If G, The set of all mappings T_a,b defined by T_a,b(r)= ar+b
and H = {T_a,b element of G,a is rational}..How do you verify that H is a normal subgroup of G?
Thanks
Sorry, I don't get what you're trying to say. Let me see if I understand it:
$\displaystyle G= \{ T_{a,b} : \mathbb{R} \rightarrow \mathbb{R} : a,b \in \mathbb{R} \}$ is this correct? If it is my question is: Let $\displaystyle T_{a,b},T_{c,d} \in G$ what is $\displaystyle T_{a,b}*T_{c,d}$? (where $\displaystyle *$ is the operation of the group $\displaystyle G$)
What you want to show is that $\displaystyle T_{a,b}*T_{c,d}*(T_{a,b})^{-1} = T_{e,f}$ where $\displaystyle e$ is a rational number for all $\displaystyle c$ a rational number.
There are three actually quite simple steps to doing this:
(1) Firstly, you must work out what the identity element is: $\displaystyle T_{a,b}*id= T_{a,b}$.
(2) Secondly, you must work out the inverse of $\displaystyle T_{a,b}$, an arbitrary element from your group.
(3) Lastly, you need to verify the result.
Which bit are you struggling with?
Thanks Swlabr for outlining the steps
so for the last step if e and f for T_e,f are rational then that means H is a normal subgroup in G. I had thought that if a subgroup is normal then every right coset is equal to every left coset i.e Ha = aH. is this also true.??
$\displaystyle aH=Ha \Leftrightarrow aHa^{-1}=H \Leftrightarrow aha^{-1} \in H \text{ } \forall \text{ } h \in H$. So what is easiest to show is that if we take an element from $\displaystyle H$, $\displaystyle h$, and an arbitrary element from $\displaystyle G$, $\displaystyle g$, then $\displaystyle ghg^{-1} \in H$. This is what we want to show here.
The elements of $\displaystyle H$ are the elements of the group of the form $\displaystyle T_{a,b}$ where $\displaystyle a$ is a rational number. So basically you have to show that if $\displaystyle a \in \mathbb{Q}$ then $\displaystyle T_{c,d}T_{a,b}T_{c,d}^{-1} \in H$. To do this you need to show that $\displaystyle T_{c,d}T_{a,b}(T_{c,d})^{-1} = T_{e,f} $ where $\displaystyle e \in \mathbb{Q}$. It does not matter what happens to $\displaystyle f$.