Re: Every subgroup with indice(index?)2 is normal!Why???
From: Sukjah Roh (dhls_DELET_Wkr8_at_gm_DELET_ail.com)
Date: 03/29/05
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Date: Tue, 29 Mar 2005 22:35:30 +0900
Evelin wrote:
> How can I show that every subgroup with indice
>
> (in French, in English perhaps index, but I am not sure,
> it is the cardinalitiy of G(the group) divided by
> the cardinality of the supgroup)
>
> 2 of a finite group is normal!!!
>
> Thank you for answering my question!!!!
(0. Analysis)
Let's look at hypothesis and conclusion.
Hypothesis: H is a subgroup of G of index 2.
Conclusion: H is normal in G.
(1. Backward)
Let's look at what conditions about H leads to the conclusion.
(1) gH = Hg for all g in G.
(2) gHg^(-1) = H for all g in G.
(3) ghg^(-1) is in H for all g in G, all h in H.
Each of the above three leads to the conclusion.
(2. Forward)
Now, let's look at what can be said from the hypothesis.
If H is of index 2. then there are exactly two right cosets H and Ha. a
is not in H.
We see that something similar to 'Ha' appears in (1).
There are also exactly two left cosets H and bH. b not in H. Because a
is also not in H, bH should equal to aH.
So H,aH are exactly two left cosets, and H,Ha are exactly two right
cosets. and aH = G-H = Ha. this is similar to something in (1).
(3. Linking)
Now we can guess that the next promising step is to try to prove (1).
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