Difference between revisions of "Chapter 9, Problem 7 MA453Fall2008walther" - Rhea

Latest revision as of 18:00, 5 October 2008

Prove that if H has index 2 in G, then H is normal in G.

With the definition of index being the number of disticnts cosets of H in G.

On first glance I don't have much on this, I am leaning toward doing it by contradiction, because I don't see any direct correlation between the two topics.

Question: Prove that if H has index 2 in G, then H is normal in G.

Answer:

Let G be a group and H be the subgroup of G.

In order of H to be normal in G, h $ \in $ H and g $ \in $ G then, gh $ g^(-1) $ $ \in $ H

So, if H = { H , ah }, and if a $ \in $ H, then aH = H = Ha.

If x is not $ \in $ H, then aH $ \in $ G but not H and Ha $ \in $ G too but not in H.

• --Mmohamad 22:57, 28 September 2008 (UTC)

What do you mean by the notation H={H,ah}, I am not familiar with what this means.

-Allen

The subgroup H of G has two distinct cosets in G: H, and aH, for some a. aH contains all elements of G that are not in H. An element x of G either lies within H or aH. If it lies within H, then xH = H = Hx by Property 2 of the lemma in Chapter 7. Now, suppose that x does not lie within H. Then, xH must be equal to aH, since they both describe the set of elements that are in G but not H. Similarly, Hx must be equal to aH. By transitivity, xH = Hx for all x belonging to G, and therefore, H is normal in G.

--Nick Rupley 00:20, 2 October 2008 (UTC)

Since ord(G)/ord(H) = 2, hence there are two right coset {H, Ha} of H in G and two left coset {H, aH} of H in G where a G but a H. Because G=H∪Ha=H∪aH , H∩Ha = $ \phi $ and H∩aH = $ \phi $. We can get that aH=Ha. So H is normal in G