# Thread: groups easy

1. ## 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

$2^x = 3~mod~11$

2. Originally Posted by Jason Bourne
(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 $x>0$.