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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- October 2nd 2009, 02:23 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 - October 2nd 2009, 03:19 PMJose27
- October 2nd 2009, 03:42 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 - October 2nd 2009, 04:40 PMJose27
- October 3rd 2009, 01:35 AMGodisgood
http://www.mathhelpforum.com/math-he...0d1ba9c1-1.gif = a(cr + d)+ b...does it make sense now

- October 3rd 2009, 09:35 AMSwlabr
What you want to show is that where is a rational number for all a rational number.

There are three actually quite simple steps to doing this:

(1) Firstly, you must work out what the identity element is: .

(2) Secondly, you must work out the inverse of , an arbitrary element from your group.

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

Which bit are you struggling with? - October 3rd 2009, 10:15 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.?? - October 3rd 2009, 10:28 AMSwlabr
. So what is easiest to show is that if we take an element from , , and an arbitrary element from , , then . This is what we want to show here.

The elements of are the elements of the group of the form where is a rational number. So basically you have to show that if then . To do this you need to show that where . It does not matter what happens to .