If K is normal in G and K is cyclic. Show that every subgroup of K is normal in G

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- May 13th 2011, 01:29 PMolgavioletNormal and Cyclic
If K is normal in G and K is cyclic. Show that every subgroup of K is normal in G

- May 13th 2011, 03:59 PMTinyboss
Let H be a subgroup of K, g an element of G, and consider the group $\displaystyle H_g=gHg^{-1}$. Since conjugation is an automorphism, $\displaystyle H_g$ is isomorphic to $\displaystyle H$. Using that and the fact that K is normal and cyclic, can you finish?

- May 14th 2011, 01:04 AMDeveno
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? - May 15th 2011, 09:19 AMolgaviolet
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?

- May 15th 2011, 09:46 AMTinyboss
Conjugation carries K to an isomorphic subgroup K' of N. How many subgroups of order |K| does the cyclic group N have?

- May 15th 2011, 10:02 AMolgaviolet
I dont understand what conjugation carries K to an isomorphic subgroup K'. I dont think we have learned that yet!

- May 15th 2011, 10:17 AMTinyboss
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, $\displaystyle k\mapsto gkg^{-1}$, is a homomorphism with an inverse, and therefore is an isomorphism.

- May 15th 2011, 02:54 PMDeveno
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. - May 16th 2011, 09:06 AMTinyboss