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Show isomorphism under specific conditions

  1. May 26, 2017 #1

    Math_QED

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    1. The problem statement, all variables and given/known data

    Let ##A,B## be subgroups of a finite abelian group ##G##

    Show that ##\langle g_1A \rangle \times \langle g_2A \rangle \cong \langle g_1,g_2 \rangle## where ##g_1,g_2 \in B## and ##A \cap B = \{e_G\}##
    where ##g_1 A, g_2 A \in G/A## (which makes sense since ##G## is abelian and all the subgroups are normal)

    (this statement is possibly false, I need it to understand a step in a proof)


    2. Relevant equations

    None
    3. The attempt at a solution

    Define ##f: \langle g_1A \rangle \times \langle g_2A \rangle \rightarrow \langle g_1,g_2 \rangle##

    by ##f(g_1^{k_1}A,g_2^{k_2}A) = g_1^{k_1}g_2^{k_2}## for integers ##k,l##

    First, we show that ##f## is well-defined:

    Therefore, let ##(g_1^{k_1}A,g_2^{k_2}A) = (g_3^{k_3}A,g_4^{k_4}A)##

    Then, ##g_1^{k_1}A = g_3^{k_3}A \iff g_1^{k_1}g_3^{-k_3} \in A##

    Since, ##g_1,g_3 \in B, g_1^{k_1}g_3^{-k_3} \in A\cap B = \{e_G\}## Hence, ##g_1^{k_1 } = g_3^{k_3}##. Completely analogue, we find ##g_2^{k_2 } = g_4^{k_4}##.

    Therefore, we conclude that ##g_1^{k_1}g_2^{k_2} = g_3^{k_3}g_4^{k_4}##. Hence, ##f## is well-defined.

    This also shows injectivity (every step we did is reversible).

    It is also clear that this mapping is surjective. If we take ##g_1^{k_1}g_2^{k_2} \in \langle g_1,g_2 \rangle##, then obviously ##(g_1^{k_1}A,g_2^{k_2}A)## gets mapped to that element.

    Last but not least, we have to show that ##f## is a homomorphism. To show this,

    ##f[(g_1^{k_1}A,g_2^{k_2}A)(g_3^{k_3}A,g_4^{k_4}A)] = f(g_1^{k_1}g_3^{k_3}A,g_2^{k_2}g_4^{k_4}A) ##

    ##= g_1^{k_1}g_3^{k_3}g_2^{k_2}g_4^{k_4} = g_1^{k_1}g_2^{k_2}g_3^{k_3}g_4^{k_4}##

    ##= f(g_1^{k_1}A,g_2^{k_2}A)f(g_3^{k_3}A,g_4^{k_4}A)##

    The result follows ##\quad \triangle##

    Can someone verify whether my attempt is correct?
     
    Last edited: May 26, 2017
  2. jcsd
  3. May 26, 2017 #2

    fresh_42

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    What do you mean by ##\langle x,y \rangle## and where are the elements of ##A## gone to? Isn't ##g_1=g_2=e_G## already a counterexample?
     
  4. May 26, 2017 #3

    Math_QED

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    Group generated by x,y, which is a subgroup of G. This is part of a proof. I filtered context.

    And what is the problem with ##g_1 = g_2 = e_G##?

    Then, we get that ##\langle A \rangle \times \langle A \rangle = \{e_{G/A}\} \times \{e_{G/A}\} \cong \{e_G\}##, which seems fine.
     
  5. May 26, 2017 #4

    fresh_42

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    O.k., then ##\langle . \rangle## depends on the group. I thought it was meant to be ##\langle gA \rangle =\langle ga\,\vert \,a \in A\rangle##.
    However, in this case ##g_1=a \in A-\{e_{G}\}\, , \,g_2=e_{G}## will be a counterexample except in case ##A=\{e_G\}##.
     
  6. May 26, 2017 #5

    Math_QED

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    But ##g_1 \in B##. If ##g_1 \in A##, then automatically ##g_1 = e_G##
     
  7. May 26, 2017 #6

    fresh_42

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    Forgotten. Seems I should take a nap. Sorry.
     
  8. May 26, 2017 #7

    Math_QED

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    No worries. I'm already glad I get some help :)
     
  9. May 26, 2017 #8

    fresh_42

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    Let's consider the first isomorphism theorem: ##AB/A \cong B/(A\cap B) \cong B## which gives us a hint where to look for the flaw. You have basically shown, that the projection followed by the multiplication onto the quotient group ##G/A## is a well-defined epimorphism. So what about injectivity? ##AB/A \times AB/A \cong B \times B \ncong B##.

    Say ##f(gA,hA)=gh = e_G##. This means ##g=h^{-1}##, but how do you get to ##g \in A##?
    (Unless I've missed something again.)
     
  10. May 26, 2017 #9

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    I see where you are going. I realised the injectivity is failing (at least my argument, since not every step is reversible). Thanks a lot for this. So probably, the statement is false.
     
  11. May 28, 2017 #10

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    I managed to prove injectivity (I could deduce from the proof's context that ##\langle y_1A\rangle \cap \langle y_2A\rangle = \{A\}##, from which the injectivity follows.)
     
  12. May 28, 2017 #11

    fresh_42

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    ##A=\langle (1),(12) \rangle\, , \,B=\langle (1),(13) \rangle\, , \,G=AB=\langle (1),(12),(13),(23) \rangle##
    The stated isomorphism now reads ##\langle (13)A \rangle \times \langle (13)A \rangle \cong B \times B \cong \langle (13),(13) \rangle = B## respectively ##f((13)A,(13)A) = (1)##, so how could ##f## be injective? ##B \times B \cong G \ncong B## by counting group elements.
     
  13. May 28, 2017 #12

    Math_QED

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    Don't worry about it. There is a lot of context missing that can be used (this statement is part of the proof that every finite abelian group is the direct product of cyclic p-groups, which is a rather large proof, which is the reason I left as much context away as possible).
     
  14. Jun 23, 2017 #13

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    I am studying for my exam group theory right now and I reread this thread. How can your counterexample make sense if ##G## is finite and we already have established a surjection between the two finite groups? This clearly must imply that the function is injective as well.
     
  15. Jun 23, 2017 #14

    fresh_42

    Staff: Mentor

    I think my misconception lied in ##\langle g_1,g_2 \rangle## which probably should be read as ##\langle (g_1,g_2) \rangle##, which makes a great difference. My example (##g_1 = g_2##) used the fact that ##B \times B \ncong B## for finite groups. So if we generate a group by ##g_1## and ##g_2##, we get ##g_iA \cong B## in the example. But if we generate it by the pairs ##\{(g_1,g_2)\,\vert \,g_i \in B\}## we get ##B \times B## and it's no counterexample. It all depends on how ##\langle g_1,g_2 \rangle## had to be understood. Usually it means "group generated by all by ##g_1## and ##g_2##" as you also suggested in the definition of ##f##. But I think we may not multiply the two and consider pairs instead.
     
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