# How to find the quotient group $Z_{1023}^*/\langle 2\rangle$?

When $m=1023$, what are the quotient groups below? $$Z_m^*/ \langle 2\rangle$$

$Z_m^*=\{1,2,4,5, \dots \}$

$\langle 2\rangle=\{1,2,4,8,16,32,64,128,256,512\}$

\begin{align*} Z_m^*/\langle 2\rangle &=\{1,2,4,8,16,32,64,128,256,512\},\\ {} & \mathrel{\hphantom{=}}\{2,3,5,9,17,33,65,129,257,513\}, &&\text{(added by 1)},\\ {} & \mathrel{\hphantom{=}}\{3,4,6,10,18,34,66,130,258,514\}, && \text{(added by 2)},\\ {} & \mathrel{\hphantom{=}}\{5,6,8,12,20,36,68,132,260,516\}, &&\text{(added by 4)}\\ & \,\, \vdots \end{align*}

Are the answers $\langle 2 \rangle$ incremented by all the elements i$\in Z_m^*$???

Am I right? Correct me if i am wrong.

• When you write $Z^*_m$, do you mean the group of multiplicative units in $\mathbb{Z}/m\mathbb{Z}$? If so, you shouldn't using addition to generate the cosets. – PeterJL Sep 6 '16 at 17:19
• Yes. So how can I find ? – mallea Sep 6 '16 at 17:23
• As a first step, I'd recommend getting a handle on the elements of $(\mathbb{Z}/m\mathbb{Z})^{\times}$. For example, you list $3$ as one such element, but $3$ isn't relatively prime to $1023$, so it isn't invertible. Have you written down a prime factorization of $1023$? Do you know how many elements are in your group to begin with? – PeterJL Sep 6 '16 at 17:27
• 1023 can be factorized as 3x11x31. the number of elements in $Z_m^*$ is 600. This means I can start with 600 different elements? – mallea Sep 6 '16 at 17:34
• I know the number of the quotient groups. That is 60. This can be led from 600/10, where 10 is the order of $\langle 2 \rangle$. – mallea Sep 6 '16 at 17:36

Since $1023 = 3*11*31$ the Chinese remainder theorem tells us that $\mathbb{Z}_{1023}^* \cong \mathbb{Z}_3^* \times \mathbb{Z}_{11}^* \times \mathbb{Z}_{31}^*$. Since $3$,$11$ and $31$ are prime, we have: $\mathbb{Z}_3^* \times \mathbb{Z}_{11}^* \times \mathbb{Z}_{31}^* \cong \mathbb{Z}_2 \times \mathbb{Z}_{10}\times\mathbb{Z}_{30}$. You found that $|\langle2\rangle| = 10$. So I'm pretty sure $\mathbb{Z}_{1023}^*/\langle2\rangle \cong \mathbb{Z}_2 \times \mathbb{Z}_{30}$. If I'm wrong hopefully one who knows better will help.

The group $(\mathbb{Z}/m)^*$ has order $\phi(m)$, so $G=(\mathbb{Z}/1023)^*$ has order $600$. It is given by $(\mathbb{Z}/3)^*\times (\mathbb{Z}/11)^* \times (\mathbb{Z}/31)^*\cong C_2\times C_{10}\times C_{30}$. Determine the subgroup $U$ generated by $2$ in $G$. By Lagrange, the order of $U$ is a divisor of $600$. Then determine the quotient group $G/U$.