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Math Help - [8]=[2]ˉ¹ ?

  1. #1
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    Lightbulb [8]=[2]ˉ¹ ?

    Why is this true? I don't understand.

    Example 2 p164
    It states that in the subgroup H={[2], [4], [6], [8]},
    [8]=[2]ˉ¹ in Z₁₀ under multiplication.

    This is what I gather:
    Since [2] ∈ H, [8]=[2]=[2]ˉ¹.

    or

    Since [8] can be written as [8]³=[12]=[2], then [2]ˉ¹ (the inverse of [2]) also equals [8].

    Please let me know if this is right and what definition or theorems support this. I think I heard something like this in class but I can't remember.
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  2. #2
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    I don't really understand what you are trying to do, but here's what I think you mean: That in the ring of integers modulo 10, 2*8=1 mod(10), but this is not true since the inverse of 2 is 5, and 2*8=16=6 mod(10). Also, I don't know what H has to do here, since operations on H are inherited from the whole ring.

    Now if you mean that 2+8=0 mod(10) then this is indeed true because 10 divides 10, and hence 2 is the inverse of 8 in the ADDITIVE subgroup H.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Hint: What is the identity of your group?

    Further hint: (Z_{10}, *) does not form a group, and neither does (Z_{10} \setminus \{0\}, *) because in one you have a zero, and in the other 5*2=0.
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