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Abstract Algebra: need a review of 1-1 and onto proof

  1. Mar 24, 2014 #1
    1. The problem statement, all variables and given/known data

    define a function f:H--> gHg[itex]^{-1}[/itex]

    2. Relevant equations
    prove if f is 1-1 and onto.


    3. The attempt at a solution
    1-1:
    f(h1)=f(h2)
    gh1g[itex]^{-1}[/itex]=gh2g[itex]^{-1}[/itex]
    h1=h2 (left and right cancellations)

    onto:
    f(g[itex]^{-1}[/itex]hg)=gg[itex]^{-1}[/itex]hgg[itex]^{-1}[/itex]=h
    so every h belonging to H has an image of g[itex]^{-1}[/itex]hg.
    However,I do not really understand the last line. I followed the example that we did in class but now I am not sure.
     
  2. jcsd
  3. Mar 24, 2014 #2

    pasmith

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    Homework Helper

    If you tell me what [itex]H[/itex] and [itex]g[/itex] are (I'm assuming there's a group [itex]G[/itex], that [itex]H[/itex] is a subgroup of [itex]G[/itex], and that [itex]g \in G[/itex]) I will know the domain and codomain of [itex]f[/itex], but you haven't said anything about what [itex]f(h)[/itex] is for [itex]h \in H[/itex], so I can't say whether [itex]f[/itex] is injective or surjective.

    That's one mystery solved: [itex]f: h \mapsto ghg^{-1}[/itex] and it is indeed 1-1.

    No. What you've written here is that if [itex]g^{-1}hg \in H[/itex] then [itex]f(g^{-1}hg) = h \in gHg^{-1}[/itex]. But unless [itex]H[/itex] is normal, and you haven't told me that it is, it may not be the case that if [itex]h \in H[/itex] then [itex]g^{-1}hg \in H[/itex].

    To prove that [itex]f[/itex] is onto, you need to start with a [itex]k \in gHg^{-1}[/itex] and show that there exists [itex]h \in H[/itex] such that [itex]f(h) = k[/itex].
     
    Last edited: Mar 24, 2014
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