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Finding a set and an injection

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##G## be a group. I need to find a set ##X## and an injective function from ##G## into ##Sym(X)##

    2. Relevant equations


    3. The attempt at a solution
    I am having difficulty with this problem, and I want to make sure I understand exactly what it is asking.
    If I understand the question correctly, we have some group ##G##, and we want to find some set ##X## and create some mapping (specifically, an injection) which associates the elements in ##G## with the elements in ##Sym(X)##, so that we can clearly see that the elements of ##G##, along with its binary operator ##\star##, form functions?

    Here are some ideas: I am pretty certain that I need to choose an ##X## such that ##Sym(X)## is at least as large as, if not greater than, ##G##. I know that the binary operator ##\star## is actually a function, so could I base my injection off of this? I have this suspicion that ##G=X## would work, but I can't quite figure it out.

    EDIT: I also know that ##(Sym(X), \circ)## forms a group, so I am basically trying to find an injection between two different groups.

    Does anyone have any suggestions?
     
  2. jcsd
  3. Nov 8, 2014 #2

    Dick

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    Yes, take G=X. Associate an element of the group g with the mapping g:X->X defined by g(x)=gx. Is that an element of Sym(G)?
     
  4. Nov 8, 2014 #3
    Wait. From which set does ##g## come from? Are you saying that it is in ##G##, or in ##Sym(G)##?
     
  5. Nov 8, 2014 #4

    Dick

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    g is in G.
     
  6. Nov 8, 2014 #5
    Okay. So, let me reiterate things, to see if I understand fully. We want to find a set ##X## and ##\pi## such that ##\pi : G \mapsto Sym(X)## is an injective map. We let ##X=G##, and so ##Sym(G)## consists of all the bijective functions acting over the elements of ##G##; that is, ##Sym(G) = \{\theta_1 , \theta_2, \theta_3,... \}##, where ##\theta_1 : G \mapsto G##, ##\theta_2 : G \mapsto G##, etc.

    Let ##g \in G## be arbitrary. I need to find an element in ##Sym(G)## to which ##\pi## can map it. If ##g \in G##, then there exists a mapping ##\theta \in Sym(G)## such that

    ##\theta : g \mapsto a## or ##\theta(g) = a##, where ##a \in G## is unique, because ##\theta## is a bijection. So,

    ##\pi : g \mapsto \theta (g)##

    ##\pi : g \mapsto a##.

    So, ##\pi## will actually map ##g## to a unique element in ##G##.

    This is where I am not certain of what I can conclude. In fact, I am not certain if any of what I said is valid. Could you pick apart my work?
     
  7. Nov 8, 2014 #6

    Dick

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    You are thinking of it way too abstractly. My hint is to point out that for any g in G the map f from G->G defined by f(x)=gx is a bijection. Prove that first. Then think about the overall problem.
     
    Last edited: Nov 8, 2014
  8. Nov 9, 2014 #7
    Are you saying that I need to demonstrate that ##f(x) = g \star x## is a bijection, where ##x \in G##, I presume? So, the function/map ##f## is defined in terms of the operator ##\star##? Well, the binary operator ##\star## is well-defined, so it maps uniquely; and because ##\star## has closure, the binary operator, and therefore the function ##f##, will map every element ##G## to some element in ##G##.
     
  9. Nov 9, 2014 #8

    Dick

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    You also have to show it's one-to-one and onto to show it's a bijection.
     
  10. Nov 9, 2014 #9
    Don't closure and being well-defined demonstrate that it's one-to-one and onto? Closure means that every element in G gets mapped to G, so f maps the elements of G onto G; and well-defined means that it takes ##g \star x## to a unique element in G.
     
  11. Nov 9, 2014 #10

    Dick

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    That just tells you f is a function. How is that supposed to imply that it's one to one or onto? Say how those are defined and spell it out in detail. You need to use that G is a group, not just that * is a binary operator. Saying f maps G TO G is not the same thing as saying f maps G ONTO G. What's the difference?
     
    Last edited: Nov 9, 2014
  12. Nov 9, 2014 #11
    Oh, okay. Let ##g \in G## be some arbitrary element. The function we have defined is ##f(x) = g \star x##, which maps elements from ##G## to the same set.

    If the function is injective, then ##f(x) = f(y) \implies x = y ~\forall x,y \in G##. Let ##x,y \in G## be arbitrary; and let us suppose that ##f(x) = f(y)##, and see whether this leads to ##x=y##.

    ##f(x) = f(y) \iff g \star x = g \star y##. Using the cancellation law, we get ##x = y##. Therefore, it is injective.

    If the function is surjective, then ##f## should take every element ##b \in G## in the range and associate it with some element in the domain. Let ##b \in G## be some arbitrary element.

    ##b = f(x) \implies b = g \star x##. Because ##g \in G##, then ##g^{-1} \in G##. Multiplying both sides by this, we get

    ##g^{-1} \star b = x##. By closure of ##\star## over ##G##, ##g^{-1} \star b## is some element in ##G##, which is the domain of ##f##; because ##x## is equal to this element, we have found that there does in fact exist an ##x## in the domain which is associated with ##b## in the range.

    How does this appear?
     
  13. Nov 9, 2014 #12

    Dick

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    Much better. Now you have a correspondence between every element g of G and a bijection of G->G. That's an element of Sym(G), yes? Now you just want to prove that correspondence is an injection.
     
  14. Nov 11, 2014 #13
    So, I need to prove that ##\pi : g \mapsto f(x)## is an injection; that is, ##\pi (g) = f(x)##?

    Let ##a,b \in G## be arbitrary. Let us suppose that ##\pi(a) = \pi(b)##.

    ##\pi(a) = \pi(b) \iff##

    ## a \star x = b \star x## Using the cancellation law,

    ##a = b##.

    Is this right?

    Would it be less ambiguous if we wrote ##f(x)## as ##f_g(x)##?
     
  15. Nov 11, 2014 #14

    Dick

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    The idea is right. The notation is pretty messy. Sure, make it ##f_g(x)=g \star x##. So ##\pi(g)=f_g##.
     
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