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| − | The subgroup H of G has two distinct cosets in G: H, and aH, for some a. | + | 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. |
:--[[User:Narupley|Nick Rupley]] 00:20, 2 October 2008 (UTC) | :--[[User:Narupley|Nick Rupley]] 00:20, 2 October 2008 (UTC) | ||
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| + | ---- | ||
| + | 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 = <math>\phi</math> and H∩aH = <math>\phi</math>. | ||
| + | We can get that aH=Ha. So H is normal in G | ||
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
