1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    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

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

  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!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Another group theory problem
  1. Another Group theory Q (Replies: 7)

Loading...