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Thread: 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 $\displaystyle \mathbb{Z}_{17}$ (non-zero elements)

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

    $\displaystyle 2^x = 3~mod~11$
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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 $\displaystyle \mathbb{Z}_{17}$ (non-zero elements)
    $\displaystyle 2^{16}\equiv 1(\bmod 17)$ thus $\displaystyle 2^{2000}\equiv 1(\bmod 17)$ by raising exponents to $\displaystyle 125$. It remains to compute $\displaystyle 2^5 \equiv 32 \equiv 15$ thus $\displaystyle 2^{2005}\equiv 15(\bmod 17)$.

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

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