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We know that A3 is all even permutations of S3, which means that |A3| is exactly half of |S3| since there are an equal number of even and odd permutations. We also know that A3 is a subgroup in S3. | We know that A3 is all even permutations of S3, which means that |A3| is exactly half of |S3| since there are an equal number of even and odd permutations. We also know that A3 is a subgroup in S3. | ||
| − | Now refer back to the first problem from this assignment. Let H = A3, G = S3. By theorem 7.1, the index (distinct right/left cosets) of H in G is | | + | Now refer back to the first problem from this assignment. Let H = A3, G = S3. By theorem 7.1, the index (distinct right/left cosets) of H in G is |G|/|H| = 2. We proved that any subgroup H in G with index = 2 is normal in G. Therefore, A3 is normal in S3. |
--[[Ysuo|Ysuo]] | --[[Ysuo|Ysuo]] | ||
Revision as of 20:28, 18 February 2009
prove that A_3 inside S_3 is normal.
Does anyone have a good idea of how to go about proving this?
I just used the fact that parity (even and oddness) is a homomorphism and respects products.
--Jzage 16:14, 18 February 2009 (UTC)
Here's an idea. I'll let you judge if it's good.
We know that A3 is all even permutations of S3, which means that |A3| is exactly half of |S3| since there are an equal number of even and odd permutations. We also know that A3 is a subgroup in S3.
Now refer back to the first problem from this assignment. Let H = A3, G = S3. By theorem 7.1, the index (distinct right/left cosets) of H in G is |G|/|H| = 2. We proved that any subgroup H in G with index = 2 is normal in G. Therefore, A3 is normal in S3.
--Ysuo
Yea I like that idea. Well done.
--Jniederh 23:09, 18 February 2009 (UTC)
Agreed, That's a brilliant finale to this assignment. --Bcaulkin 23:59, 18 February 2009 (UTC)
Indeed.
