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Proving a nonempty finite subset of a group is a subgroup

  1. Oct 3, 2007 #1


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    1. The problem statement, all variables and given/known data

    Let a non empty finite subset H of a group G be closed under the binary operation * on G. Show that H is a subgroup of H.

    2. Relevant Definitions

    Group Properties:
    G1: a*(b*c)=(a*b)*c for all a,b,c in G
    G2: e*x=x*e=x for all x in G
    G2: if x is in G then x' is in G and x*x'=x'*x=e

    A subset H of group G is a subgroup of G iff:
    S1: the identity element of G, e, is in H
    S2: the binary operation on G is closed on H
    S3: if x is in H, then x' is in H and x*x'=x'*x=e

    3. The attempt at a solution

    So I know

    H != {}
    H is finite
    H is closed, so I get the second property of a subgroup for free.
    H is subset of the elements of G and inherits the operation.

    If H only has one element, I only need to show that that element is the identity of G. I think I can do this pretty easy with the closure property.

    If H has more then one element here is my strategy,

    Assume x is in H and for every y in H x*y != e.
    This breaks associativity on G which is a contraction since G is a group.
    Therefore given x in H for some y in H x*y=e which means y is x's inverse and since it
    is in H and since H is closed e is there too.

    The reason I think this is the way to go is because I drew tables for a couple of binary operations for a small set that was a group. Then I added some elements around it with a new pseudo-identity element over the larger wanna-be group. Associativity broke on the larger operation, which means it couldn't have been a group.

    So it seems like I should be able to sketch a proof around this concept. Given H is closed and associative and assume it does not contain the identity of G. This breaks associativity of G which is a contradiction. But I am not really getting anywhere.

    Is this approach even valid? I am starting to suspect not. I am pretty stuck. Could someone perhaps suggest another direction? I am not really taking advantage of the finite piece of information. Maybe there is something there?
  2. jcsd
  3. Oct 3, 2007 #2


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    Suppose x is in H. Since H is closed under *, then x^n = x*x*...*x is in H for all n > 0.

    But H is finite, so ...
  4. Oct 4, 2007 #3


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    Right, of course.

    I think I have an outline of a proof that works.

    H={x^n,for every n in Z}
    Assume there is a y in H where y != x^i. I can show this is a contradiction
    by knowing x*c=y has a solution in G, and c cannot be a power of x, or y is a power
    of x. But x^t*y is in H due to closure and for some t, x^t*y=e otherwise H is not
    finite. But then x^t*y = x^t*x*c = e.
    A contradiction, so c is a power of x, and y = x^i.

    So now I know H is finite and has a generator x.
    So x^0=e
    And if |H|=n then x^i*x^(n-i)=e so there is an inverse for every element

    H is a subgroup of G

    I probably made that way harder than is had to be.
    Last edited: Oct 4, 2007
  5. Oct 5, 2007 #4


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    Why should it follow that x^t*y=e?

    H certainly doesn't have to be cyclic. The purpose of my hint above was to let you invoke the finiteness of H to see that not all the x^n's are distinct, i.e. there must be an s and a t with, say, s>t, such that x^s = x^t. From this, one can deduce that e is in H and that x is invertible. But since x was arbitrary, this means every element has an inverse in H, so it's a subgroup of G.
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