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    Hey guys, really stuck on this question:

    Let  G = A_4
    Let H = {1,(12)(34)}
    Let J = {1, (12)(34),(13)(24),(14)(23)}

    Show that:
    H \triangleleft J
    J \triangleleft G
    but
    H is not \triangleleft G



    I understand J consists of the conjugacy class cycles of type (2,2) in S4, but how do I deduce from this that J \triangleleft S_4 and then J \triangleleft A_4 ? Is it true in general if N \triangleleft S_4 then N \triangleleft A_4 ?
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    Well K \trianglelefteq G if and only if gKg^{-1}=K for all g \in G, so if K \le H \le G then since h \in G for all h \in H, we must also have K \trianglelefteq H.

    To show that J \trianglelefteq S_4 you need to use the fact if (a_1\ \cdots\ a_n) \in S_n and \sigma \in S_n then \sigma (a_1\ \cdots\ a_n) \sigma^{-1} = (\sigma(a_1)\ \cdots\ \sigma(a_n)); that is, conjugation preserves cycle type.
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    J also consists of the conjugacy classes of type (2,2) in A4 (since all (2,2) types are in A4), so I think you can just work directly in A4.

    If g has a certain cycle structure then obviously hgh^{-1} will have the same structure, so any (2,2) cycle will remain a (2,2) cycle under conjugacy. So \forall g\in G: gjg^{-1} \in J for all j in J and so J is normal in G.
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    (Original post by nuodai)
    Well K \trianglelefteq G if and only if gKg^{-1}=K for all g \in G, so if K \le H \le G then since h \in G for all h \in H, we must also have K \trianglelefteq H.

    To show that J \trianglelefteq S_4 you need to use the fact if (a_1\ \cdots\ a_n) \in S_n and \sigma \in S_n then \sigma (a_1\ \cdots\ a_n) \sigma^{-1} = (\sigma(a_1)\ \cdots\ \sigma(a_n)); that is, conjugation preserves cycle type.

    Thank you! It makes sense to me now that showing J \triangleleft S_4 \Rightarrow J \triangleleft A_4 .

    But about the second bit:
    Can you say: "J contains all the possibilities of cycle types (2,2), and therefore (as well as J < S4) conjugating any element of J will also remain in J (because we've got all the elements of type (2,2) ) so J is normal in S4 (and A4)."



    Thank you both for your help!
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    (Original post by sammmmmm)
    Thank you! It makes sense to me now that showing J \triangleleft S_4 \Rightarrow J \triangleleft A_4 .

    But about the second bit:
    Can you say: "J contains all the possibilities of cycle types (2,2), and therefore (as well as J < S4) conjugating any element of J will also remain in J (because we've got all the elements of type (2,2) ) so J is normal in S4 (and A4)."
    Yup, although you might want to make it a bit clearer by writing out a few more details.
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    (Original post by nuodai)
    Yup, although you might want to make it a bit clearer by writing out a few more details.
    Thank you very much for your help!
 
 
 
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