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Finding the Inverse of an Epimorphism

  1. Nov 27, 2014 #1
    1. The problem statement, all variables and given/known data
    Let ##f : G \rightarrow H## be an epimorphism from a group ##G## to ##H## and let ##h \in H##, then ##f^{-1} (h) = g ~ker(f)##.

    2. Relevant equations


    3. The attempt at a solution
    So, if I understand the problem correctly, we are trying to find a epimorphism which has a rule such that its inverse has the rule ##f^{-1} (h) = g~ker(f)##, which essentially says that ##f^{-1}## maps an element in ##H## to a left-coset of ##ker(f)##. In this light, I am rather confused why they say that ##f## maps between the groups ##G## and ##H##, rather than ##G/g~ker(f)## and ##H##.

    I am having a rather difficult time finding such a rule for the epimorphsim ##f##, so that ##f^{-1} (h) = g~ker(f)##.
     
  2. jcsd
  3. Nov 27, 2014 #2

    Stephen Tashi

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    Did you omit something that tells us what the lowercase [itex] g [/itex] is?


    My guess is that the problem wants you prove a fact that is true for any epimorphism, from [itex] G [/itex] onto [itex] H [/itex], not merely for a particular one that we must create.
     
  4. Nov 28, 2014 #3
    Oh, yes. I neglected to mention that ##g \in G##.
     
  5. Nov 28, 2014 #4

    Stephen Tashi

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    You've neglected to mention whether the statement says "then for each [itex] g \in G [/itex] ,,, or whether it says "then there there exists a [itex] g \in G [/itex] ,,, ,
     
  6. Nov 28, 2014 #5
    Sorry, again. The full statement is

    Let ##f : G \rightarrow H## be an epimorphism from the group ##G## to ##H## and let ##h \in H##, then ##f^{-1}(h) = g ~ker(f)## for some ##g \in G##.

    By the way, I am slightly confused as to what ##f^{-1} (h) = g ~ker(f)## even means. What is the rule of assignment? Obviously, ##h## gets mapped to ##g~ker(f)##, but what would, say, ##h_1## get mapped to? ##g_1 ~ker(f)##? It does not seem clear.
     
  7. Nov 28, 2014 #6
    Here is another thing with which I am concerned. Isn't ##f^{-1}## mapping an element in ##H## to an element in ##G/g~ker(f)##? And is it not true that ##f## and ##f^{-1}## have to be acting over the same sets? If we are letting ##f : G \rightarrow H##, it would not seem that they are acting over the same sets.
     
  8. Nov 28, 2014 #7

    Stephen Tashi

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    [itex] f^{-1}(h) [/itex] denotes the set of elements of elements in [itex] G [/itex] that [itex] f [/itex] maps to [itex] h [/itex].

    You should form the habit of getting your variables like [itex] h_1, g_1 [/itex] within a proper "scope". If you are familiar with computer programming, a variable like "x" in a computer program might have one meaning in one function and another meaning in another function. The syntax checking of the compiler forces you to put a variable like "x" within some "scope". If you don't, it tells you the variable is "undefined". The scope of variables in written mathematics is done less formally, but the effect must still be there to make the writing clear. The scope of a variable like [itex] h [/itex] is usually defined by a quantifier, such as "for each" or "there exists". In your problem the phrase "for some [itex] h [/itex] " is used instead of "there exists an [itex] h [/itex]". English has a great variety of phrases the have the same meaning when interpreted as pure logic.

    How are you using the variable [itex] h_1 [/itex]? What defines its scope? Do you want to say "Let [itex] h_1 [/itex] be an arbitrary element of [itex] H [/itex]"? When we want to make a statement of the form "For each [itex] h_1 \in H. [/itex] ...." we can say that in English by the phrase "Let [itex] h_1 [/itex] be an arbitrary element of [itex] H [/itex]".

    Defining the scope of variables in your writing, will help you think clearly. Using an extensive vocabulary is good style in writing fiction, but in writing mathematics, you get a clearer result by using a limited vocabulary. Take for example, your use of the word "acting". There are technical definitions for the relation "acts on" in various mathematical contexts. You haven't explained what you mean by "acting".
     
  9. Nov 28, 2014 #8
    Okay, I see. So, does ##f^{-1}(h)## denote the preimage of a single element ##h \in H##? I was confusing it with the inverse mapping of ##f##.
     
  10. Nov 28, 2014 #9
    Okay, I think I may have a proof:

    Let ##y \in ~ker(f)##, which implies that ##f(y) = e_H##, where ##e_H## is the identity of ##H##. Let ##x \in G## be such that ##f(x) = h##, where ##h \in H## is some arbitrary element.

    Note that ## x \star_G y \in x ~ker(f)##.

    ##f(xy) = f(x) f(y) \iff##

    ##f(xy) = h \star_H e_H \iff##

    ##f(xy) = h##

    So, ##f## maps the element ##x \star_G y \in x~ker(f)## to some element ##h \in H##. Therefore, the preimage of the element ##h## is ##x~ker(f)##.

    Does this seem correct?
     
  11. Nov 28, 2014 #10

    Stephen Tashi

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    No. It's ambiguous. Your aren't expressing the logical quantifiers clearly.

    Generally when you wish to prove the equality of two sets, the cleareset way is to show each is a subset of the other. Try that approach.

    Perhaps you showed for each [itex] h [/itex] and each [itex] y [/itex] and each [itex] x [/itex] that [itex] h \in H [/itex] and [itex] y \in ker(f) [/itex] and [itex] x \in f^{-1}(h) [/itex] implies [itex] xy \in x\ ker(f) [/itex]

    The problem asks us to prove:
    For each [itex] h \in H [/itex] there exists a [itex] g \in G [/itex] such that [itex] f^{-1}(h) = g \ ker(f) [/itex].

    i.e. We must show that there exists a [itex] g \in G [/itex] such that the following hold
    1) [itex] f^{-1} (h) \subset g\ ker(f) [/itex] i.e. for each [itex] x [/itex] if [itex] x \in f^{-1}(h) [/itex] then [itex] x \in g\ ker(f) [/itex]
    2) [itex] g\ ker(f) \subset f^{-1}(h) [/itex] i.e for each [itex] x [/itex] if [itex] x \in g\ ker(f) [/itex] then [itex] x \in f^{-1}(h) [/itex].


    To make the "for each" and "there exists" logical quantifiers clear we can say "Let [itex] h [/itex] be an arbitrary element of [itex] H [/itex]. There exists at least one element [itex] g [/itex] such that [itex] f(g) = h [/itex] since [itex] f [/itex] is an epimorphism." Proceed to prove 1) and 2).
     
  12. Nov 29, 2014 #11
    Okay, so in post #9 I demonstrated that ##g ~ker(f) \subset f^{-1}(h)## by showing that some arbitrary element in ##g~ker(f)## gets mapped to the element ##h##, implying that the element is also in ##f^{-1}(h)##.

    Now, I am trying to prove the statement 1) you mentioned in post #10. Here is my work:

    Let ##x \in f^{-1}(h)## be arbitrary. Then by definition of the preimage, this implies that ##f(x) = h##. Let us form the left-coset of ##ker(f)## from the element ##x##: ##x~ker(f)##. We know that ##x \in x ~ker(f)##. Therefore, the arbitrary element ##x## in ##f^{-1}(h)## was just shown to be in ##x~ker(f)## also. Therefore, ##f^{-1} \subset x~ker(f)##.

    Because the two sets are subsets of each each, they must be equal.
     
  13. Nov 29, 2014 #12

    Stephen Tashi

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    You might have shown that, depending on how you intend to incorporate the symbol [itex] g [/itex] in that work.

    The left coset [itex] x\ ker(f) [/itex] isn't relevant. You need to show [itex] x [/itex] is in the left coset [itex] g\ ker(f) [/itex].
    The symbol [itex] g [/itex] is used earlier in the proof to denote a specific element.. You should give an argument that is still within the "scope" of that use of [itex] g [/itex].
     
  14. Nov 29, 2014 #13

    Stephen Tashi

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    It's worth remembering that, in a group, first degree equations always have a unique solution. For example, to solve for [itex] y [/itex] in the equation [itex] ay = b [/itex] you can multiply both sides on the left by [itex] a^{-1} [/itex] obtaining [itex] y = a^{-1} b [/itex].

    As a consequence of this, if you are discussing a element [itex] h [/itex], you can always force another element [itex] g [/itex] into the discussion by factoring [itex] h [/itex] into a product invovling [itex] g [/itex]. To find the factorization with [itex] g [/itex] on the left, you solve equation [itex] h = gy [/itex] for [itex] y [/itex] obtaining [itex] y = g^{-1}h [/itex] and [itex] h = g( g^{-1} h) [/itex].

    If your instructor allows you do "obvious" steps without detailed explanations, you could simply say "Factor [itex] h [/itex] as follows: [itex] h = (e_G) h = (g g^{-1}) h = g (g^{-1} h) [/itex]

    We have assumed [itex]x \in f^{-1}(h) [/itex]. This implies [itex] f(x) = h [/itex]
    Using the above factorization, [itex] f(x) = h = g (g^{-1}x) [/itex] This has the form [itex] g [/itex] times something. To show [itex] x \in g\ ker(f) [/itex] you need to show the factor [itex] g^{-1} x [/itex] is in [itex] ker(f) [/itex]. To do that you need to show [itex]f(g^{-1} x) [/itex] is the identity. Show that by showing it is equal to [itex]h^{-1} h [/itex].
     
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