Hw10 Problem 3

### Problem 3

In the following, compute the indicated quantities for the given homomorphism.

(a) Ker($\phi$) and $\phi (18)$ for $\phi :\mathbb{Z} \rightarrow S_8$ such that $\phi (1)=(1426)(257)$.

(b) Ker($\phi$) and $\phi (-3,2)$ for $\phi :\mathbb{Z} \times \mathbb{Z} \rightarrow \mathbb{Z}$ where $\phi (1,0)=3$ and $\phi (0,1)=-4$.

(c)(+) Ker($\phi$) and $\phi(15)$ for $\phi : \mathbb{Z}_{24} \rightarrow S_{8}$ where $\phi(1) = (25)(1467)$.

Solution

(a) $\phi (1)=(1426)(257)=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}$
According to the homomorphism property,
$\phi (2)=\phi (1+1)=\phi (1)\phi (1)=\phi(1)^2=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 2&7&3&5&6&4&1&8\end{pmatrix}$
$\phi (3)=\phi(2+1)=\phi (2)\phi (1)=\phi(1)^3=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 2&7&3&5&6&4&1&8\end{pmatrix}\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 5&6&3&7&1&2&4&8\end{pmatrix}$
$\phi (4)=\phi(3+1)=\phi (3)\phi (1)=\phi(1)^4=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 5&6&3&7&1&2&4&8\end{pmatrix}\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 7&1&3&6&4&5&2&8\end{pmatrix}$
$\phi (5)=\phi(4+1)=\phi (4)\phi (1)=\phi(1)^5=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 7&1&3&6&4&5&2&8\end{pmatrix}\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 6&4&3&1&2&7&5&8\end{pmatrix}$
$\phi (6)=\phi(5+1)=\phi (5)\phi (1)=\phi(1)^6=\begin{pmatrix} 1&2&3&4&5&6&7&8\\ 6&4&3&1&2&7&5&8\end{pmatrix} \begin{pmatrix} 1&2&3&4&5&6&7&8\\ 4&5&3&2&7&1&6&8\end{pmatrix}=\begin{pmatrix} 1 \end{pmatrix}$
So the order of $|\langle(1426)(257)\rangle |=6$ and $\phi(1)^{6n} =(1)\ ,n\in \mathbb{Z^+}$.
Because $\phi(m) =\phi (1)^m$, when m=6n, $\phi(m) =\phi (1)^m=\phi(1)^{6n} =(1)$.
So $Ker(\phi)=\{6n|n\in\mathbb{Z}\}$
Because $18=6 \times 3$, $\phi(18)=(1)$.

(b)According to the homomorphism property,
Because $\phi(1,0)=3,\ \phi(0,1)=-4$, $\phi((4m,3m)=\phi((4m,0)+(0,3m))=\phi(4m,0)+\phi(0,3m)=4m\phi(1,0)+3m\phi(0,1)=12m-12m=0,\ m\in \mathbb{Z}$.
So $Ker(\phi)=\{(4m,3m)|m\in\mathbb{Z}\}$.
$\phi(-3,2)=\phi(-3,0)+\phi(0,2)=3\phi(-1,0)+2\phi(0,1)=-3\phi(1,0)+2\phi(0,1)=-3\times3+2\times(-4)=-17$.

(c)(+) Let $\sigma = (25)(1467)$. Now $|\sigma| = lcm(2, 4) = 4$. $Ker(\phi) = \langle 4 \rangle = \{0, 4, 8, 12, 16, 20\}$. Now $\phi(15) = \phi(12 +_{24} 3) = \sigma^{3} = (1764)(25)$.