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Projection Functions and Homomorphisms

  1. Nov 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##G##, ##H##, and ##K## be groups with homomorphisms ##\sigma_1 : K \rightarrow G## and ##\sigma_2 : K \rightarrow H##. Does there exist a homomorphism ##f: K \rightarrow G \times H## such that ##\pi_G \circ f = \sigma_1## and ##\pi_H \circ f = \sigma_2##? Is this function unique?

    If either ##\sigma_1## or ##\sigma_2## are monomorphisms, then ##f## will also be a monomorphism.

    2. Relevant equations


    3. The attempt at a solution

    Define ##f## to be the mapping ##f : K \rightarrow G \times H##.

    ##\pi_G ((g,h)) = g## and ##\pi_H((g,h)) = h##, where ##(g,h) \in G \times H##.

    ##\pi_G \circ f = \pi_G (f(k))##, where ##k \in K##.

    ##\pi_G \circ f = \pi_G(f(k)) = \pi((g,h)) = g \in G##

    There exists an ##k_1 \in K## such that ##\sigma_1 (k_1) = g##. Thus,

    ##\pi_G \circ f = \sigma_1(k_1)## or

    ##\pi_G \circ f = \sigma_1##.

    I did something similar for ##\pi_H \circ f##. However, this feels unsettling. Also, how would I even verify that ##f## is a homomorphism, if I do not know the rule of its mapping? Would I have to contrive a rule?
     
  2. jcsd
  3. Nov 19, 2014 #2

    pasmith

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    Homework Helper

    You are asked to show whether or not such a homomorphism exists. If you think it does, then you can either try to show that the non-existence of such a homomorphism leads to a contradiction, or you can exhibit a specific [itex]f[/itex].

    Unless otherwise specified, the group operation on [itex]G \times H[/itex] is [itex](g_1,h_1)(g_2,h_2) = (g_1g_2,h_1h_2)[/itex]. You are given homomorphisms [itex]\sigma_1 : K \to G[/itex] and [itex]\sigma_2 : K \to H[/itex]. Can you think of a way to use those to construct a map [itex]K \to G \times H[/itex]? Is that map a homomorphism? Does it satisfy the other conditions given in the problem?
     
  4. Nov 20, 2014 #3
    Would the rule be as simple as ##f(k) = (g,h)##? I tried this, but I ran into some difficulty when trying to verify if it is a homomorphism:

    First of all, let me define some conventions: Let ##\star_1## be the operator associated with the group ##K###, ##\star_2## with ##G##, ##\star_3## with ##H##, and ##\star## with ##G \times H##.

    ##f(k_1 \star_1 k_2)= f(k_1) \star f(k_2)##

    ##f(k_1 \star_1 k_2) = (g_1,h_1) \star (g_2,h_2)##

    ##f(k_1 \star_1 k_2) = (g_1 \star_2 g_2 , h_1 \star_3 h_2)##

    The problem I am facing is, how do I evaluate ##f(k_1 \star_1 k_2)##?
     
  5. Nov 20, 2014 #4

    pasmith

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    Homework Helper

    What is [itex]f(k_1)[/itex] in terms of the homomorphisms you are given? What is [itex]f(k_2)[/itex]? What is [itex]f(k_1k_2)[/itex]? Is it equal to [itex]f(k_1)f(k_2)[/itex]?
     
  6. Nov 21, 2014 #5
    Well, I suppose that ##f(k_1) = (g_1,h_1)## and ##f(k_2) = (g_2,h_2)##.

    That is what I am trying to demonstrate. If the equality ##f(k_1 k_2) = f(k_1) f(k_2)## is true, then I know that ##f## is a homomorphism. However, I am not sure how to evaluate ##f(k_1 k_2)##, which leads me to suspect that the rule I have constructed is not specific enough.
     
  7. Nov 29, 2014 #6
    So, does anyone have any new thoughts?
     
  8. Nov 29, 2014 #7
    I figured it out. Define the mapping ##f: K \rightarrow G \times H## to have the rule ##f(k) = \bigg(f_1(k),f_2(k) \bigg)##
     
  9. Nov 29, 2014 #8

    Stephen Tashi

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    Science Advisor

    If there are preliminary definitions needed to define [itex] f [/itex], you should give them before you say "Define [itex] f [/itex]". Aren't projections and their notation already defined in your course matherials? If so, defining [itex] \pi_G [/itex] and [itex] \pi_H [/itex] could be part of "2. Relevant equations", but you don't need to state those definitions in your proof.

    To define [itex] f [/itex], all you must do is (for each [itex] k \in K [/itex] ) is to define the element [itex] f(k) [/itex]. What element of [itex] G \times H [/itex] will it be?

    Once you have defined [itex] f [/itex] as a function, you don't need to define things like [itex] \pi_G \circ f [/itex]. The definition of that follows from the standard definition for composing functions.
     
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