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Thread: subgroup in A4 question

  1. #1
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    subgroup in A4 question

    There is only one subgroup of order 4 in A4 (Alternating group of degree 4)(This subgroup is (1), (12)(34), (13)(24), (14)(23)). Why does this imply that this subgroup must be a normal subgroup in A4? Generalize to arbitrary finite groups.
    I thought about using the fact that for some g that is an element of a group G, |gH^-1g^-1|=|H| (the order of the subgroup H equals each of its conjugates orders) in the whole group, I am not sure were to go from here. Can some help me on this one at all?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by wutang View Post
    There is only one subgroup of order 4 in A4 (Alternating group of degree 4)(This subgroup is (1), (12)(34), (13)(24), (14)(23)). Why does this imply that this subgroup must be a normal subgroup in A4? Generalize to arbitrary finite groups.
    I thought about using the fact that for some g that is an element of a group G, |gH^-1g^-1|=|H| (the order of the subgroup H equals each of its conjugates orders) in the whole group, I am not sure were to go from here. Can some help me on this one at all?
    Theorem: Let $\displaystyle G$ be a group and $\displaystyle N\leqslant G$ be the only group with order $\displaystyle |N|=n$. Then, $\displaystyle N\unlhd G$

    Proof: Define $\displaystyle \theta_g:N\to gNg^{-1}$ by $\displaystyle n\mapsto gng^{-1}$. Clearly this is a bijection. But, notice that $\displaystyle geg^{-1}=e\in gNg^{-1}$, $\displaystyle gng^{-1},gn'g^{-1}\in gNg^{-1}\implies (gng^{-1})(gn'g^{-1})=gnn'g^{-1}\in gNg^{-1}$ and $\displaystyle gng^{-1}gn^{-1}g^{-1}=gnn^{-1}g^{-1}$ and since $\displaystyle n\in N\implies n^{-1}$ we see that $\displaystyle gng^{-1}\in gNg^{-1}\implies gn^{-1}g^{-1}\in gNg^{-1}$. Thus, $\displaystyle gNg^{-1}\leqslant G$. But, since $\displaystyle \theta_g$ was a bijection we know that $\displaystyle |gNg^{-1}|=n$ and since by assumption there was only one subgroup of that order it follows that $\displaystyle N=gNg^{-1}$. The conclusion follows.
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  3. #3
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    could you explain this to me again, I am not getting it still.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by wutang View Post
    could you explain this to me again, I am not getting it still.
    $\displaystyle N$ is a subgroup of order $\displaystyle n$...the ONLY subgroup of order $\displaystyle n$. But, given any $\displaystyle g\in G$, $\displaystyle gNg^{-1}$ is a subgroup of order $\displaystyle n$. Since there is ONLY ONE subgroup of order $\displaystyle n$ they have to be the same subgroup. Thus, $\displaystyle N=gNg^{-1}$. That's the definition of normality.
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  5. #5
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    so since are subgroup of order 4 in A4 is the only subgroup of order 4 in A4 it must be normal because, Iam still lost, I thought I had it but I don't! This sucks.
    Last edited by wutang; Apr 8th 2010 at 09:35 PM.
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  6. #6
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    could you explain specifictly for order 4 in A4 for me? I think I would get it if I just understood that one example.
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  7. #7
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    never mind, I got it finally!!!!
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