Homomorphism and group problem

• Mar 31st 2008, 11:17 PM
kleenex
Homomorphism and group problem
For each $a \ne 0$ and $b$ in $GF(7) = \bold{Z}_7$ define a function $\mathit{F}_{a,b}:\bold{Z}_7\rightarrow \bold{Z}_7$ by

$\mathit{F}_{a,b}(x) = ax + b$

The set of all such function $\mathit{F}_{a,b}$ forms a group $g$ with the group multiplication given by the compostion of the functions.(Not need to verify $G$ is a group)

a. Determine $|G|$ and find $\mathit{F}_{a,b}\circ \mathit{F}_{c,d} = \mathit{F}_{a,b}(\mathit{F}_{c,d}(x))$. Hence show that $G$ is non-abelian.

b. Show that $\mathit{f}:G\rightarrow(\bold{Z}_7\setminus {0},\cdot )$,given by $\mathit{f}(\mathit{F}_{a,b})=a$, is a homomorphism.

I'm OK with the $|G|$ part but I forget how to show $\mathit{F}_{a,b}\circ \mathit{F}_{c,d} = \mathit{F}_{a,b}(\mathit{F}_{c,d}(x))$.
Also for part b, I don't really get it.

Thank you for help
• Apr 1st 2008, 08:12 AM
ThePerfectHacker
a) $F_{a,b}\circ F_{c,d} = F_{a,b}(cx+d) = a(cx+d)+b = acx + ad+b$.

b) $G$ is the set of all these linear transformations that are homomorphisms of $\mathbb{Z}_7$. Let $x\in G$, then it has the form $x=F_{a,b}$, by contrustion. We define a mapping $f:G\mapsto (\mathbb{Z}_7\setminus \{ 0 \} )$ as $f(F_{a,b})=a$, the first coefficient. You need to show $f(F_{a,b}\circ F_{c,d}) = f(F_{a,b})f(F_{c,d})$.