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Math Help - Groups

  1. #1
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    Groups

    In Z/6Z define [a] * [b] =[a+b-1]. Is that associative? If so, does this operation make Z/6Z into a group? If its not in the group explain why?


    Attempt:
    ([a] * [b]) * [c] = [a + b - 1] * [c] = [a + b + c - 2]
    [a] * ([b] * [c]) = [a] * [b + c - 1] = [a + b + c - 2] = ([a] * [b]) * [c]

    Thus, the operation is associative. Now, we just need to check identity and inverses. What's the identity? Well, we need some [e] such that:

    [a] * [e] = [e] * [a] = [a]

    Thus we need:

    [a] * [e] = [a + e - 1] = [a]

    for all a. That is, we need:

    a + e - 1 = a (mod 6)
    e - 1 = 0 (mod 6)
    e = 1 (mod 6)

    So [1] is an identity (and thus the unique identity). So, now can we compute an inverse for each element? Suppose we have some [a], and it has an inverse [b]. Then:

    [a] * [b] = [1]
    [a + b - 1] = [1]
    a + b - 1 = 1 (mod 6)
    a + b = 2 (mod 6)
    b = 2 - a (mod 6)

    Thus, if an inverse of [a] exists, it must be [2 - a]. We can check that [2 - a] is the inverse for each [a]:

    [2 - a] * [a] = [a] * [2 - a] = [a + 2 - a - 1] = [1]

    Thus, it works, and we have a group.


    2.In the dihedral group D3, are there any repetitions in the set[ I, r, r^2, t, tr, tr^2}of rigid motions?
    No. (I dont know, just a guess).



    Can someone check my work for these two problems?
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  2. #2
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    Quote Originally Posted by jjbrian View Post
    In Z/6Z define [a] * [b] =[a+b-1]. Is that associative? If so, does this operation make Z/6Z into a group? If its not in the group explain why?


    Attempt:
    ([a] * [b]) * [c] = [a + b - 1] * [c] = [a + b + c - 2]
    [a] * ([b] * [c]) = [a] * [b + c - 1] = [a + b + c - 2] = ([a] * [b]) * [c]

    Thus, the operation is associative. Now, we just need to check identity and inverses. What's the identity? Well, we need some [e] such that:

    [a] * [e] = [e] * [a] = [a]

    Thus we need:

    [a] * [e] = [a + e - 1] = [a]

    for all a. That is, we need:

    a + e - 1 = a (mod 6)
    e - 1 = 0 (mod 6)
    e = 1 (mod 6)

    So [1] is an identity (and thus the unique identity). So, now can we compute an inverse for each element? Suppose we have some [a], and it has an inverse [b]. Then:

    [a] * [b] = [1]
    [a + b - 1] = [1]
    a + b - 1 = 1 (mod 6)
    a + b = 2 (mod 6)
    b = 2 - a (mod 6)

    Thus, if an inverse of [a] exists, it must be [2 - a]. We can check that [2 - a] is the inverse for each [a]:

    [2 - a] * [a] = [a] * [2 - a] = [a + 2 - a - 1] = [1]

    Thus, it works, and we have a group.


    2.In the dihedral group D3, are there any repetitions in the set[ I, r, r^2, t, tr, tr^2}of rigid motions?
    No. (I dont know, just a guess).


    Why "guess"? Prove it! Of course, we must assume ord(r)=3\,,\,ord(t)=2 , and then t=r impossible, t=tr\Longrightarrow r=1 , impossible again, and etc.

    Tonio


    Can someone check my work for these two problems?
    .
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