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    I have this theorem in my notes:

    There is a homormophism f: D_{2n} \rightarrow G with  f(a) = x and  f(b) = y if and only if

    1. x^2=e
    2. y^n=e
    3. x^{-1}yx=y^{-1}

    Now this is what I want to prove:

    If |x|=2, |y|=n, 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.




    Thanks in advance
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    (Original post by Artus)
    I have this theorem in my notes:

    There is a homormophism f: D_{2n} \rightarrow G with  f(a) = x and  f(b) = y if and only if

    1. x^2=e
    2. y^n=e
    3. x^{-1}yx=y^{-1}
    Firstly, this isn't really a theorem. All it says is that if f\colon D_{2n} \rightarrow G is any function from the dihedral group to any other group G then it is a homomorphism if and only if it is trivial on the relations of D_{2n} where the dihedral group is tacitly given by the presentation \langle a,b \mid a^2=1,b^n=1,a^{-1}ba=b^{-1} \rangle. 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 |x|=2, |y|=n, then f is one-to-one provided n>2. Hint: y should not be its own inverse.
    [/latex]
    Consider the restriction f \colon D_{2n} \rightarrow \mathrm{im}(f) of f to its image and then use the first isomorphism theorem.
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    Thanks a lot. I think I know how to prove it, but I just have 2 more questions:

    1) y \not= x^{-1}yx, 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 y^n = e \not= x^{-1}y^nx= x^{-1}x = e. I must be doing something wrong, but I just can't see it...

    2)

    (Original post by Mark85)
    All it says is that if f\colon D_{2n} \rightarrow G is any function from the dihedral group to any other group G then it is a homomorphism if and only if it is trivial on the relations of D_{2n} where the dihedral group is tacitly given by the presentation \langle a,b \mid a^2=1,b^n=1,a^{-1}ba=b^{-1} \rangle.
    I'm not sure what "trivial on the relations" means.
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    (Original post by Artus)
    1) y \not= x^{-1}yx, 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 y^n = e \not= x^{-1}y^nx= x^{-1}x = e. 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 y \not= x^{-1}yx \implies y^n \not= x^{-1}y^nx. This statement is logically equivalent to the statement y^n = x^{-1}y^n x \implies y=x^{-1}yx [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.
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    Yeah Farhan's summed it up perfectly.
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    (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 y \not= x^{-1}yx \implies y^n \not= x^{-1}y^nx. This statement is logically equivalent to the statement y^n = x^{-1}y^n x \implies y=x^{-1}yx [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
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    (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. D_{2n} = \langle a,b \mid a^2 = 1, b^n = 1,a^{-1}ba = b^{-1}\rangle

    then you can specify any homomorphism f\colon D_{2n} \rightarrow G by saying where f maps the generators since then, for example,

    f(a^mb^ja^k) = f(a)^mf(b)^jf(a)^k

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

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

    f(a^2) = f(b^n)= f(a^-1bab) = 1_G
 
 
 
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