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Prove that if H has index 2 in G, then H is normal in G. | Prove that if H has index 2 in G, then H is normal in G. | ||
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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. | 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. | ||
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| + | Question: '''Prove that if H has index 2 in G, then H is normal in G'''. | ||
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| + | Answer: | ||
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| + | Let G be a group and H be the subgroup of G. | ||
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| + | In order of H to be normal in G, h <math>\in</math> H and g <math>\in</math> G then, gh <math>g^(-1)</math> <math>\in</math> H | ||
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| + | So, if H = { H , ah }, and if a <math>\in</math> H, then aH = H = Ha. | ||
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| + | If x is not <math>\in</math> H, then aH <math>\in</math> G but not H and Ha <math>\in</math> G too but not in H. | ||
Revision as of 17:22, 28 September 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.
