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Isomorphic direct product cyclic groups!

  1. Jan 30, 2012 #1
    Help! For p prime I need to show that

    [itex] C_{p^2} \ncong C_p \times C_p [/itex]

    where [itex] C_p [/itex] is the cyclic group of order p.



    But I've realised I don't actually understand how a group with single elements can be isomorphic to a group with ordered pairs!


    Any hints to get me started?
     
  2. jcsd
  3. Jan 30, 2012 #2

    micromass

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    What possible orders do elements in [itex]C_{p^2}[/itex] have. Are these order compatible with the elements in [itex]C_p\times C_p[/itex]?

    It certainly is possible. For example, we have [itex]C_6\cong C_2\times C_3[/itex]. Indeed, if we use additive notation ([itex]C_6=\{0,1,2,3,4,5\}[/itex]), then

    [itex]f:C_6\rightarrow C_2\times C_3[/itex]

    by f(0)=(0,0), f(1)=(1,1), f(2)=(0,2), f(3)=(1,0), f(4)=(0,1), f(5)=(1,2) is an isomorphism.
     
  4. Jan 31, 2012 #3
    But surely if again we use additive notatoin with [itex] C_9=\{0,1,2,3,4,5,6,7,8\} [/itex] then [itex] f:C_9 \to C_3 \times C_3 [/itex] with
    f(0) = (0,0), f(1) = (0,1), f(2) = (0,2), f(3) = (1,0), f(4) = (1,1), f(5) = (1,2), f(6) = (2,0),
    f(7) = (2,1), f(8) = (2,2) is as isomorphism?
     
  5. Jan 31, 2012 #4

    micromass

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    No. f(2+2)=f(4)=(1,1). But f(2+2)=f(2)+f(2)=(0,2)+(0,2)=(0,1). So it's not a homomorphism.
     
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