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Homework Help: Another group theory problem

  1. Jul 25, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove or disprove the following assertion. Let [itex]G[/itex], [itex]H[/itex], and [itex]K[/itex] be groups. If [itex]G × K \cong H × K[/itex], then [itex]G \cong H[/itex].

    2. Relevant equations
    [itex] G × H = \left\{ (g,h): g \in G, h \in H \right\} [/itex]

    3. The attempt at a solution
    I don't even know whether the statement is true or false... I tried looking in both directions but to no avail.

    For a counterexample, I figured that at least one of the sets would have to be infinite, otherwise G and H would have the same number of elements and it would be harder for them not to be isomorphic. I considered the sets:

    [itex] \left\{0\right\} × Z [/itex] and [itex] Z_{2} × Z [/itex] with the isomorphism being:
    [itex] \phi( 0, 2n ) = ( 0, n ) [/itex] and [itex]\phi( 0, 2n + 1 ) = ( 1, n ) [/itex]. Then it is well-defined, one-to-one and onto. But it does not preserve the operation because [itex] \phi( 0, 1 ) + \phi( 0, 1 ) = ( 1, 0 ) + ( 1, 0 ) = ( 0, 0 ) [/itex] while [itex] \phi( (0,1) + (0,1) ) = \phi(0,2) = (0,1) [/itex].

    I still think someone could tweak the definition a bit to make it an isomorphism, but of course I am not sure.

    As for proving the theorem is true, I don't even know how to start. I considered the following mapping [itex] \phi : G → H [/itex] defined as [itex] \phi( g ) = h [/itex] where [itex] (g,e_{K}) → (h, k') [/itex] under the isomorphism between [itex] G × K [/itex] and [itex] H × K [/itex].

    Then I proved [itex] \phi [/itex] is well-defined and preserves the operation. But I cannot prove that it is one-to-one, nor onto. The problem is that I have no reason to believe that if [itex]g_{1} ≠ g_{2} [/itex] then [itex] (g_{1},e_{K}) → (h_{1}, k_{1}) [/itex] and [itex] (g_{2}, e_{K} ) → (h_{2}, k_{2}) [/itex] where [itex] h_{1}≠h_{2} [/itex].

    On the other hand, no other natural candidate for an isomorphism comes to mind, so I am completely stuck here. If someone could, not give me the whole answer, but just point to the right direction? Because I may be looking somewhere completely unrelated :(
  2. jcsd
  3. Jul 25, 2013 #2
    The group ##\mathbb{Z}_2\times \mathbb{Z}## has an element of order ##2##, the group ##\{0\}\times \mathbb{Z}## doesn't. So the groups are not isomorphic.

    You're however correct that the statement is false. Take the following group as ##K##

    [tex]\mathbb{Z}^{\mathbb{N}} = \{(x_n)_n~\vert~x_n\in \mathbb{Z}\}[/tex]

    with the usual operations. Can you find the right ##G## and ##H##?
  4. Jul 25, 2013 #3
    Is that the set of all infinite sequences of integers?
  5. Jul 25, 2013 #4
  6. Jul 25, 2013 #5
    Ok yes! I finally got it:

    Let G be {0} and let H be Z. Then define [itex]\phi[/itex] to be:

    [itex]\phi( (0, (a_{1}, a_{2}, ... ) ) ) = ( a_{1}, (a_{2}, a_{3}, ... ) ) [/itex] obtained by left-shifting the sequence. It is obviously well-defined. It is one-to-one because if two left-shifted sequences are equal, then adding a zero in front keeps them equal. And any sequence can have a zero added to it so the mapping is onto. The operation is preserved because:

    [itex]\phi( (0, (a_{1}+b_{1}, a_{2}+b_{2}, ... ) ) ) = ( a_{1}+b_{1}, (a_{2}+b_{2}, a_{3}+b_{3}, ... ) ) [/itex] [itex] = ( a_{1}, (a_{2}, a_{3}, ... ) ) + ( b_{1}, (b_{2}, b_{3}, ... ) ) = \phi( ( 0, ( a_{1}, a_{2}, ... ) ) ) + \phi( 0, ( b_{1}, b_{2}, ... ) ) )[/itex]

    Thanks for the help. I don't think I would have thought of that anytime soon :P
  7. Jul 25, 2013 #6
    By the way, are you really 19? Because that's my age too but you seem so much more knowledgeable!
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