyond, in fact, but that's irrelevant] rather than simply D_(2n) because the only property you've used here is that y^n=e) or equivalently "if two elements of a finite group have the same order, then they must be the same". When said that way, alarm bells should be ringing as that can't possibly be true by countless counterexamples (C_3, Klein-4, Q_8 etc).

Broadly speaking, working with 'non-equalities' (for want of a better phrase) is often a bit useless/awkward as it doesn't usually give you a great deal of information in that form.

The relations are the three equations given in Mark's post. To say f is trivial on the relations is to say that it carries over the way in which the elements are combined in D_(2n) to the group G.
Thank you
7. (Original post by Artus)
I'm not sure what "trivial on the relations" means.
It means that the image of the function on the relations is the identity.

If you give a group by a presentation i.e. you specify generators and relations

e.g.

then you can specify any homomorphism by saying where maps the generators since then, for example,

using the multiplication of on the right since we are assuming to be a homomorphism.

However, this is only well defined as a homomorphism if is 'trivial on the relations' i.e. if

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 9, 2013
The home of Results and Clearing

### 2,511

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. SAE Institute
Animation, Audio, Film, Games, Music, Business, Web Further education
Thu, 16 Aug '18
2. Bournemouth University
Fri, 17 Aug '18
3. University of Bolton
Fri, 17 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams