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Math Help - normal subgroup question

  1. #1
    nhk
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    normal subgroup question

    Suppose that H is a normal subgroup of G. If | H|=4 and gH has order 3 in G/H, find a subgroup of order 12 in G.
    How would know what the subgroup of order 12 in G would look like?
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    Quote Originally Posted by nhk View Post
    Suppose that H is a normal subgroup of G. If | H|=4 and gH has order 3 in G/H, find a subgroup of order 12 in G.
    How would know what the subgroup of order 12 in G would look like?

    Look at the subgroup \left<gH\right>\leq G\slash H\Longrightarrow \left<gH\right>=K\slash K , by the correspondence theorem for some subgroup H\leq K\leq G .

    Assuming the group  G is finite, we get 3=|\left<gH\right>|=\left|K\slash H\right|=\frac{|K|}{|H|}\longrightarrow |K|=3|H|=12

    Tonio
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  3. #3
    nhk
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    Can you explain to me where you get the K subgroup and how it plays into your proof? That would be very helpful.
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  4. #4
    nhk
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    Let K=. Since |H|=4 and gH has order 3 in G/H, |K|=12. Lets now prove that K is a subgroup by the two step subgroup test.
    Obviously K is a subset of H.
    How would my elements look when I am doing a two step subgroup test when proving that the identity is in K, that the element ab is in K whenever a and b are in K, and a-1 is in K whenever a is in K
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    Quote Originally Posted by nhk View Post
    Let K=. Since |H|=4 and gH has order 3 in G/H, |K|=12. Lets now prove that K is a subgroup by the two step subgroup test.
    Obviously K is a subset of H.

    No, rather obviously it is not: for example, g\in gH\subset K , but g\notin H ...I'm assuming you meant to write gH instead of GH.

    Tonio


    How would my elements look when I am doing a two step subgroup test when proving that the identity is in K, that the element ab is in K whenever a and b are in K, and a-1 is in K whenever a is in K
    .
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    Quote Originally Posted by nhk View Post
    Can you explain to me where you get the K subgroup and how it plays into your proof? That would be very helpful.

    How K plays into my proof is rather obvious, imo. It comes from the correspondence theorem, one of the most beautiful, helpful and basic of all basic group theory theorems: Correspondence Theorem - ProofWiki

    Tonio
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  7. #7
    nhk
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    I guess we have not got to corresponedence theroem yet in class, howver the proof that you gave me for it makes perfect sense.
    Thanks a lot.
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