Hw11 Problem 1

### Problem 1

Find the order of the following elements in the factor group.

Solution

(a)

$3 + \langle 4 \rangle$ in $\mathbb{Z_{12}} / \langle 4 \rangle$

(1)
\begin{align} \langle 4 \rangle = \{ 0,4,8 \} \end{align}
(2)
\begin{align} 3 + \langle 4 \rangle \end{align}
(3)
\begin{align} 3 + \langle 4 \rangle + 3 + \langle 4 \rangle = 6 + \langle 4 \rangle \end{align}
(4)
\begin{align} 6 + \langle 4 \rangle + 3 + \langle 4 \rangle = 9 + \langle 4 \rangle \end{align}
(5)
\begin{align} 3 + \langle 4 \rangle + 9 + \langle 4 \rangle = 0 + \langle 4 \rangle \end{align}

So $|3 + \langle 4 \rangle| = 4$.

(b)

$(3,3) + \langle (1,2) \rangle$ in $(\mathbb{Z_{4}} \times \mathbb{Z_{8}} )/ \langle(1,2) \rangle$

(6)
\begin{align} \langle (1,2) \rangle = \{ (1,2),(2,4),(3,6),(0,0) \} \end{align}
(7)
\begin{align} (3,3) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (2,6) + \langle (1,2) \rangle \end{align}
(8)
\begin{align} (2,6) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (1,1) + \langle (1,2) \rangle \end{align}
(9)
\begin{align} (1,1) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (0,4) + \langle (1,2) \rangle \end{align}
(10)
\begin{align} (0,4) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (3,7) + \langle (1,2) \rangle \end{align}
(11)
\begin{align} (3,7) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (2,2) + \langle (1,2) \rangle \end{align}
(12)
\begin{align} (2,2) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (1,5) + \langle (1,2) \rangle \end{align}
(13)
\begin{align} (1,5) + \langle (1,2) \rangle + (3,3) + \langle (1,2) \rangle = (0,0) + \langle (1,2) \rangle \end{align}

$(0,0) \in \langle (1,2) \rangle$, so $|(3,3) + \langle (1,2) \rangle| = 8$