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Math Help - Normal and Cyclic

  1. #1
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    Normal and Cyclic

    If K is normal in G and K is cyclic. Show that every subgroup of K is normal in G
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  2. #2
    Senior Member Tinyboss's Avatar
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    Let H be a subgroup of K, g an element of G, and consider the group H_g=gHg^{-1}. Since conjugation is an automorphism, H_g is isomorphic to H. Using that and the fact that K is normal and cyclic, can you finish?
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  3. #3
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    or: K cyclic means that K = <k> for some k in K. so any subgroup of K is of the form <k^m> for some positive integer m.

    now, what is g(k^m)g^-1, for any g in G?
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  4. #4
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    So what you are saying is that since K is normal then K=gk^mg^-1, but since every subgroup is of the same form then every subgroup of K is normal?
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  5. #5
    Senior Member Tinyboss's Avatar
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    Conjugation carries K to an isomorphic subgroup K' of N. How many subgroups of order |K| does the cyclic group N have?
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    I dont understand what conjugation carries K to an isomorphic subgroup K'. I dont think we have learned that yet!
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  7. #7
    Senior Member Tinyboss's Avatar
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    I just used K' to denote a different (although it will turn out to be the same) group from K. To see why conjugation carries K to a group isomorphic to K, just show that the map you get from conjugating by g, k\mapsto gkg^{-1}, is a homomorphism with an inverse, and therefore is an isomorphism.
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  8. #8
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    i was hinting at a more elementary approach: K cyclic means that K = <k> for some element k in G.

    now the ONLY subgroups K can possibly have are of the form <k^m>, for some positive integer m.

    so suppose H is a subgroup of K, so that H = <k^m>. so let's look at what happens to a typical element of gHg^-1:

    we will call a typical element of H, h = (k^m)^n = k^(mn). so a typical element of gHg^-1 is ghg^-1 =

    gk^(mn)g^-1 = (gkg^-1)^(mn). now, we know that K itself is normal, so gkg^-1 is some other element of K, say k' = k^s.

    so (gkg^-1)^(mn) = (k')^(mn) = (k^s)^(mn) = k^(s(mn)) = k^((mn)s) = (k^(mn))^s = ((k^m)^n)^s = (k^m)^(ns).

    but (k^m)^(ns) is just some power of k^m, that is to say is it some element of <k^m> = H.

    so any element of gHg^-1, is in fact some element of H, that is, H is normal in G.

    @ Tinyboss: note that i have been very careful not to give any kind of order argument in the above. there is no stipulation that G is finite.
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  9. #9
    Senior Member Tinyboss's Avatar
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    Quote Originally Posted by Deveno View Post
    @ Tinyboss: note that i have been very careful not to give any kind of order argument in the above. there is no stipulation that G is finite.
    <snip>

    Edit: Bah, that was dumb. I hope nobody read it before I removed it.
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