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Math Help - Number of elements in a group

  1. #1
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    Number of elements in a group

    Let G be an abelian group of order 10. Let S={g in G | g^(-1) = g }
    Then what is the number of non identity element sis S?

    I think the anwer is 0.
    Because We cn observe that S is a normal subgroup of G.

    I define a map f : G -> G given by
    f(g) = g^2 for all g in G
    This is an onto homomorphism with kernel S.
    So G/S ~ G
    So o(G)/o(S) = o(G)
    So S has just one element 'e'

    Am i right.
    just have one doubt. Is the homomorphism srjective??
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  2. #2
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    Quote Originally Posted by poorna View Post
    Let G be an abelian group of order 10. Let S={g in G | g^(-1) = g }
    Then what is the number of non identity element sis S?

    I think the anwer is 0.
    Because We cn observe that S is a normal subgroup of G.

    I define a map f : G -> G given by
    f(g) = g^2 for all g in G
    This is an onto homomorphism with kernel S.
    So G/S ~ G
    So o(G)/o(S) = o(G)
    So S has just one element 'e'

    Am i right.
    just have one doubt. Is the homomorphism srjective??

    There is at least one non-identity element in S: to see this, pair up every element in G with its inverse....what do you get?

    By the way, the map f(g)=g^2 is a homomorphism on G iff G is abelian, so you can use it since you assume the group is abelian...but the result still is true if the group isn't abelian: there's always an element (one, at least) of order 2 in any finite group of even order.

    Tonio
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  3. #3
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    Can I not determine exactly the number of elements in S? I have been asked to.
    I can see now that by pairng every elment with their inverse, S is non trivial. So o(S)|o(G)

    So S should have 2 or 5 elements right?
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  4. #4
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    Isn't an abelian group of square-free order cyclic?
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  5. #5
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    Oh is that so? Then it becomes easy The order is 2. Thanks
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