# Normal Subgroups

• October 2nd 2009, 01:23 PM
Godisgood
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
• October 2nd 2009, 02:19 PM
Jose27
Quote:

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?
• October 2nd 2009, 02:42 PM
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
• October 2nd 2009, 03:40 PM
Jose27
Quote:

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 $G$)
• October 3rd 2009, 12:35 AM
Godisgood
http://www.mathhelpforum.com/math-he...0d1ba9c1-1.gif = a(cr + d)+ b...does it make sense now
• October 3rd 2009, 08:35 AM
Swlabr
Quote:

Originally Posted by Godisgood
http://www.mathhelpforum.com/math-he...0d1ba9c1-1.gif = 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?
• October 3rd 2009, 09:15 AM
Godisgood
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.??
• October 3rd 2009, 09:28 AM
Swlabr
Quote:

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 $f$.