what are $(\mathbb{Z}/17\mathbb{Z})^*/\langle 2 \rangle$?? Suppose modulo 17, $\langle 2 \rangle=\{1,2,4,8,9,13,15,16\}$ since
$2^1 \equiv 2 \mod 17$
$2^2 \equiv 4 \mod 17$
$2^3 \equiv 8 \mod 17$
$2^4 \equiv 16 \mod 17$
$2^5 \equiv 32 \equiv 15 \mod 17$
$2^6 \equiv 15 \times 2 \equiv 13 \mod 17$
$2^7 \equiv 13 \times 2 \equiv 9 \mod 17$
$2^8 \equiv 9 \times 2 \equiv 1 \mod 17$
Now I want to know the quotient groups  $(\mathbb{Z}/17\mathbb{Z})^*/\langle 2 \rangle$
These are $\langle 2 \rangle=\{1,2,4,8,9,13,15,16\}$ itself and $\{3,5,6,7,10,11,12,14\}$.
The latter part is the rest of $(\mathbb{Z}/17\mathbb{Z})^*$ except  $\langle 2 \rangle$
How do we find the latter part? In this example  the number of the quotient groups are 2, it seems simple.
 A: Hint: what is the order of your quotient group? How many groups exists of such an order?
A: By Lagrange's theorem, you know that 
$$|G|:=|(\mathbb{Z}/17\mathbb{Z})^*/\langle\bar{2}\rangle|=\frac{|(\mathbb{Z}/17\mathbb{Z})^*|}{|\langle\bar{2}\rangle|}.$$
As you have shown, $|\langle\bar{2}\rangle|=8,$ so $|G|=\frac{17-1}{8}=2.$ As you know too that $\langle\bar{2}\rangle\in G,$ it follows that 
$$G=\{\langle\bar{2}\rangle,(\mathbb{Z}/17\mathbb{Z})^*-\langle\bar{2}\rangle\}.$$
A: It means any unit in $(\mathbf Z/17\mathbf Z)^*$ is either a power of $2$ or $3\times$ a power of $2$. Note the square of each element in the latter part is a power of $2$.
Also, observe that, $\langle 2\rangle\simeq \mathbf Z/8\mathbf Z$ (the additive group), and  $(\mathbf Z/17\mathbf Z)^*$ is a cyclic group, isomorphic to the additive group   $ \mathbf Z/16\mathbf Z$, so the multiplicative quotient group   $(\mathbf Z/17\mathbf Z)^*/\langle 2\rangle$ is isomorphic to the additive quotient group
$$ (\mathbf Z/16\mathbf Z)/(\mathbf Z/8\mathbf Z)\simeq\mathbf Z/2\mathbf Z$$
by the third isomorphism theorem.
