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Final Group Q.

  1. Jan 13, 2008 #1
    ok ive managed to solve the other 2 questions.

    here is my final one:

    If G is a group and [itex] n \geq 1 [/itex], define G(n) = { x E G: ord(x) = n}

    If [itex] G \cong H [/itex] show that, for all [itex] n \geq 1 [/itex], |G(n)| = |H(n)|.

    Deduce that, [itex]C_3 X C_3[/itex] is not [itex]\cong C_9[/itex].
    Is it true that [itex] C_3 X C_5 \cong C_15[/itex]
    Is it true that [itex] C_2 X C_6 \cong C_12[/itex]

    What is going on here?

    any help to get me started is highly appreciated. ill attempt the questions as usual once i have some idea of what to do. thnx so much
  2. jcsd
  3. Jan 13, 2008 #2


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    Just do it. Show that there's a one-one correspondence between the elements of G of order n and the elements of H of order n. Use the fact that there is an isomorphism between G and H, and that isomorphisms preserve order.
  4. Jan 16, 2008 #3
    The best way to think of an isomorphism between two groups G and H is that it is a way to rename the elements of G as the elements of H so that you end up with exactly H. If you think of it this way, I think it should not be too hard to show (2).
  5. Jan 16, 2008 #4
    smoothman: [itex]C_3 X C_3[/itex]

    I think I can remember some of this. An example of the type of problem is the four group: [tex]C_2XC_2[/tex] What happens here?

    We have (0,0), (1,0), (0,1), (1,1). You can build up a chart of this under addition, and we get for example: (1,0)+(0,1) = (0,0), which sends us back to zero. (In fact, any element added to itself gives 0.)

    So this can not be the cyclic four group consisting, 0+1=1, 1+1=2, 1+2=3,1+3 = 0.

    In fact, it is the Kline 4 group.
    Last edited: Jan 16, 2008
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