1. ## Normal Subgroups

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

2. Originally Posted by Godisgood
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
What is the operation G?

3. under * G = T_a,b(r)= ar+b so if a =3 and b = 2 , T_3,2(r) = 3r + 2
hope this makes sense

4. Originally Posted by Godisgood
under * G = T_a,b(r)= ar+b so if a =3 and b = 2 , T_3,2(r) = 3r + 2
hope this makes sense
Sorry, I don't get what you're trying to say. Let me see if I understand it:

$G= \{ T_{a,b} : \mathbb{R} \rightarrow \mathbb{R} : a,b \in \mathbb{R} \}$ is this correct? If it is my question is: Let $T_{a,b},T_{c,d} \in G$ what is $T_{a,b}*T_{c,d}$? (where $*$ is the operation of the group $G$)

5. = a(cr + d)+ b...does it make sense now

6. Originally Posted by Godisgood
= a(cr + d)+ b...does it make sense now
What you want to show is that $T_{a,b}*T_{c,d}*(T_{a,b})^{-1} = T_{e,f}$ where $e$ is a rational number for all $c$ a rational number.

There are three actually quite simple steps to doing this:

(1) Firstly, you must work out what the identity element is: $T_{a,b}*id= T_{a,b}$.

(2) Secondly, you must work out the inverse of $T_{a,b}$, an arbitrary element from your group.

(3) Lastly, you need to verify the result.

Which bit are you struggling with?

7. 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.??

8. Originally Posted by Godisgood
Thanks Swlabr for outline 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.??
$aH=Ha \Leftrightarrow aHa^{-1}=H \Leftrightarrow aha^{-1} \in H \text{ } \forall \text{ } h \in H$. So what is easiest to show is that if we take an element from $H$, $h$, and an arbitrary element from $G$, $g$, then $ghg^{-1} \in H$. This is what we want to show here.

The elements of $H$ are the elements of the group of the form $T_{a,b}$ where $a$ is a rational number. So basically you have to show that if $a \in \mathbb{Q}$ then $T_{c,d}T_{a,b}T_{c,d}^{-1} \in H$. To do this you need to show that $T_{c,d}T_{a,b}(T_{c,d})^{-1} = T_{e,f}$ where $e \in \mathbb{Q}$. It does not matter what happens to $f$.