(New page: Category:MA453Spring2009Walther We need to show that left and right cosets agree. The left cosets of H are H and the complement of H, because there are 2 cosets and they must be disj...) |
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--[[User:Nswitzer|Nswitzer]] 17:08, 17 February 2009 (UTC) | --[[User:Nswitzer|Nswitzer]] 17:08, 17 February 2009 (UTC) | ||
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| + | To elaborate a bit on this solution: | ||
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| + | Index 2 means that there are exactly 2 distinct left cosets and 2 distinct right cosets of H in G. Furthermore, it means that |H| is half of |G|. In order to show that H is normal, we must show that xH = Hx. | ||
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| + | Suppose x is an element of G that is also in H. Then xH = H because x is in H, so xH can only produce elements already in H; furthermore, since there are only 2 distinct left and right cosets, each coset must be of order = half of |G|. We can then conclude that xH must produce exactly |H| elements already in H, which is simply H. By the same logic, Hx = H. | ||
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| + | If x is not an element of G, then xH is not in H. Therefore, xH must produce a different coset than the one above. Since there are only 2 distinct left cosets, this new coset must be exactly the elements in G but not in H, we call it H'. Using the same reasoning as above, we know that Hx = H'. | ||
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| + | From these 2 conclusions, it is clear that for any x in G, xH = Hx, which is the definition of a normal subgroup. | ||
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| + | --[[Ysuo|Ysuo]] | ||
Revision as of 09:19, 18 February 2009
We need to show that left and right cosets agree. The left cosets of H are H and the complement of H, because there are 2 cosets and they must be disjoint. The same works for the right cosets. So they are equal.
--Nswitzer 17:08, 17 February 2009 (UTC)
To elaborate a bit on this solution:
Index 2 means that there are exactly 2 distinct left cosets and 2 distinct right cosets of H in G. Furthermore, it means that |H| is half of |G|. In order to show that H is normal, we must show that xH = Hx.
Suppose x is an element of G that is also in H. Then xH = H because x is in H, so xH can only produce elements already in H; furthermore, since there are only 2 distinct left and right cosets, each coset must be of order = half of |G|. We can then conclude that xH must produce exactly |H| elements already in H, which is simply H. By the same logic, Hx = H.
If x is not an element of G, then xH is not in H. Therefore, xH must produce a different coset than the one above. Since there are only 2 distinct left cosets, this new coset must be exactly the elements in G but not in H, we call it H'. Using the same reasoning as above, we know that Hx = H'.
From these 2 conclusions, it is clear that for any x in G, xH = Hx, which is the definition of a normal subgroup.
--Ysuo
