Thread: Proof "A normal subgroup of a group G that is subset of the center of G"

Dear Friends


Could someone help me in proving the following:

Suppose H is a subgroup of G of odd order and that |H|=5. Show that H is a subset of G.

What I tried is the following:
Since |H|=5 then H is cyclic so H=<h>={e, h, h^2,h^3,h^4}. For g in G (g^-1)*h*g is in H as H is normal so (g^-1)*h*g=h^k where k=0,1,2,3,4 e is the identity element.
If K=0 the we have h=e which is a contardiction as h is a generator.
If k=1 then we are done.
For the other cases I am not able to deal with them.

G has odd order implies that g and g^-1 are distinct. What can I also get from this infromation?


Regards.

"Suppose H is a subgroup of G of odd order and that |H|=5. Show that H is a subset of G."

I am not sure I understand this. A subgroup is automatically a subset of G by definition. Am I missing something?

Dear Friends

I am sorry for that typo error I mean we want to show that H is a subset of Z(G) the center of the group

Regards

I assume you mean the order of G is odd and H is a normal subgroup of G of order 5. Recall the "N over C" theorem for any subgroup H: N(H)/C(H) is isomorphic to a subgroup of aut(H), the automorphism group of H. (The proof is easy, N(H) acts on H by conjugation.)
So for your case G/C(H) is a subgroup of the automorphism group of the cyclic group of order 5, namely the cyclic group of order 4.
But since the order of G is odd, so also is the order of G/C(H); also the order of G/C(H) divides 4. So C(H)=G; i.e. H is contained in Z(G).