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Group Homomorphism?

  1. Aug 5, 2014 #1
    1. The problem statement, all variables and given/known data
    Groups G and H are both groups in Z_7 (integer modulo), the mapping Is given by ø(g) = 2g

    is ø: G-->H a homomorphism?



    3. The attempt at a solution
    My textbook says yes, I can't understand why.

    ø(g1g2) = 2(g1g2) does not equal 2g1*2g2 = ø(g1)ø(g2)

    something missing here?
     
  2. jcsd
  3. Aug 5, 2014 #2

    HallsofIvy

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    Apparently you are missing that this is in "Z_7". (I'm not sure what you mean by "groups in Z_7". The group Z_7, with either addition as operation, has only itself and the identity as subgroups.)

    In any case, in Z_7, ø(0)= 0, ø(1)= 2, ø(2)= 4, ø(3)= 6, ø(4)= 1, ø(5)= 3, ø(6)= 5.

    I wont go through all 36 possible combinations but, for example, if g1= 3 and g2= 5, then 3+ 5= 8= 1 (mod 7) so ø(g1g2)= ø(1)= 2 while ø(3)= 6 and ø(5)= 3 and 6+ 3= 9= 2 (mod 7).
     
  4. Aug 5, 2014 #3
    Ooh thank you, I thought the group operation was multiplication
     
  5. Aug 5, 2014 #4

    jbunniii

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    ##Z_7## is the ring of integers modulo 7. Every ring is an additive group (just ignore the multiplication).

    A ring is not a multiplicative group because ##0## has no multiplicative inverse, and the same may be true of other elements. However, the set of units (elements which have multiplicative inverses), if it is nonempty, does form a group under multiplication, called the group of units. To distinguish this group from the ring, we often use a notation such as ##Z_7^\times##.

    In the case of ##Z_7##, or more generally ##Z_p## for any prime number, every nonzero element is a unit. So ##Z_7^\times## consists of ##\{1,2,3,4,5,6\}## and more generally, ##Z_p^\times## contains ##p-1## elements when ##p## is prime.

    On the other hand, if ##n## is not prime, then some nonzero elements of ##Z_n## may not be units. For example, in ##Z_4##, the group of units is ##Z_4^\times = \{1,3\}##.
     
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