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Homomorphisms of Cyclic Groups

  1. Dec 12, 2012 #1
    So this is a pretty dumb question, but I'm just trying to understand homomorphisms of infinite cyclic groups.

    I understand intuitively why if we define the homomorphism p(a)=b, then this defines a unique homorphism. My question is why is it necessarily well-defined? I think I'm confused because I'm not sure what a 'representation' of an element really means.

    For example, in the quotient group, we take any element in the equivalence class [a] to be a representative of [a], and we have to show that the homomorphism acts the same on every representative (which it doesn't in some cases).

    So, are a3 and a2a different representatives for the same element? I think I'm confused since a2 is just notational convention for aa, so I'm not sure if this really constitutes a different 'representative'. What if I defined f(x)="first power that appears in x". Would that be 3 for a3 and 2 for a2a.

    If anyone can clear me up on this that would be much appreciated.
     
    Last edited: Dec 12, 2012
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  3. Dec 12, 2012 #2

    lavinia

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    For a cyclic group you only need to know the value of the homomorphism on a generator of the group.

    If the group is infinite cyclic then one always gets a homomorphism by defining its value on a generator and then extending inductively.

    So if a generates the group and f(a) is the value of the homomorphism on a then f(a[itex]^{n}[/itex]) = f(a)[itex]^{n}[/itex]. It makes no difference how you represent a power of the generator.

    If the group is finite cyclic this may not work. For instance you can not define a homomorphism of the cyclic group of order 2 into the cyclic group of order 3 that is non identically zero.

    The infinite cyclic group is an example of a free group. It is the free group on one generator. Free means that there are no non-trivial relations among the elements of the group. For such groups one always gets a homomorphism from its values on generators of the group.

    So take the free group on two generators a and b. this group is all finite words made of powers of a and b e.g. a[itex]^{-4}[/itex]b[itex]^{15}[/itex]a[itex]^{2}[/itex]b

    If one knows the value of the homomorphism on a and b then f extends to the whole group in the same way was in the infinite cyclic group.

    the kernel of f is called the set of relations on the quotient group. In fact any group may be thought of as a free group modulo the group of relations.
     
    Last edited: Dec 12, 2012
  4. Dec 12, 2012 #3
    So how can I see rigorously that one always gets a homomorphism from the infinite cyclic group (i.e. the free group on one generator) to another group. I can see that if we write p(a)=b then this defines it all, but is there a proof to show this is well-defined? (like assuming that it isn't, and then arriving at a contradiction that shows this has a non-trivial relation, or something like that).

    Is this just a triviality I'll understand better once I understand free groups and relations better? Can an element have more than one 'representative' if there are no relations? (like the wikipedia example is 1/2 and .5, with a function that is different on each of these, but I'm not really sure what the construction of the rationals has anything to do with relating 1/2 and .5. Does .5 mean 5/10, which is a different representative of [1/2] than 1/2?)
     
    Last edited: Dec 12, 2012
  5. Dec 12, 2012 #4

    pasmith

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    The infinite cyclic group [itex]\langle a \rangle = \{a^n : n \in \mathbb{Z}\}[/itex] is trivially isomorphic to [itex]\mathbb{Z}[/itex], the group of integers under addition (given the definition of [itex]a^n[/itex] this is essentially a tautology).

    For any group G, and any [itex]g \in G[/itex], [itex]\phi_g : \mathbb{Z} \to G : n \mapsto g^n[/itex] is an obvious homomorphism (given the definition of [itex]g^n[/itex] this is essentially a tautology).
     
  6. Dec 12, 2012 #5
    I guess my main confusion is in how we necessarily know a homomorphism is well-defined. If we have to check all 'representatives' of a group, how do you know what a representative is? (like, is 1+1 a representative for 2, since 1+1=2?).

    Wikipedia says that a function is well defined if it doesn't depend on the 'form' of the input, but what does the form of the input really mean?
     
  7. Dec 12, 2012 #6

    lavinia

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    [itex]^{}[/itex]
    For free groups you only need to know the value on generators. You always get a homomorphism by extension. In a free group there are not two different words that are equal because this would be a relation.

    For instance if a[itex]^{2}[/itex] was equal to ab then you would have the relation
    a[itex]^{2}[/itex]b[itex]^{-1}[/itex]a[itex]^{-1}[/itex] = 1. That can not happen in a free group.

    Notice that in the infinite cyclic group the word aaaaa is the same as the word a[itex]^{5}[/itex]
     
  8. Dec 12, 2012 #7

    pasmith

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    There is a fundamental principle that equal objects may always be substituted for each other. It follows that if [itex]x = y[/itex] then trivially [itex]f(x) = f(y)[/itex].

    The problem with well-definedness arises if you have a function [itex]f : X \to Y[/itex] and an equivalence relation R on X, and you want to be able to define [itex]g : X/R \to Y[/itex] in terms of [itex]f[/itex], so that [itex]g : [x] \to f(x)[/itex]. For that to make sense you must get the same value of [itex]f(y)[/itex] whichever [itex]y \in [x][/itex] you choose. Thus for the definition to make sense, you must have [itex]f(x) = f(y)[/itex] whenever xRy.

    But that's not the situation when you define a homomorphism on a free group by extension from its values on generators. Instead you're defining f(x) for particular values of [itex]x \in X[/itex] in such a way that you can determine f(y) for all other [itex]y \in X[/itex] uniquely from those values.

    You may find this explanation helpful.
     
  9. Dec 12, 2012 #8
    So is this substitution an axiom? And do you mean that trivially [itex]f(x) = f(y)[/itex] for any function, or any relation at all? I see it for any function since [itex]xRf(x)[/itex], [itex]yRf(y)[/itex] and so if we are allowed to substitute we get [itex]yRf(x)[/itex] so then f(x)=f(y), but I'm not sure its clear to me for arbitrary relations.

    This is probably nonsense, but what if I tried to define a relation that was different for 1+1 than for 2? Could you say, take the function that, if it observes 1+1, takes the first summand? I would initially think this is obvious nonsense, but not having studied set theory and such enough I need some clarification (I know that 2 and 1+1 are defined to be the same things, but its not clear to me that nothing can act differently on them). This is giving me a headache.

    I read something about a formula being "well founded". Does this have anything to do with that?
     
    Last edited: Dec 12, 2012
  10. Dec 14, 2012 #9
    I think I'm still a little unsure of what being 'well-defined' really means.

    I understand that in the case of cosets we want the function to act the same on any representative. But, more generally, I've seen a definition somewhere that 'well-defined' means that [itex]a=b → f(a)=f(b)[/itex]. This seems to make sense as if we have two equivalence classes [a] and such that [a]=, the function must be the same on both.

    But then what pasmith said about the substitutability (i.e. that there is a principle that if a=b then they can always be substituted for each other) confused me further, as that would imply the above statement is always trivially true (even though there are many cases in which it is not).

    So is that definition of being well-defined, [itex]a=b[/itex] → [itex]f(a)=f(b)[/itex], right, or am I confused about pasmith's statement?
     
  11. Dec 14, 2012 #10

    lavinia

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    well defined just means that the definition is consistent. One way the definition of a function can fail is if it assigns different values to the same thing when that thing is written in two different ways.

    In the case of a free group any function on a generating set give you a well defined homomorphism. You don't need to chech anything.
     
  12. Dec 14, 2012 #11
    Ok, thanks.
     
    Last edited: Dec 14, 2012
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