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Math Help - [SOLVED] Normal Subgroups: Prove g^2 is in H for all g in G.

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    [SOLVED] Normal Subgroups: Prove g^2 is in H for all g in G.

    Can someone help me with this proof?

    Let H be a subgroup of G with index 2.
    Prove that g ∈ H for all g ∈ G.

    Thanks!
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  2. #2
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by yvonnehr View Post
    Can someone help me with this proof?

    Let H be a subgroup of G with index 2.
    Prove that g ∈ H for all g ∈ G.

    Thanks!
    A subgroup of index 2 is normal, and the factor group G/H is a group of order 2. Hence for any g\in G, g^2H=(gH)^2=H \implies g^2\in H.

    In general, if |G:H|=2, then \forall\,x,y\in G, xy\in H\ \iff\ x,y\in H\ \mbox{or}\ x,y\notin H. The result you are proving is a special case with x=y.
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    Smile Very nice!

    Thanks for the help. That was very clear!
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    (gH)^2=H ?

    Quote Originally Posted by TheAbstractionist View Post
    A subgroup of index 2 is normal, and the factor group G/H is a group of order 2. Hence for any g\in G, g^2H=(gH)^2=H \implies g^2\in H.

    In general, if |G:H|=2, then \forall\,x,y\in G, xy\in H\ \iff\ x,y\in H\ \mbox{or}\ x,y\notin H. The result you are proving is a special case with x=y.
    One more question, how did you get from

    (gH)^2=H ?

    Thanks.
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  5. #5
    Super Member Gamma's Avatar
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    In a factor group G/H the group operation is defined to be (aH)(bH)=abH. As long as H is normal, this is well defined and is a group.

    You are given that 2=[G:H]=|G/H|

    By Lagrange's theorem since  |G/H|=2 any element of G/H must have order dividing 2, so in this case g^2H=(gH)(gH)=(gH)^2=1H=H since gH \in G/H \forall g\in G.

    This means g^2H=H \Rightarrow g^2h_1=h_2 \Rightarrow g^2=h_2h_1^{-1}\in H as desired.
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