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Math Help - Isomorphism

  1. #1
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    Isomorphism

    Hey There
    I need to show that 2 groups with order 3 is isomorphic
    I basically have the answer I just wanted to double check with you guys
    I have proven that ab=identity and that simularly cd=identity. I am getting stuck just at the last piece at the end. Just showing how its isomorphic
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  2. #2
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    Re: Isomorphism

    Ok whatever I think I figured it out
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  3. #3
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    Re: Isomorphism

    suppose we have a group of order 3. suppose that's all we know. well, the first thing is: our group is of the form:

    G = {e,a,b}.

    this gives 9 possible products:

    e*e
    e*a
    e*b
    a*e
    a*a
    a*b
    b*e
    b*a
    b*b

    what must these be?

    from the definition of the identity e, we get some of these at once:

    e*e = e
    e*a = a
    e*b = b
    a*e = a
    b*e = b, which leaves just 4 more to figure out.

    so what is a*a ? well, we might guess that a*a = e (after all, that's what it is in a 2-element group). at first, there doesn't seem anything wrong with this.

    so then we look at a*b. we have 3 choices: a*b = a, a*b = b, and a*b = e.

    if a*b = a, then:

    a-1*(a*b) = a-1*a
    (a-1*a)*b = e
    e*b = e
    b = e. so that's no good.

    if a*b = b, then:

    (a*b)*b-1 = b*b-1
    a*(b*b-1) = e
    a*e = e
    a = e. so that can't be true, either.

    so we have to have a*b = e. now if a*a = e as well:

    a*b = e = a*a
    a-1*(a*b) = a-1*(a*a)
    (a-1*a)*b = (a-1*a)*a
    e*b = e*a
    b = a. huh. apparently if a*b = e (which is the only possibility that makes sense) we can't have a*a = e. since (as we saw in the 2-element case) a*a = a implies a = e, the only viable option is a*a = b

    so now we know that:

    a*a = b
    a*b = e

    so that just leaves 2 more products to figure out. but..we actually leaned something new: a*b = e means b must be a-1 (in a group, there is only ONE inverse for each element). this means b*a must equal e as well (since a*a-1 = a-1*a, and both are equal to e).

    which brings us to the last product: b*b.

    b*b = e means b*b = b*a, which forces b = a (why?)
    b*b = b means b = e (why?)

    since neither of these makes any sense, we must have:

    b*b = a.

    this means we can write G as: {e,a,a*a} = {e,a,a2} = {e,b2,b} = {e,b*b,b}.

    note that all of this follows just from the fact that G is a group with 3 elements.

    so, suppose G and H are two groups with 3 elements. we can write G = {e,a,a2}, and H = {e',x,x2}. what could be more natural than to define:

    h(e) = e'
    h(a) = x
    h(a2) = x2.

    it should be obvious that h preserves all products involving e. again there are really only 4 products to check:

    h(a*a) = h(a)*h(a) (obvious, because...?)
    h(a*a2) = h(a)*h(a2) (this is easy if you observe that a2 = a-1, and x2 = x-1, or equivalently that:

    a3 = e, and x3 = e').

    h(a2*a) = h(a2)*h(a) (you might want to convince yourself that this boils down to the same thing as the product above:

    a3 = a*(a*a) = (a*a)*a, and x3 = x*(x*x) = (x*x)*x, by associativity)

    and finally:

    h(a2*a2) = h(a2)*h(a2).

    to prove this last product: note that a4 = (a2)*(a2) = (a*a)*(a*a) = ((a*a)*a)*a = a3*a = e*a = a, and similarly for x.
    Thanks from mathrld
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  4. #4
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    Re: Isomorphism

    Wow that was exactly what i needed to be explained. Thanks )))
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