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Math Help - groups easy

  1. #1
    Member Jason Bourne's Avatar
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    groups easy

    (a) Give an example of a group which is abelian but not cyclic.

    K4 right?

    (b) Find 2^2005 in \mathbb{Z}_{17} (non-zero elements)

    (c) Find all solutions x \in \mathbb{Z} to the equation

    2^x = 3~mod~11
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post
    (a) Give an example of a group which is abelian but not cyclic.

    K4 right?
    Yes.

    (b) Find 2^2005 in \mathbb{Z}_{17} (non-zero elements)
    2^{16}\equiv 1(\bmod 17) thus 2^{2000}\equiv 1(\bmod 17) by raising exponents to 125. It remains to compute 2^5 \equiv 32 \equiv 15 thus 2^{2005}\equiv 15(\bmod 17).

    (c) Find all solutions x \in \mathbb{Z} to the equation

    2^x = 3~mod~11
    First it makes no sense to talk about negative exponents. Thus, we will solve this equation for x\in \mathbb{Z}^+. Any x can be written as 10k+r where 0\leq r\leq 9. Thus, 2^x \equiv 2^{10k+r} \equiv 2^{10k}2^r \equiv (2^{10})^k 2^r \equiv 2^r (\bmod 11).
    Thus, we require that 2^r \equiv 3(\bmod 11) for 0\leq r\leq 9. Direct computation shows that r=8 is the only such exponent. Thus all the solutions have the form 10k+8 which can be writen as x\equiv 8(\bmod 10) for x>0.
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