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Math Help - Normal Subgroups

  1. #1
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    Question 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
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  2. #2
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    Quote Originally Posted by Godisgood View Post
    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?
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by Godisgood View Post
    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)
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  5. #5
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    = a(cr + d)+ b...does it make sense now
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Godisgood View Post
    = 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?
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  7. #7
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    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.??
    Last edited by Godisgood; October 3rd 2009 at 09:25 AM.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Godisgood View Post
    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.
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