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

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- Oct 2nd 2009, 01:23 PMGodisgoodNormal 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 - Oct 2nd 2009, 02:19 PMJose27
- Oct 2nd 2009, 02:42 PMGodisgood
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 - Oct 2nd 2009, 03:40 PMJose27
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 $\displaystyle G$) - Oct 3rd 2009, 12:35 AMGodisgood
http://www.mathhelpforum.com/math-he...0d1ba9c1-1.gif = a(cr + d)+ b...does it make sense now

- Oct 3rd 2009, 08:35 AMSwlabr
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? - Oct 3rd 2009, 09:15 AMGodisgood
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.?? - Oct 3rd 2009, 09:28 AMSwlabr
$\displaystyle aH=Ha \Leftrightarrow aHa^{-1}=H \Leftrightarrow aha^{-1} \in H \text{ } \forall \text{ } h \in H$. 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 $\displaystyle f$.