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Math Help - Abelilan groups and subgroups

  1. #1
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    Abelilan groups and subgroups

    Let G be a group of order 2n and H a subgroup of G of order n. where n is and odd number. Moreover for all g in G-H and for all h in H we have ghg^-1 = h^-1.

    Now Show that H is abelian and every element of G-H has order 2.
    Last edited by Juancd08; June 11th 2009 at 10:03 PM.
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  2. #2
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    Quote Originally Posted by Juancd08 View Post
    Let G be a group of order 2n and H a subgroup of G of order n. where n is and odd number. Moreover for all g in G-H (everything in G but not in H).
    Further for all h in H we have ghg^-1 = h^-1.

    Now Show that H is abelian and every element of G-H has order 2.
    the sentence in red has no meaning because it's not complete. maybe this is what you meant:

    Moreover for all g in G-H and all h in H we have ghg^-1 = h^-1.
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  3. #3
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    Answer to ur problem

    Hi this is just same problem which herstein topics in algebra has in the section of automorphism. I restate that the problem

    Let G be a finite group of order 2n. Suppose that half the elements of G are of order 2 and the other half form a subgroup of order n. Then prove that H is of odd order and is an abelian subgroup of G.

    Soln: Suppose that H is of even order. Then there exists an element a \in H such that a^{2}=e, which implies that a should belong to S(where $S$ is the set containing elements of even order). So H is of odd order.

    Now let a \in S and b \in H then ab \in S(because if ab \in H,then a\in H which is a contradiction to the fact that a\notin H). Now ab=a^{-1}b since a \in S . Now (ab)^{2}=e since S is of order 2 and this will imply (ab)=(ab)^{-1}=b^{-1}a^{-1}.

    Now since $H$ is a normal subgroup of G therefore the mapping $\phi$ defined by \phi:H \rightarrow H by \phi(h)=ghg^{-1} is an automorphism(verify!).

    Consider any element k from the set S and h \in H,then k^{2}=e and (kh)^{2}=e which implies khk=h^{-1};where h \in H. Consider the mapping h\mapsto khk ie h \mapsto k^{-1}.h.k is that same as h \mapsto h^{-1} since $H$ is normal(therefore $ghg^{-1} \in H$ for all g \in G and h \in H). But h \mapsto h^{-1} is an automorphism iff $H$ is abelian. Hence H is of odd order and is an Abelian Subgroup of G.
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  4. #4
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by Juancd08 View Post
    Let G be a group of order 2n and H a subgroup of G of order n. where n is and odd number. Moreover for all g in G-H and for all h in H we have ghg^-1 = h^-1.

    Now Show that H is abelian and every element of G-H has order 2.
    The case |G|=2 is trivial, so we will assume that n>1.

    Since H has index 2, it is normal in G, Therefore conjugation of the elements of H is an automorphism of H. As the mapping x\mapsto x^{-1} is a group automorphism if and only if the group is Abelian, it follows that H is Abelian.

    Now, H has odd order, so if h\in H and h^2=e, it must follow that h=e (otherwise 2 would divide |H| by Lagrange). As H has index 2, it is easily proved that g\notin H\ \implies\ g^2\in H. (Proof: The element gH in the factor group G/H has order 2.) But then g^2=gg^2g^{-1}=g^{-2}\ \implies\ g^4=e. Hence (g^2)^2=e and g^2\in H. This shows that g^2=e for all g\in G-H.
    Last edited by TheAbstractionist; June 16th 2009 at 07:23 AM.
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  5. #5
    Senior Member TheAbstractionist's Avatar
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    Id just like to say that the proof I originally gave was incorrect. Ive edited my post, so hopefully my proof is correct now.
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