Let H be a subgroup of K, g an element of G, and consider the group . Since conjugation is an automorphism, is isomorphic to . Using that and the fact that K is normal and cyclic, can you finish?
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, , is a homomorphism with an inverse, and therefore is an isomorphism.
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.